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Write a shell script that removes files that contain a string

Terminal - Write a shell script that removes files that contain a string
find . | xargs grep -l "FOOBAR" | awk '{print "rm -f "$1}' > doit.sh
2014-04-06 15:48:41
User: sergeylukin
Functions: awk find grep xargs
-3
Write a shell script that removes files that contain a string

After this command you can review doit.sh file before executing it.

If it looks good, execute: `. doit.sh`

Alternatives

There is 1 alternative - vote for the best!

Terminal - Alternatives
find -type f -exec grep -q "regexp" {} \; -delete
2014-04-06 19:06:50
User: gumnos
Functions: find grep
Tags: find grep
3

Deletes files in the current directory or its subdirectories that match "regexp" but handle directories, newlines, spaces, and other funky characters better than the original #13315. Also uses grep's "-q" to be quiet and quit at the first match, making this much faster. No need for awk either.

grep -Rl "pattern" files_or_dir
2014-04-06 18:18:07
User: N1nsun
Functions: grep
Tags: awk find grep
0

Grep can search files and directories recursively. Using the -Z option and xargs -0 you can get all results on one line with escaped spaces, suitable for other commands like rm.

Know a better way?

If you can do better, submit your command here.

What others think

Hmm... You don't need find and xargs at the beginning.

grep can search recursively:

grep -rl "FOOBAR" | ...

However if any of the file/dirnames contain whitespace, everything goes badly wrong.

Here's my version (quoting the filenames):

grep -rl "FOOBAR" | sed 's/.*/rm -f "\0"/' > doit.sh
Comment by flatcap 69 weeks and 1 day ago
grep -rl "FOOBAR" | xargs -L1 echo "rm -rf $1" > doit.sh
Comment by bmx666 69 weeks and 1 day ago

Your point of view

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