如何在laravel中添加多表关系

我有以下表格，并建立了与Laravel关系的查询。在我的面板中，一旦用户登录，我将获得用户详细信息和用户业务详细信息，但现在我也不知道如何获得业务类型详细信息

用户表： 用户业务表 业务类型表 模型： 用户

<?php

namespace App;

use Illuminate\Notifications\Notifiable;
use Illuminate\Foundation\Auth\User as Authenticatable;
use Zizaco\Entrust\Traits\EntrustUserTrait;
use App\Notifications\ResetPassword as ResetPasswordNotification;
use Laravel\Passport\HasApiTokens;
use Illuminate\Database\Eloquent\SoftDeletes;
use Webpatser\Uuid\Uuid;
use App\RoleUser;
use App\UsersBusiness;

class User extends Authenticatable
{
    use Notifiable;
    use EntrustUserTrait;
    use HasApiTokens;
    use SoftDeletes, EntrustUserTrait {
        SoftDeletes::restore as sfRestore;
        EntrustUserTrait::restore as euRestore;
    }

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $fillable = [
        'business_id', 'username', 'email'
    ];

    public function usersBusiness()
    {
        return $this->belongsTo('App\UsersBusiness', 'business_id', 'id');
    }
}

我得到以下响应数据：

{
    "data": {
        "id": 1,
        "business_id": 1,
        "username": "john632",
        "email": "john632@gmail.com",
        "password": "john632",
        "users_business": {
            "id": 1,
            "user_id": 4,
            "business_type_id": 3,
            "business_name": "Honest"
        },
        "business_types": null
    },
    "token": "eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiIsImp0aSI6Ijk4YTQ5OTBmOGQxOWQ5NTg1OGFlZWU1MDY0NTBiY2Y2OWJmOWQ3NzFhZjRmN2RmMzBmMWRkZWNmNWY4OTAzM2UyNmI2MzE3MTY3MDMxOTk4In0.eyJhdWQiOiIxIiwianRpIjoiOThhNDk5MGY4ZDE5ZDk1ODU4YWVlZTUwNjQ1MGJjZjY5YmY5ZDc3MWFmNGY3ZGYzMGYxZGRlY2Y1Zjg5MDMzZTI2YjYzMTcxNjcwMzE5OTgiLCJpYXQiOjE1MzIyNjA3ODUsIm5iZiI6MTUzMjI2MDc4NSwiZXhwIjoxNTYzNzk2Nzg1LCJzdWIiOiI1Iiwic2NvcGVzIjpbXX0.f_Xv0SrtTZ9m-40oHjAglCbKv76s_bARQ74XDihhFnI-jtHKwCWiF-jai5Yt6h9QyakCTZEo1bPAJdeph7Bj0_tKJpq3sGvK4t73_LZg_OOcsmAt61a4OSAgI1pjPV0tMMwHCoHm-xLlNnriAyaLCAbTQLQkfrw53467ys6rchE5V0rzy-JswjTfmB6SvZcqXsJQo6CWDRTWYbKvJO0FSmdZfLxxO_u4i_8ah5W63qJ4MSN9q22zkZLQ-L3NZhOux2KkwWiySioL2K25Y_UZmefClYwk1h-EY_LEVht3U7Kpqn9fmM6_Q4ByD-sSzLdAixdbq4REqinSaayzfMY934nijLu7ysEIc0oIukiHYcIk9tGV6DNuQ0CWhqEn0W_308MSBU4Ffyi5SQo7ubb5uPG7l_XOdomIR9dK9KtVONbPe7iF6TuccPCWZwvqKgfFl7TqEgiUWSiAl_ekkiaUDEM3cIuIH8AOLE17UuW4W0VyR2ziIt68au8SEuP2ilMBRsRMsFGbRKQWcvLluNw_qubcdzZ4yX9kuQAvXuBrHAcXb9WMlki2votvd7RKVDwxqwsTJRoeKNtJQdEQRbRZUD6nXyzGkmtEMrfwYoLVgTX3vAgVjO_erYtI5x-NV-EnoLT352odtRDYh5gTzVbmzYAxbLf_XUCDHjvlMEvM81g",
    "status": true,
    "status_code": 1,
    "message": "Login successfully."
}

