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Handling special characters using awk

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# 1  
Old 11-08-2007
Handling special characters using awk
Hi all,
How do I extract a value without special characters? I need to extract the value of %Used from below and if its greater than 80, need to send a notification.
I am doing this right now..Its giving 17%..Is there a way to extract the value and assign it to a variable in one step?

df |grep /dev/hd9var |awk '{print $4}' > temp.out
Value=`cat temp.out`

df....
Filesystem 512-blocks Free %Used Iused %Iused Mounted on
/dev/hd9var 1048576 876448 17% 436 1% /var

-Thanks.
# 2  
Old 11-08-2007
Code:
$ echo "17%" | sed s/%//
17

# 3  
Old 11-08-2007
All in one step, eh?

Code:
df | awk '$1==F{sub("%","",$4);print $4}' F=/dev/hd9var  | read VALUE

or

Code:
df /dev/hd9var | awk 'FNR==2{sub("%","",$4);print $4}' | read VALUE

And no temp files, either. Ain't awk cool?
# 4  
Old 11-08-2007
Thanks
Yep...works great.
Thanks.
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