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How to correct this awk code without eval?


 
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# 1  
How to correct this awk code without eval?

Hi everyone,

The following piece of awk code works fine if I use eval builtin
Code:
var='$1,$2'
ps | eval "awk '{print $var}'"

But when I try to knock off eval and use awk variable as substitute then I am not getting the expected result
Code:
ps | awk -v v1=$var '{print v1}'   # output is $1,$2
ps | awk -v v1=`echo $var` '{print v1}'  # output is same as above
ps | awk -v v1=$var '{print $v1}'  # output is all the fields of ps command
ps | eval "awk -v v1=$var '{print v1}'"  # output is column of comma

How to get the desire output without using eval?

Last edited by royalibrahim; 11-18-2013 at 03:50 AM..
# 2  
The shell substitutes variables in "quotes".
Code:
var="$1,$2"

But here you maybe want
Code:
var='$1,$2'
ps | awk '{print '"$var"'}'

?

Last edited by MadeInGermany; 11-18-2013 at 04:32 AM..
# 3  
Quote:
Originally Posted by MadeInGermany
Code:
var='$1,$2'
ps | awk '{print '"$var"'}'

?
Brilliant MadeInGermany!! Also, could you please help me to how to use awk variable in this scenario?
# 4  
The $1,$2 is a piece of awk code.
You cannot run a string (as assigned to an awk variable) as awk code.
(Unless there were an eval built into awk.)
This following awk code interprets a given string as you intend:
Code:
var='1,2'
ps | awk -v v1="$var" 'BEGIN{n=split(v1,a,",")} {for(i=1;i<=n;i++) printf "%s\t",$(a[i]); print ""}'


Last edited by MadeInGermany; 11-18-2013 at 06:36 AM..
This User Gave Thanks to MadeInGermany For This Post:
# 5  
This works.
ps | awk "{print $var}"

Single quote does not expand variable.
This User Gave Thanks to Jotne For This Post:

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