Question

我知道这很简单，但我无法解决这个问题。请仔细研究。

我有一个名为notification_updates的表，它的数组是这样的：

Array
(
    [0] => common\models\NotificationUpdates Object
    (
        [_attributes:yii\db\BaseActiveRecord:private] => Array
        (
            [id] => 1
            [title] => This is the notification to inform you about this.
            [status] => 1
            [created_at] => 2017-11-20 08:29:21
        ) 
    )
    [1] => common\models\NotificationUpdates Object
    (
        [_attributes:yii\db\BaseActiveRecord:private] => Array
        (
            [id] => 2
            [title] => This is the notification to inform you about this cricket match
            [status] => 1
            [created_at] => 2017-11-20 06:24:09
        ) 
    )
    [2] => common\models\NotificationUpdates Object
    (
        [_attributes:yii\db\BaseActiveRecord:private] => Array
        (
            [id] => 3
            [title] => Inform you about this cricket match
            [status] => 1
            [created_at] => 2017-11-21 11:40:31
        )
    )
)

现在我还有一个表格，其中第一个表格的primary_key（id）在表notification_id中被称为deleted_nofitication。

这个表也有这样的数组：

Array
(
    [0] => common\models\DeletedNofitication Object
    (
        [_attributes:yii\db\BaseActiveRecord:private] => Array
        (
            [notification_id] => 1
        )
    )
    [1] => common\models\DeletedNofitication Object
    (
        [_attributes:yii\db\BaseActiveRecord:private] => Array
        (
            [notification_id] => 2
        ) 
    )
)

现在我必须检查天气notification_updates表是否具有针对该user_id的此值。如果它在那里，则不应该在JSON下显示该通知。

我在PHP（YII2）中做过这样的事情 - 请不要在此进行比较，请检查

$notifications = NotificationUpdates::find()->where([])->all();
$outt = [];  

foreach($notifications as $notification) {
    $deleted_notification = DeletedNofitication::find()
                                ->select('notification_id')
                                ->where(['user_id'=>$user_id])
                                ->all();  

    $outt[] = [
        'id' => $notification->id,
        'notification_title' => $notification->title
    ]; 
}

$out = [
    'notification'=> $outt,
    'success'     => true,
    'message'     => 'All Notification Updates'
];

Answer 1

修改

好吧，既然我得到了你想要做的事情，我可能会提供帮助。我对YII2感到不舒服（即我从未使用过它），但你的代码可能看起来像这样。底线是我希望SQL只返回相关记录，这样我们就不必用我们的php做这个逻辑了：

<?php // We're trying to select all notifications except the ones that have already been shown. // Following query uses subquery, join would be better performancewise. // $notificationsSQL = "SELECT id,title FROM NotificationUpdates WHERE id NOT in (SELECT id FROM DeletedNotifications WHERE user_id = $user_id)"; $notificationsAlreadyShown = DeletedNofitication::find()->where(['user_id' => $user_id]); $notifications = NotificationUpdates::find()->where(['not in', 'id', $notificationsAlreadyShown]); if ($notifications->count() === 0){ //Or whatever method you use to count the results. return; // Or whatever you need to do to handle this case. } $outt = []; foreach ($notifications->all() as $notification) { $outt[] = [ 'id' => $notification->id, 'notification_title' => $notification->title ]; } $out = [ 'notification' => $outt, 'success' => true, 'message' => 'All Notification Updates' ];

P.s我在删除的通知上看不到列user_id，可能需要检查。

OLD ANSWER：

对不起，您的疑问对我来说并不是很清楚。即我没有得到你想要做的事。是否仅显示尚未删除的通知？

首先引起我注意的是，您在第二次查询的where-clause中列user_id，但我不会在表结构。我对YII2不太熟悉，但你是否想尝试做一些事情：

<?php $notifications = NotificationUpdates::find()->where([])->all(); $outt = []; foreach ($notifications as $notification) { $deleted_notification = DeletedNofitication::find() ->select('notification_id') ->where(['notification_id' => $notification->id]) ->all(); // If the notification has been deleted, skip to next iteration. if (count($deleted_notification) > 0){ continue; } $outt[] = [ 'id' => $notification->id, 'notification_title' => $notification->title ]; } $out = [ 'notification' => $outt, 'success' => true, 'message' => 'All Notification Updates' ];

虽然这是您尝试做的事情，但您应该返回查询构建器并仅选择未删除的通知。或者甚至更好，使用YII2s softdeletes（如果有的话）。

Answer 2

此查询执行了我需要的所有内容..

 $sql = "select * from notification_updates where NOT EXISTS 
             (select * from deleted_nofitication where notification_id = notification_updates.id AND user_id = ".$user_id." )" ; 
            $command = Yii::$app->db->createCommand($sql);
            $notifications = $command->queryAll();

从$ notifications中获取值并将其添加到json。

比较php中数组中的值 - 如果不存在则将其添加到json数组中

2 个答案: