Script not correctly printing the correct elements of the array

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-2















So basically I have to write a bash script that examines some specific files from the working directory (which are named file.00.txt up to file.24.txt). The thing is, 3 of them are exatcly the same and my assignment is to create a script that tells me which 3 are the same.



Here is my code



#!/bin/bash 
f0=file.00.txt
f1=file.01.txt
f2=file.02.txt
f3=file.03.txt
f4=file.04.txt
f5=file.05.txt
f6=file.06.txt
f7=file.07.txt
f8=file.08.txt
f9=file.09.txt
f10=file.10.txt
f11=file.11.txt
f12=file.12.txt
f13=file.13.txt
f14=file.14.txt
f15=file.15.txt
f16=file.16.txt
f17=file.17.txt
f18=file.18.txt
f19=file.19.txt
f20=file.20.txt
f21=file.21.txt
f22=file.22.txt
f23=file.23.txt
f24=file.24.txt

array=($f0 $f1 $f2 $f3 $f4 $f5 $f6 $f7 $f8 $f9 $f10 $f11 $f12 $f13 $f14 $f15 $f16 $f17 $f18 $f19 $f20 $f21 $f22 $f23 $f24)

i=0
touch placeholder

while [ $i -lt $#array ]
do
DIFF=$(diff $array[i] $array[i+1])
if [ "$DIFF" = "" ]
then
echo "$array[i] y $array[i+1]" >> placeholder
fi
i=$((i+1))
done

cat placeholder


The idea of this code is to compare each file with the following one in the array , then store the ones that are the same in a file called placeholder and finally revealing the contents of the file with the cat command.



However, everytime I run the scrip I am getting the message



file.00.txt y file.00.txt
file.01.txt y file.01.txt
file.02.txt y file.02.txt


and so on for each file. This should not happen, since I am clearly using



echo "$array[i] y $array[i+1]" >> placeholder


to echo both positions. Why is this happening and how can I solve this ?



How can I solve this?










share|improve this question



















  • 3





    You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

    – Kusalananda
    Mar 17 at 17:16












  • Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

    – ilkkachu
    Mar 17 at 17:37











  • Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

    – pericothebig
    Mar 17 at 17:41












  • @Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

    – pericothebig
    Mar 17 at 17:46







  • 2





    Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

    – xenoid
    Mar 17 at 17:58

















-2















So basically I have to write a bash script that examines some specific files from the working directory (which are named file.00.txt up to file.24.txt). The thing is, 3 of them are exatcly the same and my assignment is to create a script that tells me which 3 are the same.



Here is my code



#!/bin/bash 
f0=file.00.txt
f1=file.01.txt
f2=file.02.txt
f3=file.03.txt
f4=file.04.txt
f5=file.05.txt
f6=file.06.txt
f7=file.07.txt
f8=file.08.txt
f9=file.09.txt
f10=file.10.txt
f11=file.11.txt
f12=file.12.txt
f13=file.13.txt
f14=file.14.txt
f15=file.15.txt
f16=file.16.txt
f17=file.17.txt
f18=file.18.txt
f19=file.19.txt
f20=file.20.txt
f21=file.21.txt
f22=file.22.txt
f23=file.23.txt
f24=file.24.txt

array=($f0 $f1 $f2 $f3 $f4 $f5 $f6 $f7 $f8 $f9 $f10 $f11 $f12 $f13 $f14 $f15 $f16 $f17 $f18 $f19 $f20 $f21 $f22 $f23 $f24)

i=0
touch placeholder

while [ $i -lt $#array ]
do
DIFF=$(diff $array[i] $array[i+1])
if [ "$DIFF" = "" ]
then
echo "$array[i] y $array[i+1]" >> placeholder
fi
i=$((i+1))
done

cat placeholder


The idea of this code is to compare each file with the following one in the array , then store the ones that are the same in a file called placeholder and finally revealing the contents of the file with the cat command.



However, everytime I run the scrip I am getting the message



file.00.txt y file.00.txt
file.01.txt y file.01.txt
file.02.txt y file.02.txt


and so on for each file. This should not happen, since I am clearly using



echo "$array[i] y $array[i+1]" >> placeholder


to echo both positions. Why is this happening and how can I solve this ?



How can I solve this?










share|improve this question



















  • 3





    You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

    – Kusalananda
    Mar 17 at 17:16












  • Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

    – ilkkachu
    Mar 17 at 17:37











  • Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

    – pericothebig
    Mar 17 at 17:41












  • @Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

    – pericothebig
    Mar 17 at 17:46







  • 2





    Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

    – xenoid
    Mar 17 at 17:58













-2












-2








-2


0






So basically I have to write a bash script that examines some specific files from the working directory (which are named file.00.txt up to file.24.txt). The thing is, 3 of them are exatcly the same and my assignment is to create a script that tells me which 3 are the same.



