PHP-如何从多维数组中回显值

问题描述

我有：

$request = Array
(
    [ID] => 2264
    [SUCCESS] => Array
        (
            [0] => Array
                (
                    [MESSAGE] => Service Details
                    [LIST] => Array
                        (
                            [retail_name] =>
                            [credit_admin] => FREE
                            [credit] => 0
                            [discount_admin] => 0
                            [discount] => 0
                            [cartdiscount] => 0

如果我这样做：

echo $request[ID];                            // it says: 2264
echo $request[SUCCESS];                       // it says: Array
echo $request[SUCCESS][0][MESSAGE];           // it says: Service Details

但是我需要回显 credit，如果我这样做：

But I need to echo "credit" and if I do:

echo $request[SUCCESS][0][LIST];           // I get ERROR
echo $request[SUCCESS][0][LIST][credit];   // I get ERROR

我不明白为什么？我该怎么做？
谢谢

I dont understand why? How can I do it?Thank you

推荐答案

以前，如果在该名称下未定义任何常量，则每个不带引号的字符串都被视为字符串，但是从 PHP 7.2 起，现在发出的错误级别为 E_WARNING 。

Previously every string without quotes was treated like string if there was no Constant defined under this name, but from PHP 7.2 now issues error of level E_WARNING. Why is $foo[bar] wrong?

PHP Warning:  Use of undefined constant test - assumed 'test' (this will throw an Error in a future version of PHP)

在此示例中，列表被视为构造。

In this example list is treated as construct list(). Using bare strings in associative arrays as keys is deprecated and should be avoided.

正确的方法是使用引号（单引号或双引号）来调用它：

Correct way is to call it with quotes (single or double):

echo $request['SUCCESS'][0]['LIST'];
echo $request['SUCCESS'][0]['LIST']['credit'];

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