Question

How can I write a bash script on Linux to determine which files in two directories have different permissions?

For example, I have two directories:

fold1 having two files:

 1- file1 (-rw-rw-r--)  
 2- file2 (-rw-rw-r--)

fold2 having same-name files with different permissions:

1- file1 (-rwxrwxr-x)  
2- file2 (-rw-rw-r--)

I need a script to output the file names that have different permissions,
so the script will print only file1

I am currently checking the permissions manually by displaying the files with:

for i in `find .`; do ls -l $i ls -l ../file2/$i; done

Answer 1

Parsing find . output with: for i in $(find .) is going to give you trouble for any filenames with spaces, newlines, or other perfectly normal characters:

$ touch "one file"
$ for i in `find .` ; do ls -l $i ; done
total 0
-rw-r--r-- 1 sarnold sarnold 0 2012-02-08 17:30 one file
ls: cannot access ./one: No such file or directory
ls: cannot access file: No such file or directory
$ 

Since permissions can also differ by owner or by group, I think you should include those as well. If you need to include the SELinux security label, the stat(1) program makes that easy to get as well via the %C directive:

for f in * ; do stat -c "%a%g%u" "$f" "../scatman/${f}" |
    sort | uniq -c | grep -q '^\s*1' && echo "$f" is different ; done

(Do whatever you want for the echo command…)

Example:

$ ls -l sarnold/ scatman/
sarnold/:
total 0
-r--r--r-- 1 sarnold sarnold 0 2012-02-08 18:00 funky file
-rw-r--r-- 1 sarnold sarnold 0 2012-02-08 18:01 second file
-rw-r--r-- 1 root    root    0 2012-02-08 18:05 third file

scatman/:
total 0
-rw-r--r-- 1 sarnold sarnold 0 2012-02-08 17:30 funky file
-rw-r--r-- 1 sarnold sarnold 0 2012-02-08 18:01 second file
-rw-r--r-- 1 sarnold sarnold 0 2012-02-08 18:05 third file
$ cd sarnold/
$ for f in * ; do stat -c "%a%g%u" "$f" "../scatman/${f}" | sort | uniq -c | grep -q '^\s*1' && echo "$f" is different ; done
funky file is different
third file is different
$