Inside a directory, how can I delete files that lack any of the words specified, so that only files that contain ALL the words are left? I tried to write a simple bash shell script using grep and rm commands, but I got lost. I am totally new to Linux, any help would be appreciated
First, remove the file-list:
rm flist
Then, for each of the words, add the file to the filelist if it contains that word:
grep -l WORD * >>flist
Then sort, uniqify and get a count:
sort flist | uniq -c >flist_with_count
All those files in flsit_with_count that don't have the number of words should be deleted. The format will be:
2 file1
7 file2
8 file3
8 file4
If there were 8 words, then file1 and file2 should be deleted. I'll leave the writing/testing of the script to you.
Okay, you convinced me, here's my script:
#!/bin/bash
rm -rf flist
for word in fopen fclose main ; do
grep -l ${word} *.c >>flist
done
rm $(sort flist | uniq -c | awk '$1 != 3 {print $2} {}')
This removes the files in the directory that didn't have all three words:
How about:
grep -L foo *.txt | xargs rm
grep -L bar *.txt | xargs rm
If a file does not contain foo
, then the first line will remove it.
If a file does not contain bar
, then the second line will remove it.
Only files containing both foo
and bar
should be left
-L, --files-without-match
Suppress normal output; instead print the name of each input
file from which no output would normally have been printed. The
scanning will stop on the first match.
See also @Mykola Golubyev's post for placing in a loop.
This will remove all files that doesn't contain words Ping or Sent
grep -L 'Ping\|Sent' * | xargs rm
list=`Word1 Word2 Word3 Word4 Word5`
for word in $list
grep -L $word *.txt | xargs rm
done
Addition to the answers above: Use the newline character as delimiter to handle file names with spaces!
grep -L $word $file | xargs -d '\n' rm
To do the same matching filenames (not the contents of files as most of the solutions above) you can use the following:
for file in `ls --color=never | grep -ve "\(foo\|bar\)"`
do
rm $file
done
As per comments:
for file in `ls`
shouldn't be used. The below does the same thing without using the ls
for file in *
do
if [ x`echo $file | grep -ve "\(test1\|test3\)"` == x ]; then
rm $file
fi
done
The -ve reverses the search for the regexp pattern for either foo or bar in the filename. Any further words to be added to the list need to be separated by \| e.g. one\|two\|three
You could try something like this but it may break if the patterns contain shell or grep meta characters:
(in this example one two three are the patterns)
for f in *; do
unset cmd
for p in one two three; do
cmd="fgrep \"$p\" \"$f\" && $cmd"
done
eval "$cmd" >/dev/null || rm "$f"
done