如何在bash中将变量的值用作另一个变量的名称 [英] How to use a variable's value as another variable's name in bash

问题描述

我想声明一个变量，其名称来自另一个变量的值，我编写了以下代码:

I want to declare a variable, the name of which comes from the value of another variable, and I wrote the following piece of code:

a="bbb"
$a="ccc"

但是没有用.完成这项工作的正确方法是什么?

but it didn't work. What's the right way to get this job done?

推荐答案

eval用于此目的，但是如果您天真地这样做，将会有令人讨厌的转义问题.这种事情通常是安全的:

eval is used for this, but if you do it naively, there are going to be nasty escaping issues. This sort of thing is generally safe:

name_of_variable=abc

eval $name_of_variable="simpleword"   # abc set to simpleword

此中断:

eval $name_of_variable="word splitting occurs"

解决方法:

eval $name_of_variable="\"word splitting occurs\""  # not anymore

最终的解决方法:将要分配的文本放入变量中.我们称之为safevariable.然后，您可以执行以下操作:

The ultimate fix: put the text you want to assign into a variable. Let's call it safevariable. Then you can do this:

eval $name_of_variable=\$safevariable  # note escaped dollar sign

转义美元符号可以解决所有逃生问题.美元符号可以逐字保存到eval函数中，该函数将有效地执行此操作:

Escaping the dollar sign solves all escape issues. The dollar sign survives verbatim into the eval function, which will effectively perform this:

eval 'abc=$safevariable' # dollar sign now comes to life inside eval!

当然，这项任务不受任何影响. safevariable可以包含*，空格，$等.(需要说明的是，我们假设name_of_variable除了有效的变量名外，不包含任何其他内容，并且我们可以自由使用它:没什么特别的.)

And of course this assignment is immune to everything. safevariable can contain *, spaces, $, etc. (The caveat being that we're assuming name_of_variable contains nothing but a valid variable name, and one we are free to use: not something special.)

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