Here is my code



#!/bin/bash 
f0=file.00.txt
f1=file.01.txt
f2=file.02.txt
f3=file.03.txt
f4=file.04.txt
f5=file.05.txt
f6=file.06.txt
f7=file.07.txt
f8=file.08.txt
f9=file.09.txt
f10=file.10.txt
f11=file.11.txt
f12=file.12.txt
f13=file.13.txt
f14=file.14.txt
f15=file.15.txt
f16=file.16.txt
f17=file.17.txt
f18=file.18.txt
f19=file.19.txt
f20=file.20.txt
f21=file.21.txt
f22=file.22.txt
f23=file.23.txt
f24=file.24.txt

array=($f0 $f1 $f2 $f3 $f4 $f5 $f6 $f7 $f8 $f9 $f10 $f11 $f12 $f13 $f14 $f15 $f16 $f17 $f18 $f19 $f20 $f21 $f22 $f23 $f24)

i=0
touch placeholder

while [ $i -lt $#array ]
do
DIFF=$(diff $array[i] $array[i+1])
if [ "$DIFF" = "" ]
then
echo "$array[i] y $array[i+1]" >> placeholder
fi
i=$((i+1))
done

cat placeholder


The idea of this code is to compare each file with the following one in the array , then store the ones that are the same in a file called placeholder and finally revealing the contents of the file with the cat command.



However, everytime I run the scrip I am getting the message



file.00.txt y file.00.txt
file.01.txt y file.01.txt
file.02.txt y file.02.txt


and so on for each file. This should not happen, since I am clearly using



echo "$array[i] y $array[i+1]" >> placeholder


to echo both positions. Why is this happening and how can I solve this ?



How can I solve this?










share|improve this question
















So basically I have to write a bash script that examines some specific files from the working directory (which are named file.00.txt up to file.24.txt). The thing is, 3 of them are exatcly the same and my assignment is to create a script that tells me which 3 are the same.



Here is my code



#!/bin/bash 
f0=file.00.txt
f1=file.01.txt
f2=file.02.txt
f3=file.03.txt
f4=file.04.txt
f5=file.05.txt
f6=file.06.txt
f7=file.07.txt
f8=file.08.txt
f9=file.09.txt
f10=file.10.txt
f11=file.11.txt
f12=file.12.txt
f13=file.13.txt
f14=file.14.txt
f15=file.15.txt
f16=file.16.txt
f17=file.17.txt
f18=file.18.txt
f19=file.19.txt
f20=file.20.txt
f21=file.21.txt
f22=file.22.txt
f23=file.23.txt
f24=file.24.txt

array=($f0 $f1 $f2 $f3 $f4 $f5 $f6 $f7 $f8 $f9 $f10 $f11 $f12 $f13 $f14 $f15 $f16 $f17 $f18 $f19 $f20 $f21 $f22 $f23 $f24)

i=0
touch placeholder

while [ $i -lt $#array ]
do
DIFF=$(diff $array[i] $array[i+1])
if [ "$DIFF" = "" ]
then
echo "$array[i] y $array[i+1]" >> placeholder
fi
i=$((i+1))
done

cat placeholder


The idea of this code is to compare each file with the following one in the array , then store the ones that are the same in a file called placeholder and finally revealing the contents of the file with the cat command.



However, everytime I run the scrip I am getting the message



file.00.txt y file.00.txt
file.01.txt y file.01.txt
file.02.txt y file.02.txt


and so on for each file. This should not happen, since I am clearly using



echo "$array[i] y $array[i+1]" >> placeholder


to echo both positions. Why is this happening and how can I solve this ?



How can I solve this?







bash shell-script diff






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 17 at 18:06







pericothebig

















asked Mar 17 at 17:13









pericothebigpericothebig

11




11







  • 3





    You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

    – Kusalananda
    Mar 17 at 17:16












  • Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

    – ilkkachu
    Mar 17 at 17:37











  • Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

    – pericothebig
    Mar 17 at 17:41












  • @Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

    – pericothebig
    Mar 17 at 17:46







  • 2





    Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

    – xenoid
    Mar 17 at 17:58












  • 3





    You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

    – Kusalananda
    Mar 17 at 17:16












  • Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

    – ilkkachu
    Mar 17 at 17:37











  • Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

    – pericothebig
    Mar 17 at 17:41












  • @Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

    – pericothebig
    Mar 17 at 17:46







  • 2





    Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

    – xenoid
    Mar 17 at 17:58







3




3





You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

– Kusalananda
Mar 17 at 17:16






You do add a comma to the filename in the call to diff. It may be worth removing that... Apart from that, is there a reason for creating all those variables and then add them to an array? Note too, that your loop is not terminated by a done and that you don't increment i, and that the length of the array is $#array[@], not 27, and that it's easier to loop over the elements of the array ("$array[@]") than to use an index (that is, if you need an array at all and can't just loop over file.*.txt).

– Kusalananda
Mar 17 at 17:16














Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

– ilkkachu
Mar 17 at 17:37





Also you could just use brace expansion: array=( file.00..24.txt ) to generate the file names. Though note that if you only diff files n and n+1, you won't notice if nonconsecutive files (e.g. 3 and 8) are the same, so the logic in the loop may need some more thought. (You'd need to look over each other file for each file.)

– ilkkachu
Mar 17 at 17:37













Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

– pericothebig
Mar 17 at 17:41






Ok so I corrected all these things but I am still not getting what I want to . I will edit to clarifi what I getting

– pericothebig
Mar 17 at 17:41














@Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

– pericothebig
Mar 17 at 17:46






@Kusalananda Ok so I am still new, so I did not know I could just loop over the files. How would I do that without an array? Thnaks to both of you guys

– pericothebig
Mar 17 at 17:46





2




2





Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

– xenoid
Mar 17 at 17:58





Very bad algorithm anyway because you will do 27*26 comparisons. Try instead: 1) get the MD5 hash of the files (md5sum command), 2) count the instances of each hash (sort | uniq -c after extracting only the hashes fro the md5sum output), 3) find the biggest count (add sort -n | tail i-1 to the pipeline), 4) extract the hash, 5) use grep $hash on the complete md5sum output to retrieve the names of the files with that hash. If paranoid, 6) use cmp to check that all three files are the same.

– xenoid
Mar 17 at 17:58










2 Answers
2






active

oldest

votes


















1














The seemingly strange output of your code could possibly be explained by the fact that your script only ever adds to the output file. This means that you might have had some error in your code previously (now corrected), but that you still see the output of that run in the output file, since the output file is never removed or emptied by the script.




You could shorten your script into



#!/bin/bash

array=( file.*.txt )

for name in "$array[@]"; do
if [ -n "$prev_name" ] && cmp -s "$prev_name" "$name"
then
printf '%s y %sn' "$prev_name" "$name"
fi

prev_name=$name
done


This uses a globbing pattern to populate the array with the filenames matching the pattern.



It then loops over the names, comparing filenames that occur adjacent to each other in the array using cmp -s. The cmp utility will exit with a true exit status if the contents of the two files that it's comparing are identical.



The loop uses $prev_name to hold the name of the previous file in the array. In the first iteration of the loop, this variable is empty, so the actual comparison of files is skipped.




What you are possibly expected to write is a double loop. Something like



for nameA in "$array[@]"; do
for nameB in "$array[@]"; do
if [ "$nameA" != "$nameB" ] && cmp -s "$nameA" "$nameB"
then
printf '%s y %sn' "$nameA" "$nameB"
fi
done
done


But this would compare A against B and B against A, and the number of calls to cmp would grow quadratically with the number of files involved, which would be both resource intensive (on disks; it would read each file as many times as there are filenames in the array) and slow.




A common way to find sets of files is with identical contents is with fdupes:



$ fdupes --sameline .
./file.1.txt ./file.2.txt ./file.7.txt


Would you want to do something similar without fdupes, you could do that by computing and comparing a checksum of each file using e.g. md5sum:



#!/bin/bash

declare -A names count

while read -r cksum name; do
names[$cksum]+=$names[$cksum]:+,$name
count[$cksum]=$(( count[$cksum] + 1 ))
done < <( md5sum file.*.txt )

for cksum in "$!count[@]"; do
if [ "$count[$cksum]" -gt 1 ]; then
printf '%sn' "$names[$cksum]"
fi
done


The first loop reads the output of md5sum which is is executed across all the relevant files. The output of md5sum may look something like



897316929176464ebc9ad085f31e7284 file.1.txt
8c9eb686bf3eb5bd83d9373eadf6504b file.10.txt
897316929176464ebc9ad085f31e7284 file.2.txt
26ab0db90d72e28ad0ba1e22ee510510 file.3.txt
84bc3da1b3e33a18e8d5e1bdd7a18d7a file.4.txt
aa6ed9e0f26a6eba784aae8267df1951 file.5.txt
6d7fce9fee471194aa8b5b6e47267f03 file.6.txt
897316929176464ebc9ad085f31e7284 file.7.txt
c30f7472766d25af1dc80b3ffc9a58c7 file.8.txt
9ae0ea9e3c9c6e1b9b6252c8395efdc1 file.9.txt


The checksum in the first column is read into cksum and filename gets read into name.



Inside the first loop, we append the name to an entry in an associative array that is indexed by the checksum. The way the assignment to names[$cksum] is done there makes sure that we add a comma before each new name if needed (which it is if the entry already contains some other names). We then update a count of the number of times that we've seen this particular checksum (this will be used in the second loop).



In the second loop, we go through the checksums ("$!count[@]" expands to a list of keys (checksums) in the count associative array), and for each checksum we test whether its count is greater than 1, meaning we've found a duplicate file (if you're looking for groups of exactly three identical files, you may want to use -eq 3 instead of -gt 1 here). If it is, we print the names associated with that checksum.



Testing it:



$ bash script.sh
file.1.txt,file.2.txt,file.7.txt





share|improve this answer
































    0














    Here's a more efficient way of doing what you're trying to do. I use a smaller sample set to keep things clearer:



    #!/bin/bash

    # clear placeholder
    printf "Files with no diff:n" > placeholder

    # set up sample data
    echo "one" > file.00.txt
    echo "one" > file.01.txt
    echo "foo" > file.02.txt
    echo "bar" > file.03.txt
    echo "two" > file.04.txt
    echo "two" > file.05.txt

    # generate array
    i=0
    while [ $i -lt 6 ]; do
    array+=( file.`printf %02d $i`.txt )
    ((i++))
    done

    i=0
    while [ $i -lt 5 ]; do
    diff --brief $array[i] $array[i+1] &&
    echo "$array[i] $array[i+1]" >> placeholder
    ((i++))
    done


    Results:



    $ sh ./test.sh 
    Files file.01.txt and file.02.txt differ
    Files file.02.txt and file.03.txt differ
    Files file.03.txt and file.04.txt differ
    $ cat placeholder
    Files with no diff:
    file.00.txt file.01.txt
    file.04.txt file.05.txt


    You don't need to generate sample data if you actually already have data.



    The code, explained:



    Building an array in a loop (in Bash) can be done by iterating the way you obviously already know, but the array+= notation appends an element.



    The ((++)) obviously increments your counter.



    Performing the diff, I use the --brief option. If you read the diff man page, it tells you that --brief only prints output when a difference is found. Therefore, the diff command succeeds if no diff is found.



    Using the && (AND) notation, this code echos the names of the files being compared into your placeholder file if and only if the diff command generates no output.



    If there is a difference between files, diff outputs the differences to the terminal. This causes the && (AND) to fail, so nothing is output to the placeholder file.



    If you have any further questions about the syntax, feel free to ask.






    share|improve this answer




















    • 2





      "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

      – ilkkachu
      Mar 17 at 19:37











    • Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

      – Klaatu von Schlacker
      Mar 22 at 9:53











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    1














    The seemingly strange output of your code could possibly be explained by the fact that your script only ever adds to the output file. This means that you might have had some error in your code previously (now corrected), but that you still see the output of that run in the output file, since the output file is never removed or emptied by the script.




    You could shorten your script into



    #!/bin/bash

    array=( file.*.txt )

    for name in "$array[@]"; do
    if [ -n "$prev_name" ] && cmp -s "$prev_name" "$name"
    then
    printf '%s y %sn' "$prev_name" "$name"
    fi

    prev_name=$name
    done


    This uses a globbing pattern to populate the array with the filenames matching the pattern.



    It then loops over the names, comparing filenames that occur adjacent to each other in the array using cmp -s. The cmp utility will exit with a true exit status if the contents of the two files that it's comparing are identical.



    The loop uses $prev_name to hold the name of the previous file in the array. In the first iteration of the loop, this variable is empty, so the actual comparison of files is skipped.




    What you are possibly expected to write is a double loop. Something like



    for nameA in "$array[@]"; do
    for nameB in "$array[@]"; do
    if [ "$nameA" != "$nameB" ] && cmp -s "$nameA" "$nameB"
    then
    printf '%s y %sn' "$nameA" "$nameB"
    fi
    done
    done


    But this would compare A against B and B against A, and the number of calls to cmp would grow quadratically with the number of files involved, which would be both resource intensive (on disks; it would read each file as many times as there are filenames in the array) and slow.




    A common way to find sets of files is with identical contents is with fdupes:



    $ fdupes --sameline .
    ./file.1.txt ./file.2.txt ./file.7.txt


    Would you want to do something similar without fdupes, you could do that by computing and comparing a checksum of each file using e.g. md5sum:



    #!/bin/bash

    declare -A names count

    while read -r cksum name; do
    names[$cksum]+=$names[$cksum]:+,$name
    count[$cksum]=$(( count[$cksum] + 1 ))
    done < <( md5sum file.*.txt )

    for cksum in "$!count[@]"; do
    if [ "$count[$cksum]" -gt 1 ]; then
    printf '%sn' "$names[$cksum]"
    fi
    done


    The first loop reads the output of md5sum which is is executed across all the relevant files. The output of md5sum may look something like



    897316929176464ebc9ad085f31e7284 file.1.txt
    8c9eb686bf3eb5bd83d9373eadf6504b file.10.txt
    897316929176464ebc9ad085f31e7284 file.2.txt
    26ab0db90d72e28ad0ba1e22ee510510 file.3.txt
    84bc3da1b3e33a18e8d5e1bdd7a18d7a file.4.txt
    aa6ed9e0f26a6eba784aae8267df1951 file.5.txt
    6d7fce9fee471194aa8b5b6e47267f03 file.6.txt
    897316929176464ebc9ad085f31e7284 file.7.txt
    c30f7472766d25af1dc80b3ffc9a58c7 file.8.txt
    9ae0ea9e3c9c6e1b9b6252c8395efdc1 file.9.txt


    The checksum in the first column is read into cksum and filename gets read into name.



    Inside the first loop, we append the name to an entry in an associative array that is indexed by the checksum. The way the assignment to names[$cksum] is done there makes sure that we add a comma before each new name if needed (which it is if the entry already contains some other names). We then update a count of the number of times that we've seen this particular checksum (this will be used in the second loop).



    In the second loop, we go through the checksums ("$!count[@]" expands to a list of keys (checksums) in the count associative array), and for each checksum we test whether its count is greater than 1, meaning we've found a duplicate file (if you're looking for groups of exactly three identical files, you may want to use -eq 3 instead of -gt 1 here). If it is, we print the names associated with that checksum.



    Testing it:



    $ bash script.sh
    file.1.txt,file.2.txt,file.7.txt





    share|improve this answer





























      1














      The seemingly strange output of your code could possibly be explained by the fact that your script only ever adds to the output file. This means that you might have had some error in your code previously (now corrected), but that you still see the output of that run in the output file, since the output file is never removed or emptied by the script.




      You could shorten your script into



      #!/bin/bash

      array=( file.*.txt )

      for name in "$array[@]"; do
      if [ -n "$prev_name" ] && cmp -s "$prev_name" "$name"
      then
      printf '%s y %sn' "$prev_name" "$name"
      fi

      prev_name=$name
      done


      This uses a globbing pattern to populate the array with the filenames matching the pattern.



      It then loops over the names, comparing filenames that occur adjacent to each other in the array using cmp -s. The cmp utility will exit with a true exit status if the contents of the two files that it's comparing are identical.



      The loop uses $prev_name to hold the name of the previous file in the array. In the first iteration of the loop, this variable is empty, so the actual comparison of files is skipped.




      What you are possibly expected to write is a double loop. Something like



      for nameA in "$array[@]"; do
      for nameB in "$array[@]"; do
      if [ "$nameA" != "$nameB" ] && cmp -s "$nameA" "$nameB"
      then
      printf '%s y %sn' "$nameA" "$nameB"
      fi
      done
      done


      But this would compare A against B and B against A, and the number of calls to cmp would grow quadratically with the number of files involved, which would be both resource intensive (on disks; it would read each file as many times as there are filenames in the array) and slow.




      A common way to find sets of files is with identical contents is with fdupes:



      $ fdupes --sameline .
      ./file.1.txt ./file.2.txt ./file.7.txt


      Would you want to do something similar without fdupes, you could do that by computing and comparing a checksum of each file using e.g. md5sum:



      #!/bin/bash

      declare -A names count

      while read -r cksum name; do
      names[$cksum]+=$names[$cksum]:+,$name
      count[$cksum]=$(( count[$cksum] + 1 ))
      done < <( md5sum file.*.txt )

      for cksum in "$!count[@]"; do
      if [ "$count[$cksum]" -gt 1 ]; then
      printf '%sn' "$names[$cksum]"
      fi
      done


      The first loop reads the output of md5sum which is is executed across all the relevant files. The output of md5sum may look something like



      897316929176464ebc9ad085f31e7284 file.1.txt
      8c9eb686bf3eb5bd83d9373eadf6504b file.10.txt
      897316929176464ebc9ad085f31e7284 file.2.txt
      26ab0db90d72e28ad0ba1e22ee510510 file.3.txt
      84bc3da1b3e33a18e8d5e1bdd7a18d7a file.4.txt
      aa6ed9e0f26a6eba784aae8267df1951 file.5.txt
      6d7fce9fee471194aa8b5b6e47267f03 file.6.txt
      897316929176464ebc9ad085f31e7284 file.7.txt
      c30f7472766d25af1dc80b3ffc9a58c7 file.8.txt
      9ae0ea9e3c9c6e1b9b6252c8395efdc1 file.9.txt


      The checksum in the first column is read into cksum and filename gets read into name.



      Inside the first loop, we append the name to an entry in an associative array that is indexed by the checksum. The way the assignment to names[$cksum] is done there makes sure that we add a comma before each new name if needed (which it is if the entry already contains some other names). We then update a count of the number of times that we've seen this particular checksum (this will be used in the second loop).



      In the second loop, we go through the checksums ("$!count[@]" expands to a list of keys (checksums) in the count associative array), and for each checksum we test whether its count is greater than 1, meaning we've found a duplicate file (if you're looking for groups of exactly three identical files, you may want to use -eq 3 instead of -gt 1 here). If it is, we print the names associated with that checksum.



      Testing it:



      $ bash script.sh
      file.1.txt,file.2.txt,file.7.txt





      share|improve this answer



























        1












        1








        1







        The seemingly strange output of your code could possibly be explained by the fact that your script only ever adds to the output file. This means that you might have had some error in your code previously (now corrected), but that you still see the output of that run in the output file, since the output file is never removed or emptied by the script.




        You could shorten your script into



        #!/bin/bash

        array=( file.*.txt )

        for name in "$array[@]"; do
        if [ -n "$prev_name" ] && cmp -s "$prev_name" "$name"
        then
        printf '%s y %sn' "$prev_name" "$name"
        fi

        prev_name=$name
        done


        This uses a globbing pattern to populate the array with the filenames matching the pattern.



        It then loops over the names, comparing filenames that occur adjacent to each other in the array using cmp -s. The cmp utility will exit with a true exit status if the contents of the two files that it's comparing are identical.



        The loop uses $prev_name to hold the name of the previous file in the array. In the first iteration of the loop, this variable is empty, so the actual comparison of files is skipped.




