我对 Laravel 模型关系有疑问。我需要让用户创建新卡车。但是,我需要将制造商的字段存储为 ID,而不是标题。所以我决定制作两个具有一对多关系的表(制造商和卡车)(制造商有多辆卡车,而一辆卡车有一个制造商)。
这是迁移文件。制造商表:
public function up()
{
Schema::create('manufacturers', function (Blueprint $table) {
$table->bigIncrements('id');
$table->string('manufacturer');
$table->timestamps();
});
}
卡车表:
public function up()
{
Schema::create('trucks', function (Blueprint $table) {
$table->bigIncrements('id');
$table->unsignedBigInteger('make_id');
$table->unsignedInteger('year');
$table->string('owner');
$table->unsignedInteger('owner_number')->nullable();
$table->text('comments')->nullable();
$table->foreign('make_id')->references('id')->on('manufacturers');
$table->timestamps();
});
}
Manufacturer.php模型:
namespace App;
use Illuminate\Database\Eloquent\Model;
class Manufacturer extends Model
{
/**
* @var string
*/
protected $table = 'manufacturers';
/**
* @var array
*/
protected $fillable = [
'manufacturer',
];
public function trucks(){
return $this->hasMany(Truck::class);
}
}
Truck.php 模型:
namespace App;
use Illuminate\Database\Eloquent\Model;
class Truck extends Model
{
/**
* @var string
*/
protected $table = 'trucks';
/**
* @var array
*/
protected $fillable = [
'make_id', 'year', 'owner', 'owner_number', 'comments',
];
public function manufacturer(){
return $this->belongsTo(Manufacturer::class);
}
}
控制器文件:
public function index()
{
$trucks = Truck::all();
return view('trucks.index')->with('trucks', $trucks);
}
索引.blade.php
@foreach($trucks as $truck)
<tbody>
<tr>
<td>{{$truck->make_id}}</td> //I need this one to show manufacturers title
<td>{{$truck->year}}</td>
<td>{{$truck->owner}}</td>
<td>{{$truck->owner_number}}</td>
<td>{{$truck->comments}}</td>
</tr>
</tbody>
@endforeach
此视图现在显示 id。我需要做什么来显示制造商标题(manufacturers.manufacturer)而不是 id?谢谢大家!
凤凰求蛊