CODEWITHSUNDEEP
shellawkwhile-loop
Arrow Cen
Arrow Cen

Reputation: 753

mix awk command in shell script while loop and the variables doensn't keep their values after the loop?

Hi guys sorry for the awkward title but I think it is a bit subtle to describe. So here is my problem, I want to keep a count (i) in a while loop which reads input from awk, and then print the value of i after the loop. However i becomes back to zero after the loop. Below is a simplified version of my program, in reality I also did some string matching inside the loop so that some lines are skipped and i does not increment.

I have tried to remove awk and do another ordinary while loop and i's value is kept after the loop, therefore I believe it's not due to some syntax error.

Any idea is great appreciated!

#!/bin/bash
arr=();
i=0;
awk -F '{print $1}' SOMEFILE | while read var
do
  echo $var;
  arr[i]=$var;
  i=$((i+1));
  echo $i;
done
echo $i;

Upvotes: 0

Views: 2081

Answers (1)

William Pursell
William Pursell

Reputation: 212198

Because the while loop is in a pipeline, it is running as a subprocess, and the value of i is local to that subprocess. There are several ways to keep the value; use a named pipe instead of running while in a pipeline, use process substitution, or use an interpolating heredoc. Here's an example of the latter:

while read var; do ... done << EOF
$( awk ... )
EOF

Upvotes: 2

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