【发布时间】:2014-10-23 03:22:20
【问题描述】:
当我使用 Laravel eloquent - 关系(一对多)时,结果为 NULL。 我已经在“work_hour”表中设置了数据,其中“workers_id”的值为 2,并且“workers”表中有一行 id=2 我按照文档“Laravel relationships”
我的架构如下所示:
public function up()
{
Schema::create('work_hour', function($table){
$table->increments('id');
$table->integer('worked_hour');
$table->time('worked_from');
$table->time('worked_to');
$table->integer('workers_id')->unsigned();
$table->foreign('workers_id')->references('id')->on('workers');
$table->integer('workday_id');
$table->integer('offer_id');
});
}
PS:列 workday_id、offer_id 尚未设置为外部!
模型如下所示:
class Worker extends Eloquent{
protected $fillable = ['first_name', 'last_name'];
}
class Workhour extends Eloquent{
protected $table = 'work_hour';
public function worker(){
return $this->belongsTo('Worker','workers_id');
}
}
控制器 PS:我在SO上搜索过同样的问题,我尝试了一些意见,然后我将它们评论出来
class WorkerController extends BaseController {
protected $table = 'workers';
....
public function show($id){
$worker = Worker::find($id);
// $workhour = Workhour::where('workers_id','=',$worker->id)->get();
// casting the result: works $workhour->toArray()
// $workhour = Worker::find($id)->with('workhour')->get();
//Call to undefined method Illuminate\Database\Query\Builder::workhour()
// $workhour = Worker::with('workhour')->where('id',2)->get();
//Call to undefined method Illuminate\Database\Query\Builder::workhour()
$workhour = Worker::find($id)->workhour;
// null
dd($workhour);
return View::make('worker.show',['worker' => $worker,'workhour'=>$workhour]);
}
public function workhour(){
return $this->hasMany('Workhour','workers_id');
}
}
我用“.../public/worker/2”调用“show.blade.php”(Laravel 4) 我究竟做错了什么?
【问题讨论】:
-
为什么在你的控制器中定义了 Eloquent 关系(
workhour()方法和表名)? -
它返回
null,因为Worker模型上没有定义关系。 -
@JarekTkaczyk_deczo_ 谢谢你,你说得对,我已将workhour() 函数添加到控制器中,不在模型中,谢谢!现在它完美地工作了,......你应该写下接受的答案
-
@kajetons 因为表名是“workers”,但模型是“Worker”
-
我的意思是为什么它在控制器而不是你的 Eloquent 模型中。无论如何,现在应该解决了。
标签: php mysql laravel-4 eloquent foreign-key-relationship