Variable value substitution issue with awk command issue


 
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# 1  
Old 03-01-2013
Code Variable value substitution issue with awk command issue

Hi All,

I am using the below script which has awk command, but it is not returing the expected result. can some pls help me to correct the command.

The below script sample.ksh should give the result if the value of last 4 digits in the variable NM matches with the variable value DAT. The script is written with awk command as shown below but the DAT variable value is not substituting in the awk command. Please help me how to fix this issue. Expected output provided below.

sample.ksh
========
Code:
#!bin/ksh
DAT="0301"
NM="xyzxyz.4560301"
echo $NM | awk '/$DAT$/  { print $1 }'

expected output:
=============
Code:
xyzxyz.4560301


Thanks in Advance
G.K.K

Last edited by Franklin52; 03-01-2013 at 03:08 AM.. Reason: Please use code tags for data and code samples
# 2  
Old 03-01-2013
Code:
echo $NM | awk -v D="$DAT" ' $0 ~ D { print $1 }'

Also fix your shebang:
Code:
#!/bin/ksh

This User Gave Thanks to Yoda For This Post:
# 3  
Old 03-01-2013
Quote:
Originally Posted by bipinajith
Code:
echo $NM | awk -v D="$DAT" ' $0 ~ D { print $1 }'

Also fix your shebang:
Code:
#!/bin/ksh

Thanks for the REply.

But that awk command will work where ever the DT variable value exists in the string NM. But i want to check against only last 4 digits in the variable NM. If it matches then it should return the entire variable value NM. Else it should return nothing.

Example: NM="xyzxyz.0301456" , in this string last four digits value is not matching with the DT value 0301. If i use the above awk command then it will still return the result. But i want to check against only las4 4 digits.
# 4  
Old 03-01-2013
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.

Last edited by Jotne; 03-01-2013 at 03:05 AM..
# 5  
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"'
# 6  
Old 03-01-2013
Code:
echo "$NM" | awk 'substr($0,lenght($0)-3,4)==D {print $1}' D="$DAT"

PS no need to quote message directly above you.
# 7  
Old 03-01-2013
All solutions above rely on the fact that DAT's length is 4. Why not use that fact and run
Code:
$ echo "$NM" | awk '$0 ~ D"$"' D="$DAT"
xyzxyz.4560301

which is close to Jotne's first proposal with some typos corrected.
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