在(bash)脚本之间传递带有空格的参数

我有下面的 bash 两个脚本

嘘:

#!/bin/bash
./b.sh 'My Argument'

B.sh:

#!/bin/bash
someApp $*

SomApp 二进制文件接收 $*作为2个参数(‘ My’和‘ Argument’) ，而不是1。

我测试了几样东西:

只通过 b.sh运行一些应用程序就可以正常工作
迭代 + 回显 b.sh中的参数，正如预期的那样
使用 $@而不是 $*并没有什么不同

bash

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最佳答案

$*, unquoted, expands to two words. You need to quote it so that someApp receives a single argument.

someApp "$*"

It's possible that you want to use $@ instead, so that someApp would receive two arguments if you were to call b.sh as

b.sh 'My first' 'My second'

With someApp "$*", someApp would receive a single argument My first My second. With someApp "$@", someApp would receive two arguments, My first and My second.