在 shell 脚本之间传递参数但保留引号 [英] pass arguments between shell scripts but retain quotes

问题描述

如何将一个 shell 脚本的所有参数传递给另一个?我已经尝试过 $*，但正如我所料，如果您引用了参数，这将不起作用.

How do I pass all the arguments of one shell script into another? I have tried $*, but as I expected, that does not work if you have quoted arguments.

示例:

$ cat script1.sh

#! /bin/sh
./script2.sh $*

$ cat script2.sh

#! /bin/sh
echo $1
echo $2
echo $3

$ script1.sh apple "pear orange" banana
apple
pear
orange

我想打印出来:

apple
pear orange
banana

推荐答案

使用 "$@" 而不是 $* 来保留引号:

Use "$@" instead of $* to preserve the quotes:

./script2.sh "$@"

更多信息:

http://tldp.org/LDP/abs/html/internalvariables.html

$*
所有的位置参数，看成一个词

$*
All of the positional parameters, seen as a single word

注意:$*"必须加引号.

Note: "$*" must be quoted.

$@
与 $* 相同，但每个参数都是一个带引号的字符串，即参数原封不动地传递，没有解释或扩展.这意味着，除其他外，参数中的每个参数list 被视为一个单独的词.

$@
Same as $*, but each parameter is a quoted string, that is, the parameters are passed on intact, without interpretation or expansion. This means, among other things, that each parameter in the argument list is seen as a separate word.

注意:当然，$@"应该被引用.

Note: Of course, "$@" should be quoted.

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