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shell command in AWK

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# 1  
Old 06-15-2010
shell command in AWK
Hi experts,

Sorry if it sounds basic question. I am trying to delete all folders except a particular and I have written the following. But its not deleting the folder.
Could you tell me whats wrong?

folders are:
Code:
1 2 3 4 5 6

Code:
ls -l | awk '{ if ($9 != 4) {system(/bin/rmdir $9);print "deleted",$9;}}'

Thanks

Amit

Moderator's Comments:
Mod Comment Use code tags please, ty.
Last edited by zaxxon; 06-15-2010 at 02:02 AM.. Reason: code tags
# 2  
Old 06-15-2010
/bin/rmdir needs to be enclosed by double quotes and leave a blank between the closing double quote and the command itself like:
Code:
ls -l | awk '{ if ($9 != 4) {system("/bin/rmdir " $9);print "deleted",$9;}}'
                                    ^          ^^

You are not omitting the line with "total" so you will get ugly output. This can be changed if you use
Code:
ls -1

instead of
Code:
ls -l

Also you can leave out the closing separator ";" after the last print since there is nothing more to delimit. Also if() is not needed since awk has testing built in like
Code:
ls -1| awk '$0 != "4" {system("/bin/rmdir " $0);print "deleted",$0}'

Last edited by zaxxon; 06-15-2010 at 02:21 AM.. Reason: added comments
# 3  
Old 06-15-2010
Hi,
Try This...

Code:
#!/bin/sh

ls -l | while read line
do
filename=`echo $line | awk '{print $9}'`
if [ "$filename" != "4" ]; then
   rmdir $filename
   if [ $? == 0 ]; then
   echo "$filename deleted"
   fi
fi
done

# 4  
Old 06-15-2010
If you want to do it by shell only, here is a compact way without expecting filenames/dirnames including a blank:
Code:
for a in *; do [ $a != "4" ] && (rmdir $a; echo "$a deleted"); done

# 5  
Old 06-15-2010
Thanks All for the help
# 6  
Old 06-15-2010
ls | grep -v 4 | xargs rm -rf

-v is the invert match and it will ignore filename/foldername when we do grep.
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