Parse a single line file and store value.

Shell Programming and Scripting
1

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 ..

2 
x=`awk '{print int($2)}' FS=: input_file`

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

Guru.

3

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 
var=$(cat infile)
var2=${var##* }

-or-

read x x x x var2 < infile

end then:

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

In ksh93 you can just do this:

if [ $var2 -gt 6 ]; then
5

this dosent work for me ...

6

What doesn't work for you?

7

I am getting this error :

line 12: [: 0.6: integer expression expected

8

That is because only integer expressions can be use in test statements in any shell other than ksh93. Did you try earlier this suggestion?

if [ ${var2%%.*} -gt 6 ]; then

$ echo $var2
6.1
$ echo ${var2%%.*}
6

Otherwise you could use the utility bc ..