在bash shell脚本中传播所有参数-Java 学习之路

630

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 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？

8 回答

26

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 .

从Bash by example起 .

回复于 2024-05-02T13:26:32+08:00
6
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 "${@}"
```
顺便说一句，我必须有这个特殊的脚本，因为我的Unix也不支持grep -r
回复于 2024-05-02T13:26:32+08:00
1
对于bash和其他类似Bourne的shell：
```
java com.myserver.Program "$@"
```
回复于 2024-05-02T13:26:32+08:00
455
工作正常，除非您有空格或转义字符 . 在这种情况下我找不到捕获参数的方法并发送到脚本内的ssh .

这可能很有用但很难看
```
_command_opts=$( echo "$@" | awk -F\- 'BEGIN { OFS=" -" } { for (i=2;i<=NF;i++) { gsub(/^[a-z] /,"&@",$i) ; gsub(/ $/,"",$i );gsub (/$/,"@",$i) }; print $0 }' | tr '@' \' )
```
回复于 2024-05-02T13:26:32+08:00

1045

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:

回复于 2024-05-02T13:26:32+08:00

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
=================================

回复于 2024-05-02T13:26:32+08:00

74
如果在带有其他字符的带引号的字符串中包含 $@ ，则当存在多个参数时，行为非常奇怪，只有第一个参数包含在引号内 .

例：
```
#!/bin/bash
set -x
bash -c "true foo $@"
```
产量：
```
$ bash test.sh bar baz
+ bash -c 'true foo bar' baz
```
但首先分配给另一个变量：
```
#!/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'
```
回复于 2024-05-02T13:26:32+08:00
1
```
#!/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
回复于 2024-05-02T13:26:32+08:00

在bash shell脚本中传播所有参数

8 回答

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