在 Laravel 关系上获取对未定义函数的错误消息调用

Question

我正在 Laravel 8 项目中查询公司详细信息。下面是结构和 MySQL 查询。

# Getting company param from request.

SELECT c.company_name, p.position_name, s.salary
FROM tbl_company as c
JOIN tbl_company_type as t ON t.id = c.company_type_id
JOIN tbl_company_position as p ON p.company_id = c.id
JOIN tbl_comapny_salary as s ON s.position_id = p.id
where c.id = 1

表结构：

tbl_company_type
id
type
description

tbl_company
id
company_type_id
company_name
description

tbl_company_position
id
position_name
company_id

tbl_comapny_salary
id
position_id
salary

公司模型

 namespace App\Models;

 use Illuminate\Database\Eloquent\Factories\HasFactory;
 use Illuminate\Database\Eloquent\Model;

class Company extends Model
{
protected $table = 'tbl_company';

public function listtype()
{
    return $this->hasOne(Type::class, 'id', 'company_type_id');
}

 public function listposition()
{
    return $this->hasOne(Type::Position, 'company_id', 'id');
}
}


namespace App\Models;

 use Illuminate\Database\Eloquent\Factories\HasFactory;
 use Illuminate\Database\Eloquent\Model;

 class Type extends Model
 {
  protected $table = 'tbl_company_type';
 }



 namespace App\Models;

 use Illuminate\Database\Eloquent\Factories\HasFactory;
 use Illuminate\Database\Eloquent\Model;

 class Position extends Model
 {
 protected $table = 'tbl_company_position';

 public function listsalary()
 {
    return $this->hasOne(Type::Salary, 'position_id', 'id');
 }
 }


  namespace App\Models;

  use Illuminate\Database\Eloquent\Factories\HasFactory;
  use Illuminate\Database\Eloquent\Model;

  class Salary extends Model
  {
  protected $table = 'tbl_comapny_salary';
  }

使用关系检索公司详细信息

   $companydetails = Company::with('listtype', 'listposition','listsalary')- 
  >whereRaw("tbl_company = 2")->get();

收到此错误

message": "调用模型上未定义的关系 [listsalary] [应用\模型\位置]

Answer 1

您的公司模型没有任何 listsalary 关系。

class Company extends Model
{
    protected $table = 'tbl_company';

    public function listtype() 
    {
        return $this->hasOne(Type::class, 'id', 'company_type_id');
     }

    public function listposition()
    {
        return $this->hasOne(Type::Position, 'company_id', 'id');
    }
}

当您调用 Company::with() 时，您仅调用该公司模型中的关系。要访问列表薪水，您可以这样做

   $companydetails = Company::with('listtype','listposition','listposition.listsalary')- 
  >whereRaw("tbl_company = 2")->get();

// or 

   $companydetails = Company::with([
      'listtype',
      'listposition' => function ($query) { 
          $query->with('listsalary'); // $query is your listposition relation
     }

   ])
  ->whereRaw("tbl_company = 2")->get();

肯定应该有更好的命名约定。