Check if any file exist in a folder when traversing through it
Clash Royale CLAN TAG#URR8PPP
I have this script:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
# do something ........
# [...........]
rm $i
done
It throws an error when the folder is empty.
*: No such file or directory
rm: cannot remove '*': No such file or directory
How can I fix that?
bash files directory
add a comment |
I have this script:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
# do something ........
# [...........]
rm $i
done
It throws an error when the folder is empty.
*: No such file or directory
rm: cannot remove '*': No such file or directory
How can I fix that?
bash files directory
add a comment |
I have this script:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
# do something ........
# [...........]
rm $i
done
It throws an error when the folder is empty.
*: No such file or directory
rm: cannot remove '*': No such file or directory
How can I fix that?
bash files directory
I have this script:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
# do something ........
# [...........]
rm $i
done
It throws an error when the folder is empty.
*: No such file or directory
rm: cannot remove '*': No such file or directory
How can I fix that?
bash files directory
bash files directory
asked Jan 8 at 5:17
JokkiJokki
6
6
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If you want to use your original code, then you just need to check if file exists, using an if condition, i.e.:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
if [[ -f "$i" ]]
then
# do something ........
# [...........]
rm $i
fi
done
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
add a comment |
try with ls
for i in $(ls)
do
# Do something
rm $i
done
2
Parsing the output ofls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.
– Jeff Schaller
Jan 8 at 11:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want to use your original code, then you just need to check if file exists, using an if condition, i.e.:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
if [[ -f "$i" ]]
then
# do something ........
# [...........]
rm $i
fi
done
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
add a comment |
If you want to use your original code, then you just need to check if file exists, using an if condition, i.e.:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
if [[ -f "$i" ]]
then
# do something ........
# [...........]
rm $i
fi
done
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
add a comment |
If you want to use your original code, then you just need to check if file exists, using an if condition, i.e.:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
if [[ -f "$i" ]]
then
# do something ........
# [...........]
rm $i
fi
done
If you want to use your original code, then you just need to check if file exists, using an if condition, i.e.:
local_dir="/some/dir1/"
cd $local_dir
for i in *.*
do
if [[ -f "$i" ]]
then
# do something ........
# [...........]
rm $i
fi
done
answered Jan 8 at 6:06
P_YadavP_Yadav
1,7563924
1,7563924
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
add a comment |
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
1
1
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
A simple -f doesn't need doubled , and don't forget that you'll miss files that don't have a . in them.
– stolenmoment
Jan 8 at 11:26
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
@stolenmoment, what's needed?
– Jokki
Jan 8 at 17:33
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
Just "if [ -f $i ]" will do, and * should get all files, not just files with a . In the name.
– stolenmoment
Jan 9 at 14:18
add a comment |
try with ls
for i in $(ls)
do
# Do something
rm $i
done
2
Parsing the output ofls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.
– Jeff Schaller
Jan 8 at 11:05
add a comment |
try with ls
for i in $(ls)
do
# Do something
rm $i
done
2
Parsing the output ofls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.
– Jeff Schaller
Jan 8 at 11:05
add a comment |
try with ls
for i in $(ls)
do
# Do something
rm $i
done
try with ls
for i in $(ls)
do
# Do something
rm $i
done
answered Jan 8 at 5:38
KamarajKamaraj
2,9441513
2,9441513
2
Parsing the output ofls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.
– Jeff Schaller
Jan 8 at 11:05
add a comment |
2
Parsing the output ofls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.
– Jeff Schaller
Jan 8 at 11:05
2
2
Parsing the output of
ls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.– Jeff Schaller
Jan 8 at 11:05
Parsing the output of
ls
is too tricky to get right. See unix.stackexchange.com/q/128985/117549 for one example.– Jeff Schaller
Jan 8 at 11:05
add a comment |
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