Php 如何在Laravel 5.3中创建两次表连接的查询？

我需要通过一个查询获取两个城市名称：

例如：

城市表：

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

模型中：from_city为THR，to_city为LON

现在，我需要在这个查询中获取from_city name和to_city name

查询无法处理一个表中的两个联接

---

如何创建此查询

使用直接SQL，您可以为每个联接的表指定一个别名，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 --

使用Eloquent可指定选择字段。还可用于使用原始SQL，例如为表提供别名，如DB:：raw“cities as toCity”

---

您还可以使用雄辩的模型来定义关系

如需了解更多详细信息，请访问

板条箱两个模型- 一是航班

<?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', 'to_city');
   }

}

---

您好，谢谢您的帮助，为什么要使用从城市到城市的函数？当您进行查询时，它会指定应该使用哪些关系进行即时加载。有关更多详细信息，请访问

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();
}

<?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', 'to_city');
   }

}

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

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