Question

我有3个表，其中一个是另外两个表之间的链接表。一个简单的多对多链接。

videos (Table fields shortened for brevity)
+----+---------------------+
| id | name                |
+----+---------------------+
|  1 | test                |
|  2 | Test Subscription 1 |
+----+---------------------+

subscriptions
+----+--------+
| id | name   |
+----+--------+
|  1 | News   |
|  2 | Sports |
|  3 | Gospel |
|  4 | Kids   |
+----+--------+

subscription_video (link table between both models)
+----+-----------------+----------+
| id | subscription_id | video_id |
+----+-----------------+----------+
|  1 |               1 |        2 |
+----+-----------------+----------+

用于创建链接表的迁移

Schema::create('subscription_video', function(Blueprint $table)
{
    $table->increments('id');
    $table->integer('subscription_id')->unsigned();
    $table->foreign('subscription_id')->references('id')->on('subscriptions');
    $table->integer('video_id')->unsigned()->nullable();
    $table->foreign('video_id')->references('id')->on('videos');
});

雄辩模式

class Video extends Model {

    use SoftDeletes;

    /**
     * @return \Illuminate\Database\Eloquent\Relations\BelongsToMany
     */
    public function subscriptions()
    {
        return $this->belongsToMany('App\Subscription');
    }

    /**
     * @return \Illuminate\Database\Eloquent\Relations\BelongsToMany
     */
    public function categories()
    {
        return $this->belongsToMany('App\Category');
    }
}

现在根据文档，我应该能够通过调用

简单地检索结果

Video::has('subscriptions')->get()

实际视图代码（有或没有orderBy时会发生相同的错误）

public function index()
{
    $videos = Video::has('subscriptions')->orderBy('id', 'desc')->get();
    return view('subscription.video.index', ['videos' => $videos ]);
}

在这种情况下应该使用video_id 2返回1个结果集。但无论我尝试什么，我都会继续

SQLSTATE[21000]: Cardinality violation: 1241 Operand should contain 9 column(s) (SQL: select `id`, `name` from `videos` where `videos`.`deleted_at` is null and (select * from `subscriptions` inner join `subscription_video` on `subscriptions`.`id` = `subscription_video`.`subscription_id` where `subscription_video`.`video_id` = `videos`.`id`) = 1)

我不知道我对这些关系的错误以及如何使Eloquent返回正确的结果？

任何帮助都将不胜感激。

1. in Connection.php line 620 2. at Connection->runQueryCallback('select * from影片where影片. deleted_at is null and (select * from订阅inner join subscription_video on订阅. ID为= subscription_video . subscription_id where subscription_video . VIDEO_ID =影片. ID为) >= 1 order by ID为desc', array(), object(Closure)) in Connection.php line 576 3. at Connection->run('select * from影片where视频{ {1}} deleted_at .订阅is null and (select * from subscription_video inner join订阅on ID为. subscription_video = subscription_id . subscription_video {{1 }} VIDEO_ID where影片. ID为= ID为.影片) >= 1 order by影片desc', array(), object(Closure)) in Connection.php line 293 4. at Connection->select('select * from deleted_at where订阅. subscription_video is null and (select * from订阅inner join ID为on subscription_video . subscription_id = subscription_video . VIDEO_ID where影片. ID为{ {1}} ID为=

Eloquent ManyToMany关系问题：基数违规

0 个答案: