解码JSON响应的不同元素之间的交叉引用

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)， Events节点参考节点从Venues和Organizers元素中分离。

下面是print_r显示JSON响应的示例:

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
                )
        )
)

实际的JSON示例如下:

{
    "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"
        }
    ]
}

下面是我用来创建表的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>';
}
?>

它可以很好地创建一个简单的表:

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

我正在努力的是如何用VenueName(场地A，场地B)取代Venue ID(100,150)，这样的结果是:

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

这有可能实现吗?

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>';
}
?>

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

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>';
}