bash - 在 Bash shell 脚本中传播所有参数

Question

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

例如，我的脚本名称是foo.sh 并调用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？

Answer 1

如果您确实希望以相同的方式传递参数，请使用 "$@" 而不是普通的 $@。

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: