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Passing parameter in sed or awk commands to print for the specific line in a file

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# 1  
Old 12-07-2011
Passing parameter in sed or awk commands to print for the specific line in a file
Hi,

I am trying to print a specific line in a file through sed or awk. The line number will be passed as a parameter from the previous step. My code looks as below.
Code:
TEMP3=`sed -n '$TEMP2p' $FILEPATH/Log.txt`

$TEMP2, I am getting from the previous step which is a numerical value(eg:3).

But, i am getting an error as below
Code:
sed: can't find label for jump to `EMP2p'

Kindly, help me on this.


Regards

Moderator's Comments:
Mod Comment Please use code tags!
Last edited by zaxxon; 12-07-2011 at 08:17 AM.. Reason: code tags, see PM
# 2  
Old 12-07-2011
Whatever $TEMP2p has as value, it can't be substituted inside single quotes '. Either change it to double quotes " or have opening and closing single quotes before and behind the variable.
# 3  
Old 12-07-2011
Thanks Zaxxon! Both the options worked.
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