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What is the best and simplest way to compare two directory structures without actually comparing the data in files. This works fine:

diff -qr dir1 dir2

But it's really slow because it's comparing files too. Is there a switch for diff or another simple cli tool to do this?

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By "directory structure", do you mean just the directory paths, or the paths of both directory and non-directory files? –  intuited Jul 22 '10 at 6:26
    
Yes, folders and files. –  Jonah Jul 22 '10 at 15:30
    
In that case you should remove the -type d option from @slartibartfast's answer, or check out my answer. –  intuited Jul 22 '10 at 17:52

9 Answers 9

up vote 16 down vote accepted

The following (if you substitute the first directory for directory1 and the second for directory2) should do what you're looking for and swiftly:

find directory1 -type d -printf "%P\n" | sort > file1
find directory2 -type d -printf "%P\n" | sort | diff - file1

The fundamental principle is that it prints out all of the directories including subdirectory paths relative to the base directoryN directories.

This could fall down (produce wierd output) if you have carriage returns in some of the directory names but not others.

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1  
+1... but I think you mean sort > file1 –  Jarvin Jul 22 '10 at 3:39
    
Whoops, earlier comment got %P confused with %p. Cleverly hiding my tracks. These aren't the droids you're looking for. –  intuited Jul 22 '10 at 6:20
    
Is a way to do that in single line/pipe ? ;) –  marioosh Oct 29 '13 at 9:33
vimdiff <(cd dir1; find . | sort) <(cd dir2; find . | sort)

will give you a nice side-by-side display of the two directory hierarchies with any common sections folded.

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+1 for nice ui solution without creating additional files –  Griddo Nov 19 '13 at 11:52

I usually use rsync for this task:

rsync -nav --delete DIR1/ DIR2

BE VERY CAREFUL to always use the -n, aka --dry-run, option, or it will synchronize (change the contents of) the directories.

This will compare files based on file modification times and sizes... I think that's what you really want, or at least you don't mind if it does that? I got the sense that you just want it to happen faster, not that you need it to ignore the difference between file contents. If you do want it to not list differing files with identical names, I think the addition of the --ignore-existing option will do that.

Also be aware that not putting a / at the end of DIR1 will cause it to compare the directory DIR1 with the contents of DIR2.

The output ends up being a bit verbose, but it will show you which files/directories differ. Files/directories present in DIR2 and not in DIR1 will be prefaced with the word deleting.

For some situations, @slartibartfast's answer may be more appropriate, though you'll need to remove the -type d option to enable the listing of non-directory files. rsync will be faster if you've got a significant number of files/directories to compare.

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This is optimum solution

diff --brief -r dir1 dir2

--brief switch reports only whether the files differ, not the details of the difference.

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The OP already has -q in the question, which is an alias for --brief. This answer doesn't provide any new information. –  Michael Dorst Aug 20 '13 at 23:54

Similar to the ls answer but if you install tree then you can

tree dir1 > out1
tree dir2 > out2
diff out1 out2
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ls > dir1.txt

ls > dir2.txt

Then just diff the two lists.

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It seems like the OP wants a heirarchy of paths. This will diff all files in the current directory. It's debatable, but possible, that he just wants directories; he might want filenames rather than the contents of files. –  intuited Jul 22 '10 at 6:24
    
@intuited - you're right. I misread it. –  MDMarra Jul 22 '10 at 13:16

I was just looking for solution for this problem. The solution that I liked the most was:

comm <(ls DIR1) <(ls DIR2)

It gives you 3 columns: 1 - files only in DIR1, 2 - files only in DIR2, 3 - files only in DIR3 For more details look at this blog post.

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Where is DIR3 specified? All I see is DIR1 and DIR2. –  Michael Dorst Aug 20 '13 at 23:59
    
I tried it, and (from what I can tell) the output was: all the files only in DIR1 in column 1, all the files only in DIR2 in column 2, and all the files shared by both in column 3. That's sort of useful, but do you know how one might strip out column 3 and leave only the differences? I have a lot of files to sort through, and most of it is identical. I don't need to see what's the same. –  Michael Dorst Aug 21 '13 at 0:14
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Also, I found that comm <(ls DIR1) <(ls DIR2) did not work recursively. For that I used comm <(ls -R1 DIR1) <(ls -R1 DIR2). ls -R crawls through directories recursively, and ls -1 (note that that is a one, not an L) makes ls print only one filename per line. –  Michael Dorst Aug 21 '13 at 0:22
    
@Michael: comm -3 (see man comm). –  Josh Jul 20 at 11:31

use "diff -qr" to get the different files and then filter out the file comparison with grep in order to only get the filenames that are only in one of the directories.

diff -qr dir1 dir2 | grep -v "Files.*differ" 
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I think only rsync is userfull. why?

diff is useful only for structures keeping files and directories. Diff does not give adequate exit codes when we use symlinks. In that situation diff can return 2 exit codes, even if src and dst are identical (times, sizes, names, timestamps, pointing softlinks etc).

dir, the filesystem does not guarantee file ordering, even if directory contents on src and dst are identical. Maybe you should filter the ls output by sorting it. But pure ls displays only node names.

maybe script including diff, cmp, test -X for node types will be usefull, but remember about overload made by many test/cmp runs. The script will be very slow.

As usual, if you want get simple info "dirs is/isn't identical", you should use rsync with the -n (dry) option. If you want to find what is different, use the diff command.

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