引用附加变量? [英] Reference an appended variable?

问题描述

如何在bash脚本中执行此操作?

How can I do this in a bash script?

#!/bin/sh

func() {
export ${NAME}_SUFFIX=`result_of_some_command`
}

NAME=my_name
func

# This variable will become my_name_SUFFIX
# but how can I reference it using $NAME?

echo ${${NAME}_SUFFIX}     # Doesn't work...

如果可以通过函数调用使用，我会使用 declare ，但似乎没有.另外，我的声明版本不支持-x.

I'd use declare if it would work through function calls, but it seems it doesn't. Also, my version of declare doesn't support -x.

推荐答案

我知道您要做什么... 警告 ...评估"是一种糟糕的方法而变量间接寻址则更好.但是，这就是您要的.

I know what you're trying to do...warning..."eval" is a terrible way to do it and variable indirection is better. But, here is what you asked for.

对脚本进行一个简单的更改:

One simple change to your script:

将导出"更改为评估"即可使用.您一直都在那里！哦，如下所示更改您的"echo"语句.

Change "export" to "eval" and it'll work. You were most of the way there! Oh, and change your "echo" statement as shown below.

示例:

#!/bin/sh

func() {
eval ${NAME}_SUFFIX=`result_of_some_command`
}

NAME=my_name
func

# This variable will become my_name_SUFFIX
# but how can I reference it using $NAME?

eval echo "\${${NAME}_SUFFIX}"

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