php - 在 PHP 中使用 foreach 数组显示最近的发布日期标题

Question

我正在制作一个游戏发布版 block ，我会在其中展示即将发布的游戏。我只处理游戏信息和发布日期。

我的数组看起来像这样(实际数组有更多信息，所以这只是一个副本):

$arr = [
    [
        'id' => 'UP0006-CUSA08724_00-BATTLEFIELDV0000',
        'attributes' => [
            'name' => 'Battlefield V [test1]',
            'thumbnail-url-base' => 'https://store.playstation.com/store/api/chihiro/00_09_000/container/US/en/999/UP0006-CUSA08724_00-BATTLEFIELDV0000/1539651459000/image'
            'release-date' => '2018-12-14T00:00:00Z'
        ],
    ],
    [
        'id' => 'UP0006-CUSA08724_00-BATTLEFIELDV0000',
        'attributes' => [
            'name' => 'Battlefield V [test2]',
            'thumbnail-url-base' => 'https://store.playstation.com/store/api/chihiro/00_09_000/container/US/en/999/UP0006-CUSA08724_00-BATTLEFIELDV0000/1539651459000/image'
            'release-date' => '2018-10-14T00:00:00Z'
        ],
    ],
    [
        'id' => 'UP0006-CUSA08724_00-BATTLEFIELDV0000',
        'attributes' => [
            'name' => 'Battlefield V [test3]',
            'thumbnail-url-base' => 'https://store.playstation.com/store/api/chihiro/00_09_000/container/US/en/999/UP0006-CUSA08724_00-BATTLEFIELDV0000/1539651459000/image'
            'release-date' => '2019-10-14T00:00:00Z'
        ],
    ],
];

我想显示离当前发布日期最近的游戏名称，例如[test1]，并跳过已经发布的游戏名称，例如[测试2].

我尝试使用这一行跳过它们:

if (strtotime(date('Y-m-d H:i:s')) > strtotime($title['attributes']['release-date'])) continue;

但出于某种原因，它似乎并没有跳过它们，只是将它们保留在里面。

此外，在尝试显示最接近当前发布日期的游戏名称时，我也不知道从哪里开始。

我的完整代码:

foreach($json['included'] as $key => $title) {
    $cusa = substr(explode('-', $title['id'], 3)[1], 0, -3);

    if($title['type'] == 'game' && substr($cusa, 0, 4) == 'CUSA') {
        // if the day of release has already passed, skip
        if (strtotime(date('Y-m-d H:i:s')) > strtotime($title['attributes']['release-date'])) continue;
            ?>
            <div class="game-banner" style="background:url(<?php echo $title['attributes']['thumbnail-url-base']; ?>)">
                <h4 class="psplus-game-name"><?php echo $title['attributes']['name']; ?></h4>
            </div>
            <?php
            if($key >= 4) break; // display only 3
        }
    }
}

Answer 1

您只需要计算发布日期的剩余秒数，如果它是正数则回显它。

foreach($arr as $game){
    $timeleft = strtotime($game['attributes']['release-date'])-time();
    if($timeleft>0) echo floor($timeleft/86400) ." days left to ".$game['attributes']['name'] ." \n";
}

//58 days left to Battlefield V [test1] 
//362 days left to Battlefield V [test3] 

https://3v4l.org/OMetR

如果您的初始数组是未排序的，而您想要排序，您可以将它们添加到一个数组中，键为 timeleft，然后使用 ksort() 对键进行排序。

foreach($arr as $game){
    $timeleft = strtotime($game['attributes']['release-date'])-time();
    if($timeleft>0) $games[$timeleft] = floor($timeleft/86400) ." days left to ".$game['attributes']['name'] ." \n";
}

ksort($games);
echo implode("", $games);

https://3v4l.org/gbLCs