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awk search and replace nth column by using a variable.

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# 1  
Old 12-29-2015
awk search and replace nth column by using a variable.
I am passing a variable and replace nth value with the variable.
I tried using many options in awk command but unable to ignore the special characters in the output and also unable to pass the actual value.

Input : "1","2","3"
Output : "1","1000","3"

Code:
TempVal=`echo 1000`
Cat file_name.txt | awk -F, -v OFS=, '{$2="\"$TempVal\""; print }

'

After running the above command - Output :

Code:
"1","$TempVal","3"

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Mod Comment This is ridiculous. Using CODE tags correctly is not that hard. Please review this tutorial that you have seen at least 10 times before:
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Last edited by Don Cragun; 12-29-2015 at 04:19 AM.. Reason: Get rid of extraneous HTML tags, add CODE and ICODE tags.
# 2  
Old 12-29-2015
Couple of things, You don't need to use cat to pass the file content to awk. It's useless use of cat.

You can specify the single quote to get the shell variable interpolate.

Code:
awk -F, '{$2="\"'$TempVal'\""; print }' OFS=, file_name.txt

(Or)

You can use -v option to assign the shell variable data into awk variable and make use of the awk variable inside the awk block.

Code:
awk -F, -v t="\"$TempVal\"" '{$2=t; print }' OFS=, file_name.txt

-Ranga
Last edited by rangarasan; 12-29-2015 at 03:51 AM.. Reason: Format
These 2 Users Gave Thanks to rangarasan For This Post:
mystition onesuri
# 3  
Old 12-29-2015
In addition to what Ranga has already suggested, you don't need to use command substitution to assign a simple string to a variable. Use:
Code:
TempVal=1000

not:
Code:
TempVal=`echo 1000`

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