Is it possible when listing a directory to view numerical unix permissions such as 644 rather than the symbolic output -rw-rw-r--
Thanks.
Is it possible when listing a directory to view numerical unix permissions such as 644 rather than the symbolic output -rw-rw-r--
Thanks.
it almost can ..
ls -l | awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwx]/) \
*2^(8-i));if(k)printf("%0o ",k);print}'
Closest I can think of (keeping it simple enough) is stat
, assuming you know which files you're looking for. If you don't, *
can find most of them:
/usr/bin$ stat -c '%a %n' *
755 [
755 a2p
755 a2ps
755 aclocal
...
@The MYYN
wow, nice awk! But what about suid, sgid and sticky bit?
You have to extend your filter with s and t, otherwise they will not count and you get the wrong result. To calculate the octal number for this special flags, the procedure is the same but the index is at 4 7 and 10. the possible flags for files with execute bit set are ---s--s--t
amd for files with no execute bit set are ---S--S--T
ls -l | awk '{
k = 0
s = 0
for( i = 0; i <= 8; i++ )
{
k += ( ( substr( $1, i+2, 1 ) ~ /[rwxst]/ ) * 2 ^( 8 - i ) )
}
j = 4
for( i = 4; i <= 10; i += 3 )
{
s += ( ( substr( $1, i, 1 ) ~ /[stST]/ ) * j )
j/=2
}
if ( k )
{
printf( "%0o%0o ", s, k )
}
print
}'
For test:
touch blah
chmod 7444 blah
will result in:
7444 -r-Sr-Sr-T 1 cheko cheko 0 2009-12-05 01:03 blah
and
touch blah
chmod 7555 blah
will give:
7555 -r-sr-sr-t 1 cheko cheko 0 2009-12-05 01:03 blah
On the command line this works fine:
ls -la | awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwx]/)*2^(8-i));if(k)printf("%0o ",k);print}'
The question is how does one put this as an alias in the .bashrc_aliases file? I MUST point out that I did not create this script, I found it on one of the numerous Linux websites and have since been unable to track it down.