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I have a pipe-delimited file and I need to grep the first column and if a pattern matched, I will print the whole line. The command below is working but when I put it on a script, I think the $1 is conflicting the command:

Command:

awk -F'|' < filename '{if ($1 == "stringtomatch") print $0}'

Script:

./scripts.sh stringtomatch

Command in script:

awk -F'|' < filename '{if ($1 == "$1") print $0}'

The $1 enclosed in the double quotes is the argument passed to the script. Any advise how to make this work?

hanabi44
  • 1

1 Answers1

5

Note that you can greatly simplify your awk. The default action if an expression evaluates to true is to print the current line. So this does the same thing:

awk -F'|' < filename '$1 == "string"'

Anyway, you can use the -v option to pass a variable. So your script can be:

#/bin/sh

if [ $# -lt 1 ]; then
  echo "At least one argument is required"
  exit
fi


Allow the script to get the filename from the 2nd argument,


default to 'filename' if no second argument is given


file=${2:-filename}


awk -F'|' -v str="$1" '$1 == str' "$file"
terdon
  • 242,166
  • awk will also take the filename - to mean stdin. So making that declaration into file="${2:--}" will use stdin if no filename is supplied. There is a variant that will pass either a list of files or - from the command line, but you would need to shift $1 off first. – Paul_Pedant Sep 25 '20 at 09:20
  • @Paul_Pedant ooh, nice, thanks. I won't add it to the answer though since the OP was already hardcoding the file name, so I am assuming they prefer to have a file. Great trick for next time though! – terdon Sep 25 '20 at 09:32