bash 将多个参数传递给shell脚本并解析它们
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时间：2020-09-18 13:06:09 来源:igfitidea点击:
Passing multiple parameters to shell script and parsing them
提问by user3834663
I am trying to run the following program, I need to pass multiple options to get the command to be executed. Here for example: I am giving the inputs
我正在尝试运行以下程序，我需要传递多个选项来获取要执行的命令。例如：我正在提供输入
/test.sh -s -n
script test.sh:
脚本test.sh：
#! /bin/bash
#set -x
while [ $# -gt 0 ]
do
case in
-s) service=
shift
;;
-n) node=
command
break
;;
*) echo "Invalid Argument"
break
;;
esac
done
回答by Kusalananda
Using the getoptsbuilt-in command in bash:
getopts在bash 中使用内置命令：
#!/bin/bash
service="default"
node="default"
while getopts 's:n:' opt; do
case $opt in
s) service="$OPTARG" ;;
n) node="$OPTARG" ;;
*) exit 1 ;;
esac
done
echo "service = '${service}'"
echo "node = '${node}'"
Testing it:
测试它：
$ ./test.sh
service = 'default'
node = 'default'
$ ./test.sh -s hello -n world
service = 'hello'
node = 'world'
$ ./test.sh -n world -s hello
service = 'hello'
node = 'world'
$ ./test.sh -e eh
./test.sh: illegal option -- e
