shell - 在 shell 脚本之间传递参数但保留引号

Question

如何将一个 shell 脚本的所有参数传递到另一个 shell 脚本中？我已经尝试过 $*，但正如我所期望的，如果您引用了参数，则该方法不起作用。

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

我希望它打印出来:

apple
pear orange
banana

Answer 1

使用 "$@" 而不是 $* 来保留引号:

./script2.sh "$@"

更多信息:

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

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

Note: "$*" must be quoted.

$@
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.

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