bash脚本编程 - 通过迭代＆QUOT;变＆QUOT;关联数组列表变量名 [英] Bash scripting - Iterating through "variable" variable names for a list of associative arrays

问题描述

我已经得到了我想要遍历并检索其键/值对关联数组变量列表。

I've got a variable list of associative arrays that I want to iterate through and retrieve their key/value pairs.

我通过列出它的所有按键和获取的值，即一个关联数组进行迭代。

I iterate through a single associative array by listing all its keys and getting the values, ie.

for key in "${!queue1[@]}" do
    echo "key : $key"
    echo "value : ${queue1[$key]}"
done

棘手的部分是，关联数组的名称都是可变的变量，例如鉴于数= 5，关联数组将被命名为队列1，队列2，队列3，队列4，队列5。

The tricky part is that the names of the associative arrays are variable variables, e.g. given count = 5, the associative arrays would be named queue1, queue2, queue3, queue4, queue5.

我试图取代基于计数上面的顺序，但到目前为止，括号和eval的每个组合还没有产生更多的再不济替代错误。例如如下：

I'm trying to replace the sequence above based on a count, but so far every combination of parentheses and eval has not yielded much more then bad substitution errors. e.g below:

for count in {1,2,3,4,5} do
    for key in "${!queue${count}[@]}" do
        echo "key : $key"
        echo "value : ${queue${count}[$key]}"
    done
done

帮助将非常AP preciated！

Help would be very much appreciated!

推荐答案

的困难在这里从这样的事实，即语法间接扩张（ $ {！nameref} ）与语法冲突从一个关联数组提取键（ $ {！数组[@]} ）。我们只能有一个或另一个，而不是两个。

The difficulty here stems from the fact that the syntax for indirect expansion (${!nameref}) clashes with the syntax for extracting keys from an associative arrays (${!array[@]}). We can have only one or the other, not both.

警惕，因为我关于使用评估，我不能看到周围用其提取的间接引用关联数组中键的方式：

Wary as I am about using eval, I cannot see a way around using it to extract the keys of an indirectly referenced associative array:

keyref="queue${count}[@]"
for key in $(eval echo '${!'$keyref'}'); do ... ; done

您可以但是要避免评估和提取给出的键数组中的值时，使用间接扩展。请注意， [重点] 后缀必须为扩展的一部分：

You can however avoid eval and use indirect expansion when extracting a value from an array given the key. Do note that the [key] suffix must be part of the expansion:

valref="queue${count}[$key]"
echo ${!valref}

要放在这一背景下：

for count in {1..5} ; do
    keyref="queue${count}[@]"
    for key in $(eval echo '${!'$keyref'}'); do
        valref="queue${count}[$key]"
        echo "key = $key"
        echo "value = ${!valref}"
    done
done

这篇关于bash脚本编程 - 通过迭代＆QUOT;变＆QUOT;关联数组列表变量名的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！