（bash）的脚本之间传递参数用空格

问题描述

限时删除！！

我有以下的bash两个脚本

I've got the following bash two scripts

a.sh：

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

b.sh：

#!/bin/bash
someApp $*

该someApp二进制接收 $ * 作为2个参数，而不是1（'我'和'参数'）

The someApp binary receives $* as 2 arguments ('My' and 'Argument') instead of 1.

我测试过的几件事情：

• 运行someApp只有直通 b.sh 按预期工作


• 迭代+回响在 b.sh 参数按预期工作


• 使用 $ @ 而不是 $ * 不有所作为

• Running someApp only thru b.sh works as expected
• Iterate+echo the arguments in b.sh works as expected
• Using $@ instead of $* doesn't make a difference

推荐答案

$ * ，不带引号的，扩展到两个词。你需要让 someApp 接收一个参数引用它。

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

someApp "$*"

这有可能是您要使用 $ @ 代替，让 someApp 将接受两个参数，如果你是调用 b.sh 为

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'

使用 someApp$ *， someApp 将接受一个参数我的第一个我的第二个。随着 someApp$ @， someApp 将接受两个参数，我的第一个和我的第二个。

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

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