我也想添加业务类型详细信息，所以请您指导我如何为该表添加关系

---

谢谢。

在您的

UserBusiness

模型中创建一个belongsTo关系，如下所示

$user = Auth::user();
$user->load(['usersBusiness.businessType']); //lazy loading
dd($user);

用户商业模式

public function businessType(){
     return $this->belongsTo(BusinessType::class, 'business_type_id');
}

之后，像这样加载它

$user = Auth::user();
$user->load(['usersBusiness.businessType']); //lazy loading
dd($user);

如果您只需要userBusiness及其类型

dd($user->usersBusiness);

旁注：我认为用户表中不需要

business\u id

。只需在userbusiness表中保留

user\u id

，然后在userbusiness的用户模型中添加

hasOne

（如果用户只能有一个业务）或

hasMany

（如果用户可以有多个业务）关系

我建议您使用

例如，在我的例子中，我有ChannelItemResource.php：

public function toArray($request)
    {
        return [
    'id' => $this->id,
    'channel_id' => $this->channel_id,
    'channel' => $this->channel->title,
    'type' => $this->type,
    'description' => $this->description,
    'price' => $this->price,
    'created_at' => $this->created_at,
    'updated_at' => $this->updated_at,
];
    }

和ChannelResource.php：

return [
    'id' => $this->id,
    'title' => $this->title,
    'user_id' => $this->user_id,
    'description' => $this->description,
    'type' => $this->type,
    'url' => $this->url,
    'ranking_total' => $this->ranking_total,
    'ranking_flag' => $this->ranking_flag,
    'ranking_country' => $this->ranking_country,
    'thumbnail' => $this->thumbnail,
    'state' => $this->state,
    'hit' => $this->hit,
    'created_at' => $this->created_at,
    'updated_at' => $this->updated_at,
    'items' => ChannelItemResource::collection($this->items),
];

使用此结构，您可以轻松地包含所需的关系，结果将以JSON格式显示

在控制器中使用：

public function all(Request $request)
    {
        return new ChannelsResource(Channel::where("state", "active")->orderBy('id', 'desc')->paginate(30));
    }

---

您有

UserBusiness

model吗？@rkj有。我已经为每个表创建了模型。我已经更新了我的问题。我已经添加了答案检查您的用户模型与业务的关系是否良好，只需在UsersBusiness中添加一个关系，您就完成了为什么您在userbusiness表中添加了

User\u id

，在User表中添加了

business\u id

？我认为用户表中不需要

business\u id

。只需在userbusiness表中保留

user\u id

，然后在用户模型中添加'hasOne'或

hasMany

关系即可

$user = Auth::user();
$user->load(['usersBusiness.businessType']); //lazy loading
dd($user);

dd($user->usersBusiness);

public function toArray($request)
    {
        return [
    'id' => $this->id,
    'channel_id' => $this->channel_id,
    'channel' => $this->channel->title,
    'type' => $this->type,
    'description' => $this->description,
    'price' => $this->price,
    'created_at' => $this->created_at,
    'updated_at' => $this->updated_at,
];
    }

return [
    'id' => $this->id,
    'title' => $this->title,
    'user_id' => $this->user_id,
    'description' => $this->description,
    'type' => $this->type,
    'url' => $this->url,
    'ranking_total' => $this->ranking_total,
    'ranking_flag' => $this->ranking_flag,
    'ranking_country' => $this->ranking_country,
    'thumbnail' => $this->thumbnail,
    'state' => $this->state,
    'hit' => $this->hit,
    'created_at' => $this->created_at,
    'updated_at' => $this->updated_at,
    'items' => ChannelItemResource::collection($this->items),
];

public function all(Request $request)
    {
        return new ChannelsResource(Channel::where("state", "active")->orderBy('id', 'desc')->paginate(30));
    }