        What you are possibly expected to write is a double loop. Something like



        for nameA in "$array[@]"; do
        for nameB in "$array[@]"; do
        if [ "$nameA" != "$nameB" ] && cmp -s "$nameA" "$nameB"
        then
        printf '%s y %sn' "$nameA" "$nameB"
        fi
        done
        done


        But this would compare A against B and B against A, and the number of calls to cmp would grow quadratically with the number of files involved, which would be both resource intensive (on disks; it would read each file as many times as there are filenames in the array) and slow.




        A common way to find sets of files is with identical contents is with fdupes:



        $ fdupes --sameline .
        ./file.1.txt ./file.2.txt ./file.7.txt


        Would you want to do something similar without fdupes, you could do that by computing and comparing a checksum of each file using e.g. md5sum:



        #!/bin/bash

        declare -A names count

        while read -r cksum name; do
        names[$cksum]+=$names[$cksum]:+,$name
        count[$cksum]=$(( count[$cksum] + 1 ))
        done < <( md5sum file.*.txt )

        for cksum in "$!count[@]"; do
        if [ "$count[$cksum]" -gt 1 ]; then
        printf '%sn' "$names[$cksum]"
        fi
        done


        The first loop reads the output of md5sum which is is executed across all the relevant files. The output of md5sum may look something like



        897316929176464ebc9ad085f31e7284 file.1.txt
        8c9eb686bf3eb5bd83d9373eadf6504b file.10.txt
        897316929176464ebc9ad085f31e7284 file.2.txt
        26ab0db90d72e28ad0ba1e22ee510510 file.3.txt
        84bc3da1b3e33a18e8d5e1bdd7a18d7a file.4.txt
        aa6ed9e0f26a6eba784aae8267df1951 file.5.txt
        6d7fce9fee471194aa8b5b6e47267f03 file.6.txt
        897316929176464ebc9ad085f31e7284 file.7.txt
        c30f7472766d25af1dc80b3ffc9a58c7 file.8.txt
        9ae0ea9e3c9c6e1b9b6252c8395efdc1 file.9.txt


        The checksum in the first column is read into cksum and filename gets read into name.



        Inside the first loop, we append the name to an entry in an associative array that is indexed by the checksum. The way the assignment to names[$cksum] is done there makes sure that we add a comma before each new name if needed (which it is if the entry already contains some other names). We then update a count of the number of times that we've seen this particular checksum (this will be used in the second loop).



        In the second loop, we go through the checksums ("$!count[@]" expands to a list of keys (checksums) in the count associative array), and for each checksum we test whether its count is greater than 1, meaning we've found a duplicate file (if you're looking for groups of exactly three identical files, you may want to use -eq 3 instead of -gt 1 here). If it is, we print the names associated with that checksum.



        Testing it:



        $ bash script.sh
        file.1.txt,file.2.txt,file.7.txt





        share|improve this answer















        The seemingly strange output of your code could possibly be explained by the fact that your script only ever adds to the output file. This means that you might have had some error in your code previously (now corrected), but that you still see the output of that run in the output file, since the output file is never removed or emptied by the script.




        You could shorten your script into



        #!/bin/bash

        array=( file.*.txt )

        for name in "$array[@]"; do
        if [ -n "$prev_name" ] && cmp -s "$prev_name" "$name"
        then
        printf '%s y %sn' "$prev_name" "$name"
        fi

        prev_name=$name
        done


        This uses a globbing pattern to populate the array with the filenames matching the pattern.



        It then loops over the names, comparing filenames that occur adjacent to each other in the array using cmp -s. The cmp utility will exit with a true exit status if the contents of the two files that it's comparing are identical.



        The loop uses $prev_name to hold the name of the previous file in the array. In the first iteration of the loop, this variable is empty, so the actual comparison of files is skipped.




        What you are possibly expected to write is a double loop. Something like



        for nameA in "$array[@]"; do
        for nameB in "$array[@]"; do
        if [ "$nameA" != "$nameB" ] && cmp -s "$nameA" "$nameB"
        then
        printf '%s y %sn' "$nameA" "$nameB"
        fi
        done
        done


        But this would compare A against B and B against A, and the number of calls to cmp would grow quadratically with the number of files involved, which would be both resource intensive (on disks; it would read each file as many times as there are filenames in the array) and slow.




        A common way to find sets of files is with identical contents is with fdupes:



        $ fdupes --sameline .
        ./file.1.txt ./file.2.txt ./file.7.txt


        Would you want to do something similar without fdupes, you could do that by computing and comparing a checksum of each file using e.g. md5sum:



        #!/bin/bash

        declare -A names count

        while read -r cksum name; do
        names[$cksum]+=$names[$cksum]:+,$name
        count[$cksum]=$(( count[$cksum] + 1 ))
        done < <( md5sum file.*.txt )

        for cksum in "$!count[@]"; do
        if [ "$count[$cksum]" -gt 1 ]; then
        printf '%sn' "$names[$cksum]"
        fi
        done


        The first loop reads the output of md5sum which is is executed across all the relevant files. The output of md5sum may look something like



        897316929176464ebc9ad085f31e7284 file.1.txt
        8c9eb686bf3eb5bd83d9373eadf6504b file.10.txt
        897316929176464ebc9ad085f31e7284 file.2.txt
        26ab0db90d72e28ad0ba1e22ee510510 file.3.txt
        84bc3da1b3e33a18e8d5e1bdd7a18d7a file.4.txt
        aa6ed9e0f26a6eba784aae8267df1951 file.5.txt
        6d7fce9fee471194aa8b5b6e47267f03 file.6.txt
        897316929176464ebc9ad085f31e7284 file.7.txt
        c30f7472766d25af1dc80b3ffc9a58c7 file.8.txt
        9ae0ea9e3c9c6e1b9b6252c8395efdc1 file.9.txt


        The checksum in the first column is read into cksum and filename gets read into name.



