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 $4
如何在不显式指定每个参数的情况下做到这一点?
当前回答
我知道这个问题已经得到了很好的回答，但这里有一个“$@”$@“$*”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 $ARG
done
echo "================================="
现在，运行带有各种参数的测试脚本:
# ./test.sh "arg with space one" "arg2" arg3
=================================
Quoted DOLLAR-AT
arg with space one
arg2
arg3
=================================
NOT Quoted DOLLAR-AT
arg
with
space
one
arg2
arg3
=================================
Quoted DOLLAR-STAR
arg with space one arg2 arg3
=================================
NOT Quoted DOLLAR-STAR
arg
with
space
one
arg2
arg3
=================================
其他回答
如果你确实希望传递相同的参数，请使用“$@”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:
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。
工作正常，除非您有空格或转义字符。我没有找到在这种情况下捕获参数并发送到脚本中的ssh的方法。
这可能是有用的，但如此丑陋
_command_opts=$( echo "$@" | awk -F\- 'BEGIN { OFS=" -" } { for (i=2;i<=NF;i++) { gsub(/^[a-z] /,"&@",$i) ; gsub(/ $/,"",$i );gsub (/$/,"@",$i) }; print $0 }' | tr '@' \' )
bash和其他类似bourne的炮弹:
bar "$@"
"${array[@]}"是在bash中传递任何数组的正确方式。我想提供一个完整的备忘单:如何准备参数，绕过和处理它们。
Pre.sh -> foo.sh -> bar.sh。
#!/bin/bash
args=("--a=b c" "--e=f g")
args+=("--q=w e" "--a=s \"'d'\"")
./foo.sh "${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'"
