Re: Bitwise comparision of files

Here's a non-Perl variant.

In directory 1, run

$ find . -type f -print0 | xargs -0 md5sum >checksums
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to create a list of checksums. Then, form within directory 2, verify the list of checksums

$ md5sum -c checksums 2>&1 | grep -q FAILED\$ && echo FAIL
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(The advantage of using the checksumming technique is that the directories don't need to be accessible from the same machine (as a direct block-by-block comparison would require). You only need to copy the checksums file.)

This assumes that directory 2 isn't a subdirectory of directory 1. Also, you haven't specified what is supposed to happen, if there are additional files in directory 2, which are not present in directory 1 (they would be ignored by this approach).

Comment on Re: Bitwise comparision of files Select or Download Code

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Re^2: Bitwise comparision of files by JavaFan (Canon) on Feb 28, 2012 at 19:06 UTC
I'd use `diff -r directory1 directory2`.	[reply] [d/l]
Re^3: Bitwise comparision of files by Eliya (Vicar) on Feb 28, 2012 at 19:11 UTC
As I mentioned in parentheses, the advantage of using the checksumming technique is that the directories don't need to be accessible from the same machine. Whether that matters here, I don't know, but I figured it might perhaps be useful to someone, sometime.	[reply]
Re^4: Bitwise comparision of files by JavaFan (Canon) on Feb 28, 2012 at 21:17 UTC
As I mentioned in parentheses, the advantage of using the checksumming technique is that the directories don't need to be accessible from the same machine. Then I'd use something like `rsync -rcn directory1 remote:directory2`	[reply] [d/l]
Re^5: Bitwise comparision of files by Eliya (Vicar) on Feb 28, 2012 at 21:34 UTC