How to correct this awk code without eval?

Shell Programming and Scripting
1

Hi everyone,

The following piece of awk code works fine if I use eval builtin

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

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?

2

The shell substitutes variables in "quotes".

var="$1,$2"

But here you maybe want

var='$1,$2'
ps | awk '{print '"$var"'}'

?

3

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:

var='1,2'
ps | awk -v v1="$var" 'BEGIN{n=split(v1,a,",")} {for(i=1;i<=n;i++) printf "%s\t",$(a); print ""}'
1 Like
5

This works.
ps | awk "{print $var}"

Single quote does not expand variable.

1 Like