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。
例如，我的脚本名为foo.sh，调用bar.sh。
foo.sh:
bar $1 $2 $3 $4
如何在不显式指定每个参数的情况下做到这一点?
当前回答
使用“$@”(适用于所有POSIX兼容程序)。
[…)， bash提供了“$@”变量，它扩展为所有用空格分隔的命令行参数。
以Bash为例。
其他回答
bash和其他类似bourne的炮弹:
bar "$@"
使用“$@”(适用于所有POSIX兼容程序)。
[…)， bash提供了“$@”变量，它扩展为所有用空格分隔的命令行参数。
以Bash为例。
"${array[@]}"是在bash中传递任何数组的正确方式。我想提供一个完整的备忘单:如何准备参数，绕过和处理它们。
Pre.sh -> foo.sh -> bar.sh。
#!/bin/bash
args=("--a=b c" "--e=f g")
args+=("--q=w e" "--a=s \"'d'\"")
./foo.sh "${args[@]}"
#!/bin/bash
./bar.sh "$@"
#!/bin/bash
echo $1
echo $2
echo $3
echo $4
结果:
--a=b c
--e=f g
--q=w e
--a=s "'d'"
如果你确实希望传递相同的参数，请使用“$@”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:
#!/usr/bin/env bash
while [ "$1" != "" ]; do
echo "Received: ${1}" && shift;
done;
只是认为在尝试测试args如何进入脚本时，这可能更有用
