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,我需要将参数从当前脚本传播到我正在执行的脚本.

例如,我的脚本名称是foo.sh和调用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

如果您真的希望您的参数传递相同,请使用"$@"而不是plain $@.

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:

Answer 2

对于bash和其他类似Bourne的shell:

java com.myserver.Program "$@"

Answer 3

使用"$@"(适用于所有POSIX兼容机).

[...],bash具有"$ @"变量,该变量扩展为由空格分隔的所有命令行参数.

以Bash为例.

Answer 4

我意识到这已经得到了很好的解答,但这里是"$ @"$ @"$*"和$*之间的比较

测试脚本的内容:

# 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,使用各种参数运行测试脚本:

# ./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
=================================

Answer 5

这里有很多答案推荐$@或$*带引号和不带引号，但是似乎没有人解释这些真正的作用以及为什么你应该这样做。所以让我从这个答案中窃取这个优秀的总结：

+--------+---------------------------+
| Syntax |      Effective result     |
+--------+---------------------------+
|   $*   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|   $@   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|  "$*"  |    "$1c$2c$3c...c${N}"    |
+--------+---------------------------+
|  "$@"  | "$1" "$2" "$3" ... "${N}" |
+--------+---------------------------+

请注意，引号使一切变得不同，如果没有它们，它们都具有相同的行为。

出于我的目的，我需要按原样将参数从一个脚本传递到另一个脚本，为此最好的选择是：

# file: parent.sh
# we have some params passed to parent.sh 
# which we will like to pass on to child.sh as-is

./child.sh $*

注意没有引号，$@应该在上述情况下也能正常工作。

Answer 6

#!/usr/bin/env bash
while [ "$1" != "" ]; do
  echo "Received: ${1}" && shift;
done;

只是认为在尝试测试args如何进入脚本时可能会更有用

Answer 7

我的SUN Unix有很多限制,甚至"$ @"也没有被解释为所希望的.我的解决方法是$ {@}.例如,

#!/bin/ksh
find ./ -type f | xargs grep "${@}"

顺便说一句,我必须有这个特殊的脚本,因为我的Unix也不支持grep -r

Answer 8

如果$@在带引号的字符串中包含其他字符，则当有多个参数时，行为非常奇怪，引号内仅包含第一个参数。

例：

#!/bin/bash
set -x
bash -c "true foo $@"

产量：

$ bash test.sh bar baz
+ bash -c 'true foo bar' baz

但是首先要分配一个不同的变量：

#!/bin/bash
set -x
args="$@"
bash -c "true foo $args"

产量：

$ bash test.sh bar baz
+ args='bar baz'
+ bash -c 'true foo bar baz'

Answer 9

有时你想传递所有参数，但前面有一个标志（例如--flag）

$ bar --flag "$1" --flag "$2" --flag "$3"

您可以通过以下方式执行此操作：

$ bar $(printf -- ' --flag "%s"' "$@")

注意：为了避免额外的字段拆分，您必须引用%sand $@，并且为了避免单个字符串，您不能引用printf.

Answer 10

bar "$@"将相当于bar "$1" "$2" "$3" "$4"

请注意，引号很重要！

"$@"、$@、"$*"或$*每个在转义和连接方面的行为略有不同，如本stackoverflow 答案中所述。

一个密切相关的用例是将所有给定的参数传递到一个参数中，如下所示：

bash -c "bar \"$1\" \"$2\" \"$3\" \"$4\""。

我使用@kvantour的答案的变体来实现这一点：

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