Question

以下是我在laravel的关系，我无法使用用户的对象访问公司，但是当我尝试访问它时，我得到 null ，请参阅下面的代码获取图片

以下是我的模特

    IF if exists(select * from INFORMATION_SCHEMA.VIEWS where TABLE_NAME='ACTIVE_GROUPS')
BEGIN
 ALTER  view ACTIVE_GROUPS AS
    SELECT id, name, status
    FROM test1.group
    WHERE status != 2 group by

END
ELSE
    CREATE  view ACTIVE_GROUPS AS
    SELECT id, name, status
    FROM test1.group
    WHERE status != 2 group by id
BEGIN
END

以下是我的公司模特

<?php

namespace App\Models;

use Illuminate\Notifications\Notifiable;
use Illuminate\Foundation\Auth\User as Authenticatable;
use Zizaco\Entrust\Traits\EntrustUserTrait;
use Session;
use Illuminate\Support\Facades\DB;

class User extends Authenticatable
{
    use Notifiable;
    use EntrustUserTrait;

    protected $table = 'tbl_users';
    protected $primaryKey = 'id';
    protected $guarded = ['id'];
    const API = 'api';
    const WEB = 'web';

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $fillable = [
        'name', 'email', 'password', 'last_login', 'Address', 'Age', 'DateOfBirth', 'created_by', 'deleted_by'
    ];

    /**
     * The attributes that should be hidden for arrays.
     *
     * @var array
     */
    protected $hidden = [
        'password', 'remember_token',
    ];

    protected $casts = [
        'is_admin' => 'boolean',
    ];

    public function isAdmin()
    {
        return $this->is_admin;
    }

    static function GetUserNamebyID($id)
    {
        $name = User::select("name")->where(["id" => $id])->pluck('name');
        if (isset($name[0])) {
            return $name[0];
        } else {
            return '';
        }
    }



    public function loginlogout()
    {
        $this->hasMany("App\Models\LoginLogoutLogs", 'userID');
    }

    public function company()
    {
        $this->hasMany("App\Models\Company", "company_id");
    }
}

这就是我试图访问公司的方式，但我得到 null 。

<?php

namespace App\Models;

use Illuminate\Database\Eloquent\Model;
use DB;
use App\Models\CarePlanData;
use Session;

class Company extends Model
{
    protected $table = 'companies';
    protected $primaryKey = 'id';
    protected $guarded = ['id'];

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $fillable = [
        'name', 'phone_no', 'address', 'password', 'description', 'city', 'company_logo', 'country', 'email'
    ];

    static public function fetchAllActiveCompanies()
    {

        return DB::table("companies")->where(['is_active' => 1])->pluck('name', 'id');
    }

// change company to hasmany

    public function users()
    {
        return $this->hasmany('App\Models\User');
    }

}

Answer 1

首先，你正在做什么

public function fetchCompany(){
 $User = User::find($user->id);
        dd($User->company());
}

但它应该是

   public function fetchCompany(){
 $User = User::find($user->id);
        dd($User->company()->get());
}

或

 public function fetchCompany(){
 $User = User::find($user->id);
        dd($User->company);
}

请同时检查User::find($user->id)是否给予您反对。如果没有，那么你就无法获得非对象的公司

Answer 2

首先，如果用户属于1家公司，那么它应该是：

public function company()
{
    $this->belongsTo("App\Models\Company", "company_id");
}

然后fetchCompany()应

public function fetchCompany(){
   $User = User::with('company')->find($user->id);
   dd($User->company);
}

您需要使用with来加载关系。您将User模型中定义关系的函数名称传递给with with('function_name')。

Answer 3

实际上，如果您的company_id字段位于用户模型上，那么您的关系应为

public function company()
    {
        $this->belongsTo("App\Models\Company", "company_id");
    }

除非用户可以拥有多家公司？

laravel belongsTo on User Model无法正常工作

3 个答案: