utilizing a variable in awk then enter output into another variable

Question

#!/bin/bash

cat /home/user/list
read -p "enter a number: " LISTNUMBER
USERNAME=$(awk '$LISTNUMBER {
  match($0, $LISTNUMBER); print substr($0, RLENGTH + 2); }' /home/user/list)

echo "you chose $USERNAME."

This script will use awk to search another file that has a list of numbers and usernames: 1 bob 2 fred etc... I only want the username not the corresponding number which is why I tried using: print substr($0, RLENGTH + 2)

Unfortunately, the output of the awk won't attach to $USERNAME.

I have attempted to grep for this but could not achieve the answer. I then read about awk and got here, but am stuck again. Utilizing grep or awk is all the same for me.

use awk -v variable... : stackoverflow.com/questions/15786777/… — Jayan
– Jayan, Commented Dec 13, 2015 at 2:01

mklement0 · Accepted Answer · 2015-12-13 01:59:36Z

4

Single-quoted strings in POSIX-like shells are invariably literals - no interpolation of their contents is performed. Therefore, $LISTNUMBER in your code will not expand to its value.

To pass shell variables to Awk, use -v {awkVarName}="${shellVarName}".

Also, Awk performs automatic splitting of input lines into fields of by runs of whitespace (by default); $1 represents the 1st field, $2 the 2nd, ...

If we apply both insights to your scenario, we get:

#!/bin/bash

read -p "enter a number: " LISTNUMBER
USERNAME=$(awk -v LISTNUMBER="$LISTNUMBER" '
  $1 == LISTNUMBER { print $2 }' /home/user/list)

echo "you chose $USERNAME."

answered Dec 13, 2015 at 1:59

mklement0

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