        Inside the first loop, we append the name to an entry in an associative array that is indexed by the checksum. The way the assignment to names[$cksum] is done there makes sure that we add a comma before each new name if needed (which it is if the entry already contains some other names). We then update a count of the number of times that we've seen this particular checksum (this will be used in the second loop).



        In the second loop, we go through the checksums ("$!count[@]" expands to a list of keys (checksums) in the count associative array), and for each checksum we test whether its count is greater than 1, meaning we've found a duplicate file (if you're looking for groups of exactly three identical files, you may want to use -eq 3 instead of -gt 1 here). If it is, we print the names associated with that checksum.



        Testing it:



        $ bash script.sh
        file.1.txt,file.2.txt,file.7.txt






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 17 at 20:14

























        answered Mar 17 at 19:04









        KusalanandaKusalananda

        142k18265440




        142k18265440























            0














            Here's a more efficient way of doing what you're trying to do. I use a smaller sample set to keep things clearer:



            #!/bin/bash

            # clear placeholder
            printf "Files with no diff:n" > placeholder

            # set up sample data
            echo "one" > file.00.txt
            echo "one" > file.01.txt
            echo "foo" > file.02.txt
            echo "bar" > file.03.txt
            echo "two" > file.04.txt
            echo "two" > file.05.txt

            # generate array
            i=0
            while [ $i -lt 6 ]; do
            array+=( file.`printf %02d $i`.txt )
            ((i++))
            done

            i=0
            while [ $i -lt 5 ]; do
            diff --brief $array[i] $array[i+1] &&
            echo "$array[i] $array[i+1]" >> placeholder
            ((i++))
            done


            Results:



            $ sh ./test.sh 
            Files file.01.txt and file.02.txt differ
            Files file.02.txt and file.03.txt differ
            Files file.03.txt and file.04.txt differ
            $ cat placeholder
            Files with no diff:
            file.00.txt file.01.txt
            file.04.txt file.05.txt


            You don't need to generate sample data if you actually already have data.



            The code, explained:



            Building an array in a loop (in Bash) can be done by iterating the way you obviously already know, but the array+= notation appends an element.



            The ((++)) obviously increments your counter.



            Performing the diff, I use the --brief option. If you read the diff man page, it tells you that --brief only prints output when a difference is found. Therefore, the diff command succeeds if no diff is found.



            Using the && (AND) notation, this code echos the names of the files being compared into your placeholder file if and only if the diff command generates no output.



            If there is a difference between files, diff outputs the differences to the terminal. This causes the && (AND) to fail, so nothing is output to the placeholder file.



            If you have any further questions about the syntax, feel free to ask.






            share|improve this answer




















            • 2





              "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

              – ilkkachu
              Mar 17 at 19:37











            • Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

              – Klaatu von Schlacker
              Mar 22 at 9:53















            0














            Here's a more efficient way of doing what you're trying to do. I use a smaller sample set to keep things clearer:



            #!/bin/bash

            # clear placeholder
            printf "Files with no diff:n" > placeholder

            # set up sample data
            echo "one" > file.00.txt
            echo "one" > file.01.txt
            echo "foo" > file.02.txt
            echo "bar" > file.03.txt
            echo "two" > file.04.txt
            echo "two" > file.05.txt

            # generate array
            i=0
            while [ $i -lt 6 ]; do
            array+=( file.`printf %02d $i`.txt )
            ((i++))
            done

            i=0
            while [ $i -lt 5 ]; do
            diff --brief $array[i] $array[i+1] &&
            echo "$array[i] $array[i+1]" >> placeholder
            ((i++))
            done


            Results:



            $ sh ./test.sh 
            Files file.01.txt and file.02.txt differ
            Files file.02.txt and file.03.txt differ
            Files file.03.txt and file.04.txt differ
            $ cat placeholder
            Files with no diff:
            file.00.txt file.01.txt
            file.04.txt file.05.txt


            You don't need to generate sample data if you actually already have data.



            The code, explained:



            Building an array in a loop (in Bash) can be done by iterating the way you obviously already know, but the array+= notation appends an element.



            The ((++)) obviously increments your counter.



            Performing the diff, I use the --brief option. If you read the diff man page, it tells you that --brief only prints output when a difference is found. Therefore, the diff command succeeds if no diff is found.



            Using the && (AND) notation, this code echos the names of the files being compared into your placeholder file if and only if the diff command generates no output.



