Propagate all arguments in a bash shell script
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正在执行的脚本。
例如，我的脚本名是
SH：
1 | bar $1 $2 $3 $4 |
如何在不显式指定每个参数的情况下完成此操作？
如果您希望传递相同的参数，请使用
观察：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 | $ 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: |
对于巴什和其他类似波恩的贝壳：
1 | java com.myserver.Program"$@" |
使用
[...] , bash features the"$@" variable, which expands to all command-line parameters separated by spaces.
一个接一个的例子。
1 2 3 4 | #!/usr/bin/env bash while ["$1" !="" ]; do echo"Received: ${1}" && shift; done; |
只是觉得在测试args如何进入脚本时，这可能会更有用一些。
我知道这个问题得到了很好的回答，但是这里比较了"$@"、"$@"、"$*"和$*
测试脚本内容：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | # cat ./test.sh #!/usr/bin/env bash echo"=================================" echo"Quoted DOLLAR-AT" for ARG in"$@"; 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 $ARG done echo"=================================" |
现在，使用各种参数运行测试脚本：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | # ./test.sh "arg with space one""arg2" arg3 ================================= Quoted DOLLAR-AT arg with space one arg2 arg3 ================================= NOT Quoted DOLLAR-AT arg with space one arg2 arg3 ================================= Quoted DOLLAR-STAR arg with space one arg2 arg3 ================================= NOT Quoted DOLLAR-STAR arg with space one arg2 arg3 ================================= |
我的Sun Unix有很多限制，即使"$@"也没有按需要解释。我的解决方法是@。例如，
1 2 | #!/bin/ksh find ./ -type f | xargs grep"${@}" |
顺便说一下，我必须有这个特定的脚本，因为我的Unix也不支持grep-r
如果在带其他字符的带引号的字符串中包含
例子：
1 2 3 | #!/bin/bash set -x bash -c"true foo $@" |
产量：
1 2 | $ bash test.sh bar baz + bash -c 'true foo bar' baz |
但首先分配给不同的变量：
1 2 3 4 | #!/bin/bash set -x args="$@" bash -c"true foo $args" |
产量：
1 2 3 | $ bash test.sh bar baz + args='bar baz' + bash -c 'true foo bar baz' |
工作正常，除非您有空格或转义字符。在这种情况下，我找不到捕获参数并将其发送到脚本内部的ssh的方法。
这可能有用，但太难看了
1 | _command_opts=$( echo"$@" | awk -F\- 'BEGIN { OFS=" -" } { for (i=2;i<=NF;i++) { gsub(/^[a-z] /,"&@",$i) ; gsub(/ $/,"",$i );gsub (/$/,"@",$i) }; print $0 }' | tr '@' \' ) |