php - 如何在 Laravel 中显示模型关系？

Question

我有这个数据库表:job、department、pcn，并且该表pcn具有属性job_id 和 department_id 。所以在 Laravel 中，我在其等效模型上有这样的定义:

class Job extends Model
{
    public function pcns(){return $this->hasMany('App\Pcn', 'id');}
}

class Department extends Model
{
    public function pcns(){return $this->hasMany('App\Pcn', 'id');}
}

class Pcn extends Model
{
    public function job(){return $this->belongsTo('App\Job');}

    public function department(){return $this->belongsTo('App\Department');}
}

我现在的问题是我的 pcn 索引显示的 Pcn 列表给了我这个错误:

Trying to get property 'name' of non-object (View: C:\wamp\www\bookersystem\resources\views\pcn\index.blade.php)

其中我的index.blade.php有这个:

@foreach($pcns as $key => $value)
    <td>{{ $value->control_number }}</td>
    <td>{{ $value->job->name }}</td>
    <td>{{ $value->department->name }}</td>
@endforeach

在我的 Pcn Controller 上:

public function index()
{

    $pcns = Pcn::paginate(50);

    return view('pcn.index', compact('pcns'));

}

对于我的迁移，我有这样的定义:

public function up()
{
    Schema::create('pcn', function (Blueprint $table) {
        $table->engine = "InnoDB";
        $table->charset = 'utf8mb4';
        $table->collation = 'utf8mb4_general_ci';
        $table->bigIncrements('id');
        $table->unsignedBigInteger('department_id');
        //$table->foreign('department_id')->references('id')->on('department');
        $table->unsignedBigInteger('job_id');
        //$table->foreign('job_id')->references('id')->on('job');
        $table->string('control_number', 45);
        $table->string('center', 5);
        $table->string('status', 8)->nullable();
        $table->unsignedBigInteger('mrf_id')->nullable();
        $table->string('degree', 25)->default('Not Recruiting');
        $table->timestamps();
    });
}

我在这里做错了吗？

Answer 1

首先，最好从关系定义中删除 id 或声明右 外键:

class Job extends Model
{
    //this
    public function pcns(){return $this->hasMany('App\Pcn');}
    //or this
    public function pcns(){return $this->hasMany('App\Pcn', 'job_id', 'id');}
}

class Department extends Model
{
    //this
    public function pcns(){return $this->hasMany('App\Pcn');}
    //or this 
    public function pcns(){return $this->hasMany('App\Pcn', 'department_id', 'id');}
}

第二步:最好eager load减少所需查询数量的关系:

public function index()
{
    $pcns = Pcn::with(['job', 'department'])->paginate(50);

    return view('pcn.index', compact('pcns'));
}

之后，在您看来，您实际上并不需要 @foreach 中的 $key=>$value:

@foreach($pcns as $pcn)
    <td>{{ $pcn->control_number }}</td>
    <td>{{ $pcn->job->name }}</td>
    <td>{{ $pcn->department->name }}</td>
@endforeach