Bash 中的动态变量名 [英] Dynamic variable names in Bash

问题描述

我对 bash 脚本感到困惑.

I am confused about a bash script.

我有以下代码:

function grep_search() {
    magic_way_to_define_magic_variable_$1=`ls | tail -1`
    echo $magic_variable_$1
}

我希望能够创建一个变量名，其中包含命令的第一个参数并带有例如的值ls 的最后一行.

I want to be able to create a variable name containing the first argument of the command and bearing the value of e.g. the last line of ls.

为了说明我想要什么:

$ ls | tail -1
stack-overflow.txt

$ grep_search() open_box
stack-overflow.txt

那么，我应该如何定义/声明 $magic_way_to_define_magic_variable_$1 以及我应该如何在脚本中调用它?

So, how should I define/declare $magic_way_to_define_magic_variable_$1 and how should I call it within the script?

我已经尝试过eval、${...}、$${...}，但我还是很困惑.

I have tried eval, ${...}, $${...}, but I am still confused.

推荐答案

使用关联数组，以命令名称作为键.

Use an associative array, with command names as keys.

# Requires bash 4, though
declare -A magic_variable=()

function grep_search() {
    magic_variable[$1]=$( ls | tail -1 )
    echo ${magic_variable[$1]}
}

如果你不能使用关联数组(例如，你必须支持bash 3)，你可以使用declare来创建动态变量名:

If you can't use associative arrays (e.g., you must support bash 3), you can use declare to create dynamic variable names:

declare "magic_variable_$1=$(ls | tail -1)"

并使用间接参数扩展来访问值.

and use indirect parameter expansion to access the value.

var="magic_variable_$1"
echo "${!var}"

请参阅 BashFAQ:间接 - 评估间接/引用变量.

See BashFAQ: Indirection - Evaluating indirect/reference variables.

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