如何将一个shell脚本的所有参数传递给另一个？我试过$ *，但正如我所料，如果你引用的参数，这不工作。
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
使用“$ @”而不是$ *来保留引号：
- ./script2.sh "$@"
更多信息：
http://tldp.org/LDP/abs/html/internalvariables.html
$*
All of the positional parameters,seen as a single wordNote: “$*” 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.
