Correct recursive chmod, separate for files & dirs

Question

I'd like to recursively chmod a directory so that:

• Files are 0664
• Directories are 0775

How to do it better, shorter, fancier? :) Maybe, use umask somehow?

All find solutions are too long: I always end with Copy-Paste :)

Answer 1

Depending on your version of chmod, you may be able to do this:

chmod -R . ug+rwX,o+rX,o-w

Note the capital X. This sets the executable bit on directories and files that already have any of the execute bit already set.

Note that you can only use capital X with '+', not '=' or '-'.

Answer 2

Better, shorter, fancier than what ?

cd /directory
find . -type d -exec chmod 0755 {} +
find . -type f -exec chmod 0664 {} +

Answer 3

Adding a oneliner to the mix

find -type f -exec chmod 0644 {} + -o -type d -exec chmod 0755 {} +

Answer 4

find /dir -type f -print0 |xargs -0 chmod 0644
find /dir -type d -print0 |xargs -0 chmod 0775

Test it before use it in real environment

Answer 5

I am using this for anything copied from FAT filesystems:

chmod -R a-x+X .

If it does not work, for example on Mac OS X, try the GNU version of the command chmod:

gchmod -R a-x+X .

Answer 6

Without knowing more about why you're trying to do this, the most common reasons people tend to use are either:

• Files and directories are other-writeable

  chmod -R o-w /path/to/dir

• Files and directories are not group-writeable

  chmod -R g+w /path/to/dir

Or, combine the two:

chmod -R o-w,g+w /path/to/dir

Alternately, if you want files and directories to the correct permissions by default, modify the creating process's umask.

Basically, it's a rare day that it's correct to use numeric modes with chmod; typically, directories already have the executable bit set and files that don't need it lack it, so why muck around with it at all when the + and - operators obviate the need to do so?