传播的所有参数在bash shell脚本 [英] Propagate all arguments in a bash shell script

问题描述

我写一个非常简单的脚本调用另一个脚本，我需要的参数从我当前脚本传播到我执行脚本。

I am writing a very simple script that calls another script, and I need to propagate the parameters from my current script to the script I am executing.

例如，我的脚本名称为 foo.sh 键，通话 bar.sh

For instance, my script name is foo.sh and calls bar.sh

foo.sh：

bar $1 $2 $3 $4

我怎样才能做到这一点没有明确指定每个参数？

How can I do this without explicitly specifying each parameter?

推荐答案

使用$ @而不是简单的 $ @ 如果你真的希望你的参数传递相同的。

Use "$@" instead of plain $@ if you actually wish your parameters to be passed the same.

注意：

$ cat foo.sh
#!/bin/bash
baz.sh $@

$ cat bar.sh
#!/bin/bash
baz.sh "$@"

$ cat baz.sh
#!/bin/bash
echo Received: $1
echo Received: $2
echo Received: $3
echo Received: $4

$ ./foo.sh first second
Received: first
Received: second
Received:
Received:

$ ./foo.sh "one quoted arg"
Received: one
Received: quoted
Received: arg
Received:

$ ./bar.sh first second
Received: first
Received: second
Received:
Received:

$ ./bar.sh "one quoted arg"
Received: one quoted arg
Received:
Received:
Received:

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