[英]Laravel Eloquent join table realtionships
我在這里瀏覽了多個示例和以前的文章,但是找不到我做錯的事情。 我有兩個表:
用戶
id username
1 test1
2 test2
3 test3
分數表
id user_id points
1 1 52
2 2 62
3 3 12
現在,我想獲得所有分數並將user_id替換為實際的用戶名。 這是我的代碼:
ScoresController
<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Score;
use App\Http\Resources\Score as ScoreResource;
class ScoresController extends Controller
{
public function index()
{
$scores = Score::with('user')->paginate(4);
return ScoreResource::collection($scores);
}
}
Score.php
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
use App\User;
class Score extends Model
{
public function user(){
return $this->belongsTo(User::class,'user_id');
}
}
用戶
<?php
namespace App;
use Laravel\Passport\HasApiTokens;
use Illuminate\Notifications\Notifiable;
use Illuminate\Contracts\Auth\MustVerifyEmail;
use Illuminate\Foundation\Auth\User as Authenticatable;
use App\Score;
class User extends Authenticatable
{
use HasApiTokens, Notifiable;
protected $fillable = [
'username', 'email', 'password', 'avatar'
];
protected $hidden = [
'password', 'remember_token',
];
public function scores(){
return $this->hasMany(Score::class,'user_id');
}
}
和分數資源
<?php
namespace App\Http\Resources;
use Illuminate\Http\Resources\Json\JsonResource;
class Score extends JsonResource
{
public function toArray($request)
{
return [
'username' => $this->user_id,
'points' => $this->points,
];
}
}
目前,我正在顯示user_id和積分,我嘗試了
$this->username OR $this->user-id->username
但是它們都不起作用,我知道我可以使用數據庫或原始SQL,但是我想這樣做
閱讀此內容: https : //laravel.com/docs/5.7/eloquent-relationships
foreach ($scores as $score) {
echo $score->user->username;
}
或在Blade中:
@foreach ($score as $score)
<p>{{ $score->user->username }}</p>
@endforeach
您必須首先訪問“屬性”,它是您定義用於處理關系的函數的名稱。 您的情況就是user
。
之后,將返回該User
類,然后您可以在其中訪問所需的任何屬性。
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