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I am trying to transpose a content of a file into another.

Input file Test.txt:

HLRSN = 3
IMSI = 404212109727229
KIVALUE = A24AD11812232B47688ADBF15CE05CA9
K4SNO = 1
CARDTYPE = SIM
ALG = COMP128_3

HLRSN = 3
IMSI = 404212109727230
KIVALUE = A24AD11812232B47688ADBF15CE05CB8
K4SNO = 1
CARDTYPE = SIM
ALG = COMP128_3

HLRSN = 3
IMSI = 404212109727231
KIVALUE = A24AD11812232B47688ADBF15CE05CD6
K4SNO = 1
CARDTYPE = SIM
ALG = COMP128_3

Output needed in another text file:

3,404212109727229,A24AD11812232B47688ADBF15CE05CA9,1,SIM,COMP128_3
3,404212109727230,A24AD11812232B47688ADBF15CE05CB8,1,SIM,COMP128_3
3,404212109727231,A24AD11812232B47688ADBF15CE05CD6,1,SIM,COMP128_3

A bash solution:

declare -a out

EOF=false
IFS=$'='

until $EOF; do
 read -r skip val || EOF=true
 if [ ! -z "$val" ]
 then
 out+=("$val//[[:space:]]/")
 else
 tmp="$out[@]"
 printf '%sn' "$tmp// /,"
 out=()
 fi
done < file

How does this work

• Declare array out for holding output line, set variable EOF to keep track end of file, IFS for input field separator for read.


• Until we read end of file, we read each line of file, set value of last field to variable val.


• 
  if [ ! -z "$val" ]: check if length of variable $val is not zero, we remove space in $val, push it to array out.


• if length $val is zero, meaning we get blank line or end of file, we assign all element of array out to variable tmp, then replace all space variable tmp by ,, our designed output recode separator.


• Set out to null for next work.

Another solution, more concise, shorter for you is using perl:

$ perl -F'=' -anle '
 BEGIN $, = "," 
 push @out,$F[-1] if @F;
 print @[map s/s// && $_ @out] and @out = ()
 if /^$/ or eof;
' file
3,404212109727229,A24AD11812232B47688ADBF15CE05CA9,1,SIM,COMP128_3
3,404212109727230,A24AD11812232B47688ADBF15CE05CB8,1,SIM,COMP128_3
3,404212109727231,A24AD11812232B47688ADBF15CE05CD6,1,SIM,COMP128_3

Simply:

awk -v RS= -v OFS=, 'print $3,$6,$9,$12,$15,$18'

An empty record separator (RS=) enables the paragraph mode whereby records are separated by sequences of empty lines. Inside a record, the default field separator applies (records are separated by blanks) so in each record, the fields we are interested in are the 3rd, 6th, 9th...

We change the output field separator to a comma character (OFS=,) and print the fields we're interested in.

Save the following to a file (eg split.awk)

BEGIN 
RS="nn";
FS="n";
ORS=",";



 for (i=1;i<=NF;i++)
 
 split($i, sf, "= ")
 print sf[2]
 
 printf "n"
 

Then run:

awk -f split.awk Test.txt

Or run the whole command as one line:

awk 'BEGIN RS="nn";FS="n";ORS=",";for(i=1;i<=NF;i++)split($i, sf, "= ")print sf[2]printf "n"' Test.txt

It works as follows:

• The BEGIN block runs once at the start and sets the record seperator (RS) to two newlines and the field seperator (FS) to a single newline. The output record seperator (ORS) is set to a comma.




• It then loops through each field in the record (NF is the number of fields in the current record) and splits it on "= ".




• It then prints the right hand of this split with a comma between each (the ORS)




• After each line it prints a newline to give you the CSV format.

Based on the other answers... I needed a slight modification, because I was in need of the column header in the final CSV-file. This is my awk-solution:

awk -v RS= -v FS='n' '
 NR==1 
 for (i=1; i<=NF; i++) 
 split($i,line,"= "); 
 printf "%s%s", line[1], (i==NF ? "n" : ",");
 
 
 
 for (i=1; i<=NF; i++) 
 split($i,line,"= "); 
 printf "%s%s", line[2], (i==NF ? "n" : ",");
 
 ' 

As in the other answers, this reads the entire block (until a blank line) in fields, then splits each field again with split(). The only difference is that for the first record, it prints the part before the '=' and in a second for-loop, it prints the right-hand-side for all records.

If all the blocks have exactly the same format (same field names, in the same order), then you can use awk in “paragraph mode”, and print the desired field number in each block. If there are always spaces around the equal signs and the values never contain spaces, you can rely on whitespace-separated fields.

awk -v RS= -v ORS=',' 'print $3, $6, $9, $12, $15, $18'

If you can rely on the order and presence of fields but not on the whitespace, then you'll need a bit of parsing to split at the equal signs.

awk -v RS= -F 'n' '
 for (i = 1; i <= NF; i++) 
 sub(/[^=]*= */, "", $i);
 printf "%s%s", $i, (i==NF ? "n" : ",");
 
 '

Here's a Perl method:

perl -000 -ne '
 $, = ","; $ = "n";
 @kv = split /n| *= */;
 print @kv[grep $_%2 0..$#kv];
 '

If the fields in the blocks can come out of order and you always want a specific order as output, you'll need to store the fields and print them out in the right order at the end of each paragraph. In awk, this is easier to do in line mode than in paragraph mode.

awk -v OFS=',' '
 match($0, / *= */) a[substr($0,1,RSTART-1)] = substr($0,RSTART+RLENGTH)
 /^$/ print a["HLRSN"], a["IMSI"], a["KIVALUE"], a["K4SNO"], a["CARDTYPE"], a["ALG"]; split("", a)
 END print a["HLRSN"], a["IMSI"], a["KIVALUE"], a["K4SNO"], a["CARDTYPE"], a["ALG"]
'

This one is a one-liner in Perl.

perl -000 -F'/n|s*=s*/' -ane '%F = @F; $ = "n"; $, = ","; print @Fqw(HLRSN IMSI KIVALUE K4SNO CARDTYPE ALG)'