按名称查找shell变量，间接地 [英] Lookup shell variables by name, indirectly

问题描述

让我们说我有一个变量的名字保存在另外一个变量：

  MYVAR = 123
VARNAME = MYVAR
 

现在，我想只要使用$ varname的变量获得123。
是否有一个直接的方法？我发现内置由名称查找没有这样的庆典，所以想出了这个：

 函数VAR {V =\\ $$ 1; EVAL回声$ V; }
 

所以

 变量$＃VARNAME 123提供
 

看起来并不太坏到底，但我想知道如果我错过了一些比较明显的。在此先感谢！

解决方案

从的庆典手册页：

  $ {！VARNAME}
 

  

如果参数的第一个字符是一个感叹号，水平
         可变间接介绍。 bash使用从参数作为变量的名称的其余部分形成的变量的值;
         然后这个变量被扩展并且该值是在的其余部分中使用
         取代，而不是参数本身的值。这是
         被称为间接扩张。

Let's say I have a variable's name stored in another variable:

myvar=123
varname=myvar

now, I'd like to get 123 by just using $varname variable. Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:

function var { v="\$$1"; eval "echo "$v; }

so

var $varname  # gives 123

Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious. Thanks in advance!

解决方案

From the man page of bash:

${!varname}

If the first character of parameter is an exclamation point, a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion.

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