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Parse a single line file and store value.

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# 1  
Old 02-24-2012
Parse a single line file and store value.
I have a single line file like this :

Average Fragmentation Quotient : 3.084121

Now I want to store the value which comes after ":" i,e 3.084121 into a variable.

And if this variable crosses above 6 i want to call another script...

can any one help me on this please ..
Last edited by vbe; 02-24-2012 at 10:03 AM.. Reason: typos...
# 2  
Old 02-24-2012
Code:
x=`awk '{print int($2)}' FS=: input_file`

if [ $x -gt 6 ]
then
   <call your script here>
fi

Guru.
# 3  
Old 02-24-2012
Hi Guru,

Thanks for the reply..

I want to make 3 more more updation like

1]f x is >0.6
2] less 7
3] less 0.8 ..
please suggest for same...
# 4  
Old 02-24-2012
Code:
var=$(cat infile)
var2=${var##* }

-or-

Code:
read x x x x var2 < infile

end then:
Code:
if [ ${var2%%.*} -gt 6 ]; then
  echo "do stuff with \$var2: $var2"
fi

In ksh93 you can just do this:
Code:
if [ $var2 -gt 6 ]; then

Last edited by Scrutinizer; 02-24-2012 at 10:52 AM..
# 5  
Old 02-24-2012
this dosent work for me ...
# 6  
Old 02-24-2012
What doesn't work for you?
# 7  
Old 02-24-2012
I am getting this error :

line 12: [: 0.6: integer expression expected
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