Laravel雄辩的表联接

时间:2018-09-19 14:07:33

标签: php laravel eloquent

我在这里浏览了多个示例和以前的文章,但是找不到我做错的事情。我有两个桌子:

用户

id    username
1     test1
2     test2
3     test3

得分表

id    user_id    points
1     1          52
2     2          62
3     3          12

现在我想获得所有分数并将user_id替换为实际的用户名。这是我的代码:

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');
    }
}

用户

<?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');
    }
}

得分资源

<?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,
        ];
    }
}

目前我正在显示user_id和积分,我尝试过

$this->username OR $this->user-id->username

但是它们都不起作用,我知道我可以使用数据库或原始SQL,但我想这样做

1 个答案:

答案 0 :(得分:1)

阅读:https://laravel.com/docs/5.7/eloquent-relationships

foreach ($scores as $score) {
    echo $score->user->username;
}

或者在Blade中:

@foreach ($score as $score)
    <p>{{ $score->user->username }}</p>
@endforeach

您必须首先访问“属性”,即您定义的用于处理关系的函数的名称。您的情况是user
之后,将返回该用户的User类,然后您可以在其中访问所需的任何属性。