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pass value from awk back to bash shell

Hi all,

Here is my bash script:

VERSION=""
awk 'BEGIN {
file='"'$SOURCE_PATH'"'
while ( (getline aline < file) > 0 )
{
if ($aline ~ /Id=[0-9]*.[0-9]*/)
{
print aline
VERSION=aline
break
}
}
}'
echo "$VERSION"

What I tried to do is to use awk get a line matching a pattern and pass the line back to bash shell script. I got "print aline" outputing the matched line. However, I was unable to get VERSION out of awk. So, echo "$VERSION" has empty. Could anyone tell me how to get a value back to bash shell from awk?

Thank you

You can not set a bash variable directly from awk. But you can have bash execute awk and assign its output to a variable with $(command).

Like this:

Code:

#!/bin/bash



SOURCE_PATH=file.txt



VERSION=$(awk 'BEGIN {

    file="'$SOURCE_PATH'"

    while ( (getline aline < file) > 0 ) {

      if (aline ~ /Id=[0-9]*.[0-9]*/) {

        print aline

        break

      }

    }

  }')



echo "Version line: $VERSION"

But in your case (just print 1 matching line) I'd use grep instead of awk:

Code:

#!/bin/bash



SOURCE_PATH=file.txt

VERSION=$(grep 'Id=[0-9]*.[0-9]*' "$SOURCE_PATH")

echo "Version line: $VERSION"

Or at least simplify the awk script:

Code:

#!/bin/bash



SOURCE_PATH=file.txt

VERSION=$(awk '/Id=[0-9]*.[0-9]*/ {print $0}'  "$SOURCE_PATH")

echo "Version line : $VERSION"

Hi HKO,

Thank you. The short form of awk is working perfectly with a minor change on wildcard (* -> +) to make an exact match in my case.

Thank you once more
Chris