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Regular expression query in AWK

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# 1  
Old 06-05-2006
Question Regular expression query in AWK
Hi,
I have a string like this-->"After Executing service For 10 Request"
in this string i need to extract "10".
the contents of the string is variable and "10" appears before "For" and after "Request" i.e, in this format "For x Request"
I need to extract the value of x. How to do this in AWK?

Regards,
Omprasad
# 2  
Old 06-05-2006
Try:
Code:
echo "${string}"|awk '/After Executing service For .* Request/ {VAL=$5;print VAL}'

# 3  
Old 06-06-2006
Quote:
req_no=`echo "${string}" | awk { print $5}`
Try this .....
# 4  
Old 06-06-2006
Hi,
The solution that have been proposed may not solve my problem because the position of x i.e 10 in this case may not always be 5th word in the string.
Only thing that is constant here is the pattern "For X Request".
First I need to find this pattern and then extract x from it.
# 5  
Old 06-06-2006
Use:
Code:
echo "${string}"|awk '/For .* Request/ {for (i=0;i<=NF;i++){if ($i == "For" && $(i+2) == "Request"){VAL=$(i+1)}}print VAL}'

# 6  
Old 06-06-2006
That works....thanks Klashxx.
But how to do the same inside a script file. I have the entire string in a variable. Smilie
Last edited by omprasad; 06-06-2006 at 06:38 AM..
# 7  
Old 06-06-2006
Try using expr...
Code:
a="After Executing service For 10 Request"
b=$(expr match "$a" ".*For \(.*\) Request.*")
echo $b
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