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How can I list the file/directory names in a directory recursively that are the same, but with different capitalization/case? ex.:
INPUT (not the ls command, the directories):
[user@localhost ~/a] ls -R
.:
b
./b:
ize Ize
./b/ize:
./b/Ize:
[user@localhost ~/a]
OUTPUT:
/b/ize
If you have GNU uniq, you can sort case insensitively (-i), and use -d to print only duplicate lines:
find . | sort -f | uniq -di
As @StephaneChazelas mentioned in his answer, this might not do what you expect if you can have duplicate paths that only differ in case (like a/b/foo and A/b/foo).
sort -f here. Also note that GNU uniq has the same limitation as GNU tr as in it doesn't work for matching case of multi-byte characters.
Aug 3, 2013 at 22:55
sort -f? If uniq can deal with the case, why would I also need to make sort case insensitive? And what do you mean by multi-byte characters? Things like \n,\r etc? How can they have different cases?
export LC_ALL=C; printf '%s\n' a A b B | sort | uniq -di. Some locales sort case-insensitively, some others (like C) don't. uniq needs a sorted input, its duplicate lines must be adjacent.
Aug 4, 2013 at 7:25
Assuming file names don't contain newline characters, you could do something like:
find . | tr '[:upper:]' '[:lower:]' | sort | uniq -d
Note that some tr implementations like GNU tr don't change the case of multi-byte characters.
Also note that the path it reports may not be the paths of any file. For instance, if there's a ./a/b/fOo and a ./A/b/fOo file, it will report ./a/b/foo. If it's not what you want, you may want to refine your requirements.
All of these ideas are bad. Use checksums and be sure the files are the same. Then the task becomes easy.
find . -type f -exec md5sum {} + |
sort |
perl -a -nE'push(@{$db{$F[0]}},$F[1]);END{for(keys%db){say"Dupe detected @{@db{$_}}"if scalar@{$db{$_}}>1}}'
This will sha1sum every file in the directory and all subdirectorys and output all dupes of that file, if there are any. I made the pipeline multiline for readability.
