I have a lot of zip files. Some are not downloaded correctly and are corrupted. I want to remove them.
Is there a way to find the corrupted archives in bash?
2 Answers
With GNU (for -readable and -iname) find:
find . -iname '*.zip' -type f -readable ! -exec unzip -t {} \; -exec rm -i {} \;
answered Nov 12, 2012 at 15:28
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1@Nima It's a logical not operator, ie. if
unzip -t {}fails, execrm -i {}. It has to be escaped because by default!is used for history expansion inbash. Nov 12, 2012 at 16:02 -
2@ChrisDown Is there actually a shell that doesn't treat
!literally when it's followed by a space? I have the habit of writing\!too, but I thought it was out of habit (because!has to be quoted in so many places in bash), not out of necessity. Nov 12, 2012 at 23:35 -
@Gilles - Good point! I don't know of any. I guess this goes also to the manual pages saying that
{}needs escaping... Nov 13, 2012 at 2:21 -
maybe @Nima wants to remove the files manualy then using
-print0instead ofexec rm -i {} \;will display the files that are corrupted without deleting them. Nov 13, 2012 at 3:06
The following will print the name of all corrupted zip files in the current directory and its subdirectories:
#!/bin/bash
shopt -s dotglob nullglob globstar
for file in ./**/*.zip; do
[[ -r $file ]] || continue
unzip -t "$file" >/dev/null 2>&1 || printf '%s\n' "$file"
done
If you wish to remove them, simply replace printf '%s\n' "$file" with rm -f "$file".
answered Nov 12, 2012 at 15:12
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1Beware though that it's not the same permissions that apply to read a file and to delete it, so you may end up deleting a file just because you don't have the right to read it. Also note that bash recursive globbing (contrary to other shells) descends into symlinks to directories which is generally not what you want. Also, you may want to write it
./**/*.zipto avoid problems with file/dir names starting with "-". Nov 12, 2012 at 15:21 -