How to get permission number by string : -rw-r--r--

Question

I need to set the same chmod, how to get number for -rw-r--r-- ?

Answer 1

Please check stat output:

# stat .xsession-errors 
  File: ‘.xsession-errors’
  Size: 839123          Blocks: 1648       IO Block: 4096   regular file
Device: 816h/2070d      Inode: 3539028     Links: 1
Access: (0600/-rw-------)  Uid: ( 1000/     lik)   Gid: ( 1000/     lik)
Access: 2012-05-30 23:11:48.053999289 +0300
Modify: 2012-05-31 07:53:26.912690288 +0300
Change: 2012-05-31 07:53:26.912690288 +0300
 Birth: -

Answer 2

r=4
w=2
x=1

in every group. Your example is 6(r+w=4+2)4(r=4)4(r=4).

Answer 3

The full permissions mode number is a 4-digit octal number, though most of the time, you only use the 3 least-significant digits. Add up each group in the permissions string, taking r=4, w=2, x=1. For example:

 421421421
-rwxr-xr--
 \_/        -- r+w+x = 4+2+1 = 7
    \_/     -- r+_+x = 4+0+1 = 5
       \_/  -- r+_+_ = 4+0+0 = 4     => 0754

Now, sometimes you'll see an odd modestring like this:

-rwsr-xr-T

The fourth digit is overloaded onto the x bits in the modestring. If you see a letter other than x there, then it means one of these "special" fourth-digit bits is set, and if the letter is lower case, then x for that position is also set. So the translation for this one is:

   4  2  1
 421421421
-rwsr-xr-T
   +  +  +  -- s+_+T = 4+0+1 = 5  
 \_/        -- r+w+s = 4+2+1 = 7  (s is lowercase, so 1)
    \_/     -- r+_+x = 4+0+1 = 5
       \_/  -- r+_+T = 4+0+0 = 4  (T is uppercase, so 0)   => 05754

The standard UNIX way to show that a number is octal is to start it with a zero. GNU chmod will assume the mode you're giving it is octal anyway, but it's safest to prepend the zero.

Finally, if you see a + at the end of the modestring:

-rwxr-xr-x+

then that means the file has extended permissions, and you'll need more than chmod. Look into the setfacl and getfacl commands, for starters.