how to run shell script to display command line arguments?-CodePudding

#! /usr/bin/bash
for var in "$@"
do
    echo $var
done

simple shell script which displays each of the command line arguments, one at a time and stops displaying command line arguments when it gets an argument whose value is “stop”.?

CodePudding user response：

Process args with a while loop:

#!/usr/bin/env bash

while test -n "$1"; do
    test "$1" != 'stop' && echo "$1" && shift || break
done

CodePudding user response：

IIUC, try this:

#!/bin/bash
for var in "$@"
do 
    if [[ "$var" == 'stop' ]]; then
        exit 0
    else
        echo "$var"
    fi
done

CodePudding user response：

#!/usr/bin/env bash

fn_EXE(){

  VARIABLE=$(echo $@)

  for var in $VARIABLE;do
      if [[ "$var" == "STOP" ]]; then
        echo "encontre el stop=$var Adios !!"
        break
      else
        echo $var
      fi
  done
}

fn_EXE "A" "B" "C" "D" "E" "STOP" "F"

Saldría del bucle al encontrar un stop en la lista de argumentos..