在（bash）脚本之间传递带有空格的参数
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Passing arguments with spaces between (bash) script
提问by John Fear
I've got the following bash two scripts
我有以下 bash 两个脚本
a.sh:
灰：
#!/bin/bash
./b.sh 'My Argument'
b.sh:
b.sh：
#!/bin/bash
someApp $*
The someApp binary receives $*as 2 arguments ('My' and 'Argument') instead of 1.
someApp 二进制文件接收$*2 个参数（“My”和“Argument”）而不是 1。
I've tested several things:
我已经测试了几件事：
- Running someApp only thru
b.shworks as expected - Iterate+echo the arguments in
b.shworks as expected - Using
$@instead of$*doesn't make a difference
- 仅通过运行 someApp 可按
b.sh预期工作 b.sh按预期迭代+回显工作中的参数- 使用
$@而不是$*没有区别
回答by chepner
$*, unquoted, expands to two words. You need to quote it so that someAppreceives a single argument.
$*，不加引号，扩展为两个词。您需要引用它以便someApp接收单个参数。
someApp "$*"
It's possible that you want to use $@instead, so that someAppwould receive two arguments if you were to call b.shas
这有可能是您要使用$@代替，这样someApp如果你要打电话将获得两个参数b.sh为
b.sh 'My first' 'My second'
With someApp "$*", someAppwould receive a single argument My first My second. With someApp "$@", someAppwould receive two arguments, My firstand My second.
使用someApp "$*",someApp将接收一个参数My first My second。使用someApp "$@",someApp将接收两个参数，My first和My second。
