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 $4
如何在不显式指定每个参数的情况下做到这一点?
当前回答
如果你确实希望传递相同的参数，请使用“$@”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:
其他回答
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 $0 }' | tr '@' \' )
如果你确实希望传递相同的参数，请使用“$@”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:
这里有很多答案推荐带引号或不带引号的$@或$*，但似乎没有人解释这些参数的真正作用以及为什么你应该这样做。所以让我从这个答案中偷取一个很好的总结:
+--------+---------------------------+
| Syntax | Effective result |
+--------+---------------------------+
| $* | $1 $2 $3 ... ${N} |
+--------+---------------------------+
| $@ | $1 $2 $3 ... ${N} |
+--------+---------------------------+
| "$*" | "$1c$2c$3c...c${N}" |
+--------+---------------------------+
| "$@" | "$1" "$2" "$3" ... "${N}" |
+--------+---------------------------+
请注意，引号会造成所有的不同，如果没有引号，两者的行为是相同的。
出于我的目的，我需要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 $*
注意，在上述情况下，没有引号和$@也可以工作。
我的SUN Unix有很多限制，甚至“$@”也没有按预期解释。我的变通方法是${@}。例如,
#!/bin/ksh
find ./ -type f | xargs grep "${@}"
顺便说一下，我必须有这个特定的脚本，因为我的Unix也不支持grep -r
我知道这个问题已经得到了很好的回答，但这里有一个“$@”$@“$*”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
=================================
