Bash 将变量作为带引号的参数传递
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Bash pass variable as argument with quotes
提问by user2249516
Assume ./programis a program that just prints out the parameters;
假设./program是一个只打印参数的程序；
$ ./program "Hello there"
Hello there
How can I properly pass arguments with quotes in from a variable? I am trying to do this;
如何从变量中正确传递带引号的参数？我正在尝试这样做；
$ args='"Hello there"'
$ echo ${args}
"Hello there"
$ ./program ${args}
Hello there # This is 1 argument
but instead, when I go through a variable the quotes in argsseem to be ignored so I get;
但是相反，当我通过一个变量时，引号中的引号args似乎被忽略了，所以我得到了；
$ args='"Hello there"'
$ echo ${args}
"Hello there"
$ ./program ${args}
"Hello there" # This is 2 arguments
Is it possible to have bash treat the quotes as if I entered them myself in the first code block?
是否可以让 bash 将引号视为我自己在第一个代码块中输入？
采纳答案by redneb
I don't know where you got programfrom, but it appears that it's broken. Here's a correct way to write it in bash:
我不知道你program从哪里得到的，但它似乎坏了。这是在 bash 中编写它的正确方法：
#!/bin/bash
for arg in "$@"; do
echo "$arg"
done
This will print each argument in a separate line to make them easier to distinguish (it will of course have a problem with arguments that contain a line break, but we will not pass such an argument).
这将在单独的行中打印每个参数，使它们更容易区分（当然，包含换行符的参数会有问题，但我们不会传递这样的参数）。
After you have saved the above as programand gave it the execute permission, try this:
将上述内容另存为program并授予执行权限后，请尝试以下操作：
$ args='"Hello there"'
$ ./program "${args}"
"Hello there"
whereas
然而
$ args='"Hello there"'
$ ./program ${args}
"Hello
there"
