雄辩的模型关系返回为空

Posted 2023-02-16

【中文标题】雄辩的模型关系返回为空

【英文标题】：Eloquent model relationship returning as null

【发布时间】：2013-04-10 15:28:34

【问题描述】：

我有一个 laravel 模型

class Project extends Eloquent 

    public static $timestamps = true;

    public $includes = array('members','members.memberdata');

    public function tasks() 

        return $this->has_many('Usertask','project_id');
    

    public function members() 

        return $this->has_many('Projectmember','project_id');
    

及相关模型

class Projectmember extends Eloquent 

    public static $table = "project_members";
    public static $timestamps = true;

    public function project() 

        return $this->belongs_to('Project');
    

    public function memberdata() 

        return $this->has_one('Usermetadata','user_id');
    


class Usermetadata extends Eloquent 

    public static $table = "users_metadata";

    public function user() 

        return $this->belongs_to('User');
    

    public function member() 

        return $this->belongs_to('Projectmember','user_id');
    

当我尝试像这样检索单个项目模型时

 $project = Project::find($id);

 return Response::eloquent($project);

我的 json 输出如下所示

"id":1,"user_id":1,"name":"UberWork","description":"Web based project management \/ task management app","target_date":"2013-11-15 00:00:00","budget":null,"status":0,"created_at":"2013-04-16 20:13:59","updated_at":"2013-04-16 20:13:59","members":["id":1,"project_id":1,"user_id":1,"created_at":"2013-04-16 20:13:59","updated_at":"2013-04-16 20:13:59","memberdata":"user_id":1,"first_name":"Tamarakuro","last_name":"Foh","photo":"","company_name":null,"phone":null,"package":"free","subscription_start":"0000-00-00 00:00:00","subscription_end":"0000-00-00 00:00:00","api_key":"12b14a7d3ca48c53bb5b1a88fa3eca3b","id":3,"project_id":1,"user_id":3,"created_at":"2013-04-16 20:13:59","updated_at":"2013-04-16 20:13:59","memberdata":"user_id":3,"first_name":"Ebere","last_name":"Uche","photo":"","company_name":"Chronotech Labs","phone":null,"package":"free","subscription_start":"0000-00-00 00:00:00","subscription_end":"0000-00-00 00:00:00","api_key":"ab446bd9ffbb898e818a892c7401e0f6","id":4,"project_id":1,"user_id":2,"created_at":"2013-04-17 08:13:00","updated_at":"2013-04-17 08:13:00","memberdata":null]

我的数据库是这样的；

用户

身份证 电子邮件 密码 ip_address 有效 ...

users_metadata

身份证 user_id 名字 姓氏 个人资料照片 ...

项目

身份证 user_id 姓名 说明 状态 ...

项目成员

身份证 project_id user_id

我的问题是为什么项目的最后一个成员的 memberdata 为“null”，而其他成员不为 null。我是不是做错了什么？

【问题讨论】：

您确定您的最后一个成员在 memberdata 表中有匹配的记录吗？您最后一个成员的id 和user_id 不匹配，我不知道这是否重要，但关系将匹配id 到另一个表的user_id？

是的，users_metadata 表上有匹配记录

您能详细解释一下您的数据库设计吗？ user_id 到处指的是什么？

我已经编辑了我的问题以包含数据库设计

问题是因为你想要project_members.user_id = users_metadata.user_id，但是关系总是使用一个表的id和另一个表的fk（user_id）。这适用于您的大部分数据，因为 project_members.id 和 project_members.user_id 是相同的；当他们不是（您的最后一个项目成员）时，它会中断。

【参考方案1】：

Eloquent 中的关系总是将主键 (pk) 与外键 (fk) 联系起来。但是，您正在尝试基于两个外键建立关系，从而跳过了关系。唯一的 Eloquent 解决方案是包含额外的关系步骤。这里有一些模型（我已经省略了我们在这个例子中不需要的关系）......

class Project extends Eloquent 

    public static $timestamps = true;

    public $includes = array('members','members.user', 'members.user.metadata');

    public function members()
    
        return $this->has_many('Projectmember','project_id');
    


class Projectmember extends Eloquent 

    public static $table = "project_members";
    public static $timestamps = true;

    public function user()
    
        return $this->belongs_to('User');
    


class User extends Eloquent 

    public static $timestamps = true;

    public $hidden = array('password', 'ip_address');

    public function metadata()
    
        return $this->has_one('Usermetadata');
    



class Usermetadata extends Eloquent 

    public static $table = "users_metadata";

    public function user() 

        return $this->belongs_to('User');
    

我明白你为什么要跳过这段关系，但遗憾的是，这不可能在关系中实现。