在 bash shell 脚本中传播所有参数 [英] Propagate all arguments in a bash shell script

问题描述

我正在编写一个调用另一个脚本的非常简单的脚本，我需要将参数从当前脚本传播到我正在执行的脚本中.

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，我的脚本名称是 foo.sh 并调用 bar.sh

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

foo.sh:

bar $1 $2 $3 $4

如何在不明确指定每个参数的情况下执行此操作?

How can I do this without explicitly specifying each parameter?

推荐答案

如果您确实希望参数一样通过.

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

观察:

$ cat no_quotes.sh
#!/bin/bash
echo_args.sh $@

$ cat quotes.sh
#!/bin/bash
echo_args.sh "$@"

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

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

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

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

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

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