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:

我知道这个问题已经得到了很好的回答，但这里有一个“$@”$@“$*”和$*之间的比较

测试脚本内容:

# cat ./test.sh
#!/usr/bin/env bash
echo "================================="

echo "Quoted DOLLAR-AT"
for ARG in "$@"; do
    echo $ARG
done

echo "================================="

echo "NOT Quoted DOLLAR-AT"
for ARG in $@; do
    echo $ARG
done

echo "================================="

echo "Quoted DOLLAR-STAR"
for ARG in "$*"; do
    echo $ARG
done

echo "================================="

echo "NOT Quoted DOLLAR-STAR"
for ARG in $*; do
    echo $ARG
done

echo "================================="

现在，运行带有各种参数的测试脚本:

# ./test.sh  "arg with space one" "arg2" arg3
=================================
Quoted DOLLAR-AT
arg with space one
arg2
arg3
=================================
NOT Quoted DOLLAR-AT
arg
with
space
one
arg2
arg3
=================================
Quoted DOLLAR-STAR
arg with space one arg2 arg3
=================================
NOT Quoted DOLLAR-STAR
arg
with
space
one
arg2
arg3
=================================

bash和其他类似bourne的炮弹:

bar "$@"

Bar "$@"将等价于Bar "$1" "$2" "$3" "$4"

注意，引号很重要!

"$@"， $@， "$*"或$*将在此stackoverflow回答中描述的转义和连接方面各自表现略有不同。

一个密切相关的用例是在一个参数中传递所有给定的参数，就像这样:

Bash -c“bar \”$1\“$2\”$3\“$4\””。

我使用@kvantour的答案的一个变体来实现这一点:

Bash -c "bar $(printf - ' ' %s ' ' ' $@")"

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。

#!/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