Question

我有此数据库表：job，department，pcn，此表pcn的属性为job_id和department_id。因此，在Laravel中，我在等效模型上有以下定义：

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');}
}

我现在的问题是显示pcn列表的pcn索引给我这个错误：

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

我的index.blade.php具有以下内容：

@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控制器上：

public function index()
{

    $pcns = Pcn::paginate(50);

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

}

对于我的迁移，我有以下定义：

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

更恰当的做法是，从关系定义中删除id或声明 right foreign key：

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：最好是建立eager load关系以减少所需的查询数量：

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

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

在那之后，您实际上并不需要$key=>$value中的@foreach：

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

Answer 2

paginate()方法不返回Pcn的集合，它返回一个LengthAwarePaginator，当转换为数组时，它具有作为索引（total，{{1 }}，per_page ....）

您需要从next_link恢复

LengthAwarePaginator

顺便说一句，您应该在查询中预加载@foreach($pcns as $pcn) <td>{{ $pcn->control_number }}</td> <td>{{ $pcn->job->name }}</td> <td>{{ $pcn->department->name }}</td> @endforeach和job关系，以便仅向数据库发送 3个请求，而不是 101 >。

department

如何在Laravel中显示模型关系？

2 个答案: