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

The numbers are calculated by adding the binary values represented by r, w, and x

r = 100b = 4
w = 010b = 2
x = 001b = 1

in every group. In your case, -rw-r--r-- would be represented by

6(r+w=4+2)4(r=4)4(r=4)

so the relevant command is

chmod 644 path/to/file

Answer 2

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 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.

Answer 4

This might be straightforward

-bash-3.2$ stat --format=%a sample_file
755

Answer 5

Permissions are just the string representation of a binary number.
The 0 is mostly represented by -, the rest are letters.

basic

For basic permissions:

Convert all - and caps S or T to 0, the rest should represent 1.
The resulting binary number so constructed should be printed as octal:

$ a=-rw-r--r--
$ b=${a//[ST-]/0}
$ b=${b//[!0]/1}
$ printf '%04o\n' $((2#$b))
0644

In one line:

$ b=${a//[ST-]/0}; b=${b//[!0]/1}; printf '%04o\n' $((2#$b))
0644

Error correction and detecting the other 3 bits 1000, 2000 or 4000 require some more code:

#!/bin/bash

Say     (){ printf '%s\n' "$@"; }
SayError(){ a=$1; shift; printf '%s\n' "$@" >&2; exit "$a"; }

e1="Permission strings should have 10 characters or less"
e2="Assuming first character is the file type"
e3="Permission strings must have at least 9 characters"
e4="Permission strings could only contain 'rwxsStT-'"

a=$1

((${#a}>10))  &&   SayError 1 "$e1"
((${#a}==10)) && { Say        "$e2"; a=${a#?}; }
((${#a}<9))   &&   SayError 2 "$e3"
a=${a//[^rwxsStT-]}
((${#a}<9))   &&   SayError 3 "e4"
b=${a//[ST-]/0}; b=${b//[!0]/1}; c=0
[[ $a =~ [sS]......$ ]] && c=$((c|4)) 
[[ $a =~    [sS]...$ ]] && c=$((c|2)) 
[[ $a =~       [tT]$ ]] && c=$((c|1))

printf '%04o\n' "$((2#$b|c<<9))"

Answer 6

Get the list of files with their string and hex permission values. Putting %N at the end so the output can be put into Excel easier.

stat -c "%A %a %N" *

-rw-r--r-- 644 `file2.txt'
-rw-r--r-- 644 `file3.txt'
-rw-r--r-- 644 `file4.txt'
-rw-r--r-- 644 `file.txt'
drwxr-xr-x 755 `hsperfdata_root'
-rw-r--r-- 644 `junk.txt'
drwx------ 700 `vmware-root'

This will find all files with a specific hex permission.

find /tmp1 -user root -perm 644