Mysql join 三表/ Laravel 查询答案

【问题标题】：Mysql join three table/ Laravel queryMysql join 三表/ Laravel 查询
【发布时间】：2019-10-18 19:13:51
【问题描述】：

我有三个表受益人，问题，响应..受益人

Benficiary
id name
1  user1
2 user2
3 user3

Questions
id question             type
1  what is your hobby   hobby
2  address              contact
3  phone                contact
..questions upto 500 and i need to select according to its type

Response
id beneficiary_id question_id response
1  1              1           football
2  1              3           5354353
3  2              1           basketball
4  3              2           cityname
5  3              3           432423

Beneficiary 表和 Question 表是静态的，但响应表连接了它们。我试图从 laravel 查询中做到这一点，但它只给了我回复 sql查询

select response,questions.question, beneficiarys.name 
from responses RIGHT join (questions,beneficiarys) 
on beneficiarys.id = responses.beneficiary_id && questions.id = responses.question_id 
where questions.sheet_type = 'SCDS' OR questions.sheet_type = 'SMM'

laravel 查询

$beneficiaries =       Beneficiary::with('quests','response')->whereHas('response', function ($query)use($sheet_type) {
             $query->where('beneficiary_id', '>', 0);
        })
->whereHas('quests',function($query)use($sheet_type){
                $query->where('questions.sheet_type','=',$sheet_type);
                $query->orWhere('questions.sheet_type','=','SCDS');
            })
        ->where('updated_at','>=',!empty($this->request->from_date)?$this->request->from_date:$from)->where('updated_at','<=',!empty($this->request->to_date)?$this->request->to_date:$date)->get();

受益人模式

public function response()
{
return $this->hasMany('App\Model\Response');
}

public function quests(){
return $this->belongsToMany('App\Model\Question',   'responses','beneficiary_id','question_id');

}

问题模型

public function response(){
    return $this->hasMany('App\Model\Response');
}

public function bens(){
    return $this->belongsToMany('App\Model\Bebeficiary', 'responses','question_id','beneficiary_id');
}

响应模型

public function beneficiary(){
    return $this->belongsTo('App\Model\Beneficiary', 'beneficiary_id');
}

public function question(){
    return $this->belongsTo('App\Model\Question', 'question_id');
}

我需要根据其类型的所有问题（类型由用户从视图中选择），我需要这样的输出

beneficary  hobby     address phono  ....
user1       football  null    4244   ....
user2       basketball null   null   ....

有没有什么办法可以用sql或者laravel查询实现*

【问题讨论】：

你能发布 Eloquent 模型之间的关系吗？
一种解决方案是使用querybuilder..连接表。另一种选择是使用关系概念...
如果您在查询构建器中需要它，我可以在答案中发布它..
@jitheshjose 如果可以使用 querybuilder 那么请帮助我。

标签： php mysql laravel

【解决方案1】：

这是 mysql 选择，也许是更好的方法，但这项工作

SELECT b.name,hobby.response as hobby,address.response as addres,phone.response as 
phone FROM beneficiary as b
left join (select *  from response where question_id=1 ) as hobby
      on hobby.beneficiary_id = b.id
left join (select *  from response where question_id=2 ) as address
      on address.beneficiary_id = b.id
left join (select *  from response where question_id=3 ) as phone
      on phone.beneficiary_id = b.id

【讨论】：

谢谢..我有 400 个问题，我也可以动态获取问题吗？和已回复的受益人
在 mysql 中你可以制作数据透视表，但我不确定你是否可以动态制作它，所以如果我在 laravel 中创建它，只调用一个字段并为每个字段进行搜索问题，然后使用联合。我认为这可以发挥作用

【解决方案2】：

您将从该查询中获得结果。 Laravel 也支持自定义。

$sSQL = "SELECT * FROM RESPONSE 
INNER JOIN QUESTION ON QUESTION.id=RESPONSE.QUESTION_id
INNER JOIN  BENEFICIARY ON BENEFICIARY.id=RESPONSE.BENEFICIARY_id";
DB::query($sSQL);

【讨论】：

【解决方案3】：

试试

 $res = DB::table('responses')
   ->join('questions','responses.question_id','=','questions.id')
   ->join('beneficiarys','responses.beneficiary_id','=','beneficiarys.id')
   ->select('responses.*','questions.question','beneficiarys.name')
   ->where('questions.sheet_type', '=','SCDS')
   ->orWhere('questions.sheet_type','=','SMM')
   ->get();

将结果用作

 foreach($res as $res)
 {
   echo $res->name;
 }

【讨论】：

【解决方案4】：

如果您想以 Eloquent 方式进行操作，您可以尝试在嵌套关系上使用 whereHas。尝试类似：

Beneficiary::with('response.question')->whereHas('response.question')

更多参考请尝试这个链接Eloquent with nested relationships

【讨论】：