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# 1  
Old 03-16-2009
substr
Quote:
awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwxs]/)*2^(8-i));if(k)printf(" %0o ",k);print}'
can anybody explain this code?

thanks in advance..Smilie
# 2  
Old 03-17-2009
can you provide the some data of the file on which this awk command has been run..
that will help us to explain it in much better way
# 3  
Old 03-17-2009
in ls -l to list files in numerals like 766 755 etc
ls -l | awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwxs]/)*2^(8-i));if(k)printf(" %0o ",k);print}'
# 4  
Old 03-17-2009
this will convert the file permission from rwxs to octal values..
this simple perl code will also do that
Code:
perl -e 'for (@ARGV) { printf "%04o %s",  (stat)[2] & 07777, $_ }'

# 5  
Old 03-17-2009
I know that it will convert that.. but i cant understand how does it do that..
It will be better if u explain the code.
# 6  
Old 03-17-2009
Quote:
Originally Posted by janani_kalyan
I know that it will convert that.. but i cant understand how does it do that..
It will be better if u explain the code.
dear janani_kalyan,
I think this much explanation is enough.
I have explained a lot of code like this here so first time i said what it will do but i think you are very new to unix..
regards,
vidya
Code:
 
 awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwxs]/)*2^(8-i));if(k)printf(" %0o ",k);print}'
ls -l give a o/p like below
-rwxr-xr-x    1 fnsonlu1 fnsur             0 Mar 17 12:03 vidya
where $1=-rwxr-xr-x
first iteration of for loop when i=1
substr($1,2,1)=r  ##it will get the second char of $1 that is r
then it matches with either of r,w,x,s
if it matches with one of four then it adds 2 to the power 7
printf(" %0o ",k) then prints the k value in octal that is 7
the same continue till i reaches 8 that is last char of $1

# 7  
Old 03-17-2009
thanks vidhya. yes i'm new to unix so required lot of explanation thanks for your explaining me in detail.
 
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