如何用Laravel 5.3中的表連接兩次創建查詢？

Question

我需要一個查詢得到兩個城市名稱：

例如：

市表：

+---------+----------+ 
| Pana | Name | 
+---------+----------+ 
| THR | Tehran | 
| LON | London | 
+---------+----------+ 

在型號：FROM_CITY是THR和TO_CITY是LON

public function scopePrintQuery($query, $id) 
{ 
    $join = $query 
     -> join('cities', 'cities.pana', 'flights.from_city') 
     -> join('cities', 'cities.pana', 'flights.to_city') 
     -> where('flights.id', $id) 
     ->get([ 
      'flights.*', 
      'cities.name as from_city' 
      ??? for to_city? 
     ]); 
    return $join; 
} 

現在，我需要獲得from_city名稱和to_city在此查詢中的名稱。

該查詢不適用於一個表中的兩個連接！

如何創建此查詢？

Answer 1

對於直接的SQL，您可以爲每個連接的表賦予一個別名 - 例如，

SELECT flights.* 
FROM flights as f 
JOIN cities as fromCity on fromCity.pana = f.from_city 
JOIN cities as toCity on toCity.pana = f.to_city 
WHERE f.id = 3 -- 

With Eloquent，使用select()指定選擇字段。還可以使用DB::raw()使用原始SQL（如給一個別名錶像DB::raw('cities as toCity')。

public function scopePrintQuery($query, $id) 
{ 
    $join = $query 
    -> join(DB::raw('cities as fromCity'), 'fromCity.pana', 'flights.from_city') 
    -> join(DB::raw('cities as toCity'), 'toCity.pana', 'flights.to_city') 
    -> where('flights.id', $id) 
    ->select([ 
     'flights.*', 
     DB::raw('fromCity.name as from_city') 
     DB::raw('toCity.name as to_city') 
    ]); 
    return $join->get(); 
} 

Answer 2

，你也可以用雄辯的模型定義的關係。

也爲更多詳情，請登錄https://laravel.com/docs/5.3/eloquent-relationships

箱兩型 - 月1日是機票

<?php 


class Flights extends Model 
{ 
    protected $table = 'flights'; 

    /** 
    * Get the From City detail. 
    */ 
    public function fromCity() 
    { 
     return $this->hasOne('App\Models\City', 'Pana', 'from_city'); 
    } 

    /** 
    * Get the To city state. 
    */ 
    public function toCity() 
    { 
     return $this->hasOne('App\Models\City', 'Pana', 'from_city'); 
    } 

} 

第二個模式是

<?php 
class City extends Model 
{ 
    protected $table = 'city'; 
} 

現在對於獲取

Flights::where(id, $id)->with('toCity', 'fromCity')->get();