            If there is a difference between files, diff outputs the differences to the terminal. This causes the && (AND) to fail, so nothing is output to the placeholder file.



            If you have any further questions about the syntax, feel free to ask.






            share|improve this answer




















            • 2





              "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

              – ilkkachu
              Mar 17 at 19:37











            • Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

              – Klaatu von Schlacker
              Mar 22 at 9:53













            0












            0








            0







            Here's a more efficient way of doing what you're trying to do. I use a smaller sample set to keep things clearer:



            #!/bin/bash

            # clear placeholder
            printf "Files with no diff:n" > placeholder

            # set up sample data
            echo "one" > file.00.txt
            echo "one" > file.01.txt
            echo "foo" > file.02.txt
            echo "bar" > file.03.txt
            echo "two" > file.04.txt
            echo "two" > file.05.txt

            # generate array
            i=0
            while [ $i -lt 6 ]; do
            array+=( file.`printf %02d $i`.txt )
            ((i++))
            done

            i=0
            while [ $i -lt 5 ]; do
            diff --brief $array[i] $array[i+1] &&
            echo "$array[i] $array[i+1]" >> placeholder
            ((i++))
            done


            Results:



            $ sh ./test.sh 
            Files file.01.txt and file.02.txt differ
            Files file.02.txt and file.03.txt differ
            Files file.03.txt and file.04.txt differ
            $ cat placeholder
            Files with no diff:
            file.00.txt file.01.txt
            file.04.txt file.05.txt


            You don't need to generate sample data if you actually already have data.



            The code, explained:



            Building an array in a loop (in Bash) can be done by iterating the way you obviously already know, but the array+= notation appends an element.



            The ((++)) obviously increments your counter.



            Performing the diff, I use the --brief option. If you read the diff man page, it tells you that --brief only prints output when a difference is found. Therefore, the diff command succeeds if no diff is found.



            Using the && (AND) notation, this code echos the names of the files being compared into your placeholder file if and only if the diff command generates no output.



            If there is a difference between files, diff outputs the differences to the terminal. This causes the && (AND) to fail, so nothing is output to the placeholder file.



            If you have any further questions about the syntax, feel free to ask.






            share|improve this answer















            Here's a more efficient way of doing what you're trying to do. I use a smaller sample set to keep things clearer:



            #!/bin/bash

            # clear placeholder
            printf "Files with no diff:n" > placeholder

            # set up sample data
            echo "one" > file.00.txt
            echo "one" > file.01.txt
            echo "foo" > file.02.txt
            echo "bar" > file.03.txt
            echo "two" > file.04.txt
            echo "two" > file.05.txt

            # generate array
            i=0
            while [ $i -lt 6 ]; do
            array+=( file.`printf %02d $i`.txt )
            ((i++))
            done

            i=0
            while [ $i -lt 5 ]; do
            diff --brief $array[i] $array[i+1] &&
            echo "$array[i] $array[i+1]" >> placeholder
            ((i++))
            done


            Results:



            $ sh ./test.sh 
            Files file.01.txt and file.02.txt differ
            Files file.02.txt and file.03.txt differ
            Files file.03.txt and file.04.txt differ
            $ cat placeholder
            Files with no diff:
            file.00.txt file.01.txt
            file.04.txt file.05.txt


            You don't need to generate sample data if you actually already have data.



            The code, explained:



            Building an array in a loop (in Bash) can be done by iterating the way you obviously already know, but the array+= notation appends an element.



            The ((++)) obviously increments your counter.



            Performing the diff, I use the --brief option. If you read the diff man page, it tells you that --brief only prints output when a difference is found. Therefore, the diff command succeeds if no diff is found.



            Using the && (AND) notation, this code echos the names of the files being compared into your placeholder file if and only if the diff command generates no output.



            If there is a difference between files, diff outputs the differences to the terminal. This causes the && (AND) to fail, so nothing is output to the placeholder file.



            If you have any further questions about the syntax, feel free to ask.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 22 at 9:54

























            answered Mar 17 at 18:00









            Klaatu von SchlackerKlaatu von Schlacker

            2,337710




            2,337710







            • 2





              "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

              – ilkkachu
              Mar 17 at 19:37











            • Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

              – Klaatu von Schlacker
              Mar 22 at 9:53












            • 2





              "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

              – ilkkachu
              Mar 17 at 19:37











            • Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

              – Klaatu von Schlacker
              Mar 22 at 9:53







            2




            2





            "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

            – ilkkachu
            Mar 17 at 19:37





            "Because Bash interprets silence as success...", noo, not really. && and if check the exit code/exit status of the command. diff just happens to return a truthy exit code (i.e. zero) if the files are identical, and a falsy exit code (a one) if they differ. But you can have commands that produce output and return a truthy value, or the other way around, e.g. echo foo && echo "it's true" or false && echo "it's true"

            – ilkkachu
            Mar 17 at 19:37













            Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

            – Klaatu von Schlacker
            Mar 22 at 9:53





            Yes, you're right, my comment was unintentionally broad there. I'll edit for accuracy. Thanks.

            – Klaatu von Schlacker
            Mar 22 at 9:53

















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