Laravel 如何在 Eloquent 上设置条件关系

发布于04月28日

我有这个(简化的)表格 struct :

users
- id
- type (institutions or agents)

institutions_profile
- id
- user_id
- name

agents_profile
- id
- user_id
- name

And I need to create a profile relationship on the Users model, but the following doesn't work:

class User extends Model
{
    public function profile()
    {
        if ($this->$type === 'agents')
            return $this->hasOne('AgentProfile');
        else
            return $this->hasOne('InstitutionProfile');
    }    
}

我怎样才能实现这样的目标？

推荐答案

让我们用不同的方法来解决你的问题.首先，让我们分别为各种模型设置关系.

class User extends Model
{
    public function agentProfile()
    {
        return $this->hasOne(AgentProfile::class);
    }    

    public function institutionProfile()
    {
        return $this->hasOne(InstitutionProfile::class);
    }

    public function schoolProfile()
    {
        return $this->hasOne(SchoolProfile::class);
    }

    public function academyProfile()
    {
        return $this->hasOne(AcademyProfile::class);
    }

    // create scope to select the profile that you want
    // you can even pass the type as a second argument to the 
    // scope if you want
    public function scopeProfile($query)
    {
        return $query
              ->when($this->type === 'agents',function($q){
                  return $q->with('agentProfile');
             })
             ->when($this->type === 'school',function($q){
                  return $q->with('schoolProfile');
             })
             ->when($this->type === 'academy',function($q){
                  return $q->with('academyProfile');
             },function($q){
                 return $q->with('institutionProfile');
             });
    }
}

现在你可以像这样访问你的个人资料了