Question

我正在解码JSON响应并将其输出到表中。 JSON响应分为三个元素（Events，Organizers和Venues）和Events个节点引用Venues和Organizers元素的节点。< / p>

以下是print_r为JSON响应显示的示例：

Array
(
    [Events] => Array
        (
            [0] => Array
                (
                    [EventTitle] => Concert One                 
                    [Details] => Array
                        (
                            [VenueID] => 100
                            [EventDate] => 2016-01-01
                        )
                )

            [1] => Array
                (
                    [EventTitle] => Concert Two
                    [Details] => Array
                        (
                            [VenueID] => 150
                            [EventDate] => 2016-01-02
                        )
                )

          )

    [Venues] => Array
        (
            [0] => Array
                (
                    [HallID] => 100
                    [VenueName] => Venue A
                )

            [1] => Array
                (
                    [HallID] => 150
                    [VenueName] => Venue B
                )

        )

)

实际JSON示例如下所示：

{
    "Events": [
        {
            "EventTitle": "Concert One",
            "Details": {
                "VenueID": 100,
                "EventDate": "2016-01-01"
            }
        },
        {
            "EventTitle": "Concert Two",
            "Details": {
                "VenueID": 150,
                "EventDate": "2016-01-02"
            }
        }
    ],
    "Venues": [
        {
            "HallID": 100,
            "VenueName": "Venue A"
        },
        {
            "HallID": 150,
            "VenueName": "Venue B"
        }
    ]
}

这是我用来创建表的foreach循环：

<?php

foreach($results['Events'] as $values)

{
        echo '<tr><td>' . $values['EventTitle'] . '</td>';
        echo '<td>' . $values['Details']['VenueID'] . '</td>';
        echo '<td>' . $values['Details']['EventDate'] . '</td></tr>';
}

?>

效果很好并创建了一个简单的表格：

Event title | Event venue | Event date
Concert One | 100 | 2016-01-01
Concert Two | 150 | 2016-01-02

我正在努力解决的问题是如何用Venue ID（地点A，地点B）替换VenueName（100,150），结果是：

Event title | Event venue | Event date
Concert One | Venue  A | 2016-01-01
Concert Two | Venue  B | 2016-01-02

有可能实现这个目标吗？

Answer 1

我要做的是将您的Organizers和Venue解析为自己的数组，并将ID作为密钥。所以你会得到一个这样的数组（其中$arr代表保存完整数组的变量）

$venue = [];
foreach($arr['Venue'] as $vals) {
    $venue[$vals['HallID']] = $vals['VenueName'];
}

您可以使用foreach循环构建它。然后，您将使用密钥迭代Events，以获取正确的关系数据（将$organizers构建为$venue同样的事情）

foreach($results['Events'] as $values) {
     echo '<tr><td>' . $values['EventTitle'] . '</td>';
     echo '<td>' . $venue[$values['Details']['VenueID']] . '</td>';
     echo '<td>' . $organizer[$values['Details']['OrganizerID']] . '</td>';
     echo '<td>' . $values['Details']['EventDate'] . '</td></tr>';
}

Answer 2

试试这个：

foreach($results['Events'] as $values)

{
        echo '<tr><td>' . $values['EventTitle'] . '</td>';
        echo '<td>' . $results['Venues'][array_search($values['Details']['VenueID'], array_column($results['Venues'], 'HallID'))]['VenueName'] . '</td>';
        echo '<td>' . $values['Details']['EventDate'] . '</td></tr>';
}

?>

一切都在一行！

解码JSON响应的不同元素之间的交叉引用

2 个答案: