击：传递变量作为一个参数/壳报价参数 [英] Bash: pass variable as a single parameter / shell quote parameter

问题描述

我正在写有传递到另一个程序变量的bash脚本：

I'm writing a bash script which has to pass a variable to another program:

./program $variable

的问题是，这是绝对必要的$变量作为一个单一的参数，如果它包含空白这不是的情况下通过。

The problem is, it is absolutely necessary for $variable to be passed as a single parameter, which isn't the case if it contains whitespace.

variable=Hello World
./program $variable
-> program receives two arguments: 'Hello' 'World'

引用它不会做所有事情（做得很不错，bash的开发者）：

Quoting it doesn't do anything at all (well done, bash devs):

variable=Hello World
./program "$variable"
-> program receives: 'Hello' 'World'

双引号它疯狂的东西（做得很好，bash的开发者）：

Double quoting it does crazy stuff (well done, bash devs):

variable=Hello World
./program "'$variable'"
-> program receives: "'Hello" "World'"

有没有一种简单的方法来做到这一点？哎呀，有没有办法做到这一点呢？

Is there an easy way to do this? Heck, is there a way to do this at all?

更新：好吧，因为这个问题似乎并没有被bash的，这里的一些额外的信息。
我传递参数给程序是一个python脚本。不以任何方式修改参数，我得到

Update: Okay, since the problem doesn't seem to be bash, here's some additional info. The program I'm passing arguments to is a python script. Without modifying the arguments in any way, I get

print sys.argv
-> ['/usr/bin/program', "'Hello", "World'"]

我该如何解决呢？

How can I fix that?

编辑：否，我没试过

variable="Hello World"

因为我从来没有申报$变量。它没有被宣告我的bash函数里面，我不允许修改它。

because I never declare $variable. It's not being declared inside my bash function and I'm not allowed to modify it.

编辑：好吧，我得到它的工作方式

Okay, I got it to work that way.

local temp="$variable"
./program "$temp"

我想知道它为什么这样工作，而不是任何其他方式，但。

I'd like to know why it works that way and not any other way, though.

推荐答案

这个问题似乎是方案在

variable="Hello World"    # quotes are needed because of the space
./program "$variable"     # here quotes again

和里面的程序

echo "program received $# arguments:" 
i=1
for arg in "$@"      # again quotes needed
do echo "arg $((i=i+1)): '$arg'"    # and again
done

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