Question

我对Laravel有点陌生，所以我希望这个问题很清楚。

我有一个包含Users的表，另一个有Tasks的表。

在我的用户模型中，我有以下内容：

public function tasks() {
    return $this->hasMany('App\User' , 'id');
}

我可以执行以下操作从数据库中检索单个用户

$users = \App\User::find(1)->tasks()->paginate();

但是我明白了

{"current_page":1,"data":[{"id":1,"name":"Ed","email":"mail@weqweqeq.com","email_verified_at":null,"created_at":null,"updated_at":null,"tasks":[{"id":1,"name":"Ed","email":"mail@weqweqeq.com","email_verified_at":null,"created_at":null,"updated_at":null}]},{"id":2,"name":"Alyse","email":"mail@rxygewe.com","email_verified_at":null,"created_at":null,"updated_at":null,"tasks":[]}],"first_page_url":"http:\/\/127.0.0.1:8000?page=1","from":1,"last_page":1,"last_page_url":"http:\/\/127.0.0.1:8000?page=1","next_page_url":null,"path":"http:\/\/127.0.0.1:8000","per_page":15,"prev_page_url":null,"to":2,"total":2}

我也尝试过：

    $users = \App\User::with(['tasks' => function($q) {
        $q->first();
    }])->paginate();

但是task属性为空

我的问题是我如何才能获得所有用户，但只有第一个任务才能分页工作？

任务表

1   id(Primary) bigint(20)      UNSIGNED    No  None        AUTO_INCREMENT
2   created_at  timestamp           Yes NULL        
3   updated_at  timestamp           Yes NULL        
4   task_name   varchar(255)    utf8mb4_unicode_ci      No  None        
5   user_id(Index)  bigint(20)      UNSIGNED    No  None

Answer 1

您的模型关系定义不正确。您已将一个用户与许多用户相关联。
https://laravel.com/docs/master/eloquent-relationships#one-to-many

public function tasks() {
    // a User ($this) has many Tasks
    return $this->hasMany('App\Task');
}

另外，使用->first()时要小心，因为除非指定，否则不能保证顺序。

$users = \App\User::with(['tasks' => function($q) {
    // get the most recently created task
    $q->latest()->first();
}])->paginate();

Laravel仅以第一结果分页

1 个答案: