#!/usr/bin/env python3ĭef compare_dirs(d1: "old directory name", d2: "new directory name"):
So the output is quite concise and the script works fast with large directories. Also it doesn't go inside subdirectories which are missing in one of the directories. Unlike many other solutions it doesn't compare contents of the files. Inspired by Sergiy's reply, I wrote my own Python script to compare two directories. u, -update skip files that are newer on the receiverĪ one-liner: rsync -rtOvcsu -progress -n /dir1/ /dir2/ & rsync -rtOvcsu -progress -n /dir2/ /dir1/ In case you do that, maybe a good option is to use -u, to avoid overwriting newer files.
That is copying the list of files to the second folder. You can delete the -n option to undergo the changes. s, -protect-args no space-splitting only wildcard special-chars c, -checksum skip based on checksum, not mod-time & size O, -omit-dir-times omit directories from -times n, -dry-run perform a trial run with no changes made The same for dir2 #from the rsync -help : With the previous line, you will get files that are in dir1 and are different (or missing) in dir2. Maybe one option is to run rsync two times: rsync -rtOvcs -progress -n /dir1/ /dir2/ Or as a single command using process substitution: diff every file is listed in output then): git diff -no-index dir1/ dir2/ Then compare the result two files with diff: diff -u dir1.txt dir2.txt Use find to list all the files in the directory then calculate the md5 hash for each file and pipe it sorted by filename to a file: find /dir1/ -type f -exec md5sum + | sort -k 2 > dir2.txt A good way to do this comparison is to use find with md5sum, then a diff.
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