如何在 Laravel 中显示模型关系？

Question

我有这样的数据库表：job，department，pcn和该表pcn具有性能job_id及department_id在其上。因此，在 Laravel 中，我对其等效模型有以下定义：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 控制器上：public function index(){    $pcns = Pcn::paginate(50);    return view('pcn.index', compact('pcns'));}至于我的迁移，我有这个定义：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id从关系定义中删除或声明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：最好预先加载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'));

}

之后，在你看来，你并不真正需要$key=>$value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()方法不返回的集合Pcn，它返回一个LengthAwarePaginator，当变成一个数组时，它有作为索引 ( total, per_page, next_link....)

您需要从 LengthAwarePaginator

@foreach($pcns as $pcn)

    <td>{{ $pcn->control_number }}</td>

    <td>{{ $pcn->job->name }}</td>

    <td>{{ $pcn->department->name }}</td>

@endforeach

顺便说一下，您应该在查询中预加载job和department关系以仅向数据库发送3 个请求而不是101。

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