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bash variable

ShellScript

Posted at 2023-12-26

#!/bin/bash

# Parse command line arguments
while getopts "f:" opt; do
  case $opt in
    f)
      file_name="$OPTARG"
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      exit 1
      ;;
    :)
      echo "Option -$OPTARG requires an argument." >&2
      exit 1
      ;;
  esac
done

# Check if the file_name is provided, exit if not
if [ -z "$file_name" ]; then
  echo "Error: File name not provided. Use -f option to specify the file name."
  exit 1
fi

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