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!
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凤凰求蛊
TA贡献1825条经验 获得超4个赞
卡车表中制造商的外键不是 manufacturer_id。在这种情况下,您需要在模型中声明它:
return $this->belongsTo(Manufacturer::class, 'make_id' )
和
return $this->hasMany(Truck::class, 'make_id' )
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