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Full Discussion: Variable value substitution issue with awk command issue
Top Forums Shell Programming and Scripting Variable value substitution issue with awk command issue Post 302773943 by G.K.K on Friday 1st of March 2013 02:08:04 AM
Old 03-01-2013
Quote:
Originally Posted by Jotne
As bipinajith write, you can not use a variable directly in your awk code
You need to declare a variable. This can be done before the code with -v or after the code like this.
Code:
echo $NM | awk -v D="$DAT" '$0~"D$" {print $1}'
echo $NM | awk '$0~"D$" {print $1} D="$DAT"'

If NM do contain multiple lines, you should add quotes to it:
Code:
echo "$NM" | awk '$0~"D$" {print $1} D="$DAT"'

PS after posting more than 40 post here, you should know how to use code tags to.
Thanks I understood.

But how to check the variable value DAT against the last 4 digits of variable value NM using awk command.

---------- Post updated at 11:08 PM ---------- Previous update was at 11:06 PM ----------

Quote:
Originally Posted by Jotne
As bipinajith write, you can not use a variable directly in your awk code
You need to declare a variable. This can be done before the code with -v or after the code like this.
The D$ should march only end of line
Code:
echo $NM | awk -v D="$DAT" '$0~"D$" {print $1}'
echo $NM | awk '$0~"D$" {print $1} D="$DAT"'

If NM do contain multiple lines, you should add quotes to it:
Code:
echo "$NM" | awk '$0~"D$" {print $1} D="$DAT"'

PS after posting more than 40 post here, you should know how to use code tags to.
Thanks , I understood.

But here i want to check the last 4 digits only of a variable value NM. So, how to correct the syntax for below code:
echo "$NM" | awk '$0~"D$" {print $1} D="$DAT"'
 

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