如何在bash中使用变量的值作为另一个变量的名称 [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，它的名字来自另一个变量的值，我写了如下一段代码:

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，但如果您天真地这样做，将会出现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

这中断了:

eval $name_of_variable="word splitting occurs"

修复:

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

终极修复:将您要分配的文本放入变量中.我们称之为safevariable.然后你可以这样做:

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转义问题.美元符号逐字保留到 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，这个任务不受任何影响.safevariable 可以包含 *、空格、$ 等(需要注意的是，我们假设 name_of_variable 包含只是一个有效的变量名，我们可以自由使用:没有什么特别的.)

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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