Laravel Eloquent: hasManyThrough 还是其他?

Laravel Eloquent: hasManyThrough or Something Else?

在 Laravel 5.7 我已经阅读了 Has Many Through documentation 但我仍然不能正确地使用它来处理我的情况。

这是数据库:

Analytics

    id
    data
    subscriber_id

Subscribers

    id
    city_id

Cities

    id
    name

我需要 Analytics 模型从 Analyticssubscribers.idcities.name

获取数据

我做了什么:

连接的分析和订阅者模型

<?php

class Analytics extends Model
{
    public function subscriber()
    {
        return $this->belongsTo('App\Models\Subscriber');
    }
}

class Subscriber extends Model
{
    public function analytics()
    {
        return $this->hasMany('App\Models\Analytics');
    }
}

发出一个请求,从 Analytics table 获取数据,订阅者数据:

$results = Analytics::where('survey_id', '4')
    ->with('subscriber')
    ->whereDate('created_at', '>=', $last_survey_date)
    ->orderBy('data')
    ->get();

如果有人对如何获取城市名称有任何想法,请分享。

  // maybe this will work for you?
  class Analytics extends Model
    {
        public function subscriber()
        {
            return $this->belongsTo('App\Models\Subscriber');
        }

        public function cities() {
            return $this->hasManyThrough('App\City', 'App\Subscriber');
        }   

    }

我不确定我是否正确理解了您的要求。您想获取给定 CitySubscribers 中的所有 Analytics 吗?或者您想要分析订阅者的城市名称吗?无论哪种方式,这里都有两种解决方案。

获取给定城市订阅者的所有分析:

$city = 'Vienna';

$results = Analytics::query()
    ->with('subscriber')
    ->whereHas('subscriber', function ($query) use ($city) {
        $query->whereHas('city', function ($query) use ($city) {
            $query->where('name', $city);
        });
    })
    ->where('survey_id', '4')
    ->whereDate('created_at', '>=', $last_survey_date)
    ->orderBy('data')
    ->get();

或者要获取分析记录的城市名称,您有两种选择。一种是对关系使用 Laravel 预加载,这可行,但可能会将大量不必要的数据加载到内存中:

$results = Analytics::query()
    ->with('subscriber.city') // you can nest relationships as far as they go
    ->where('survey_id', '4')
    ->whereDate('created_at', '>=', $last_survey_date)
    ->orderBy('data')
    ->get();

foreach ($results as $analytics) {
    $city = $analytics->subscriber->city->name;
}

另一种是自己加入table并且select只需要必要的数据:

$results = Analytics::query()
    ->join('subscribers', 'subscribers.id', '=', 'analytics.subscriber_id')
    ->join('cities', 'cities.id', '=', 'subscribers.city_id')
    ->where('analytics.survey_id', '4')
    ->whereDate('analytics.created_at', '>=', $last_survey_date)
    ->orderBy('analytics.data')
    ->select('analytics.*', 'cities.name as city_name')
    ->get();

foreach ($results as $analytics) {
    $city = $analytics->city_name;
}

请注意,您可以使用 select('analytics.*', 'cities.name'),但这将覆盖 analytics table 的 name 列 selected(如果存在)。所以最好使用 as col_alias.

的列别名

感谢 Nick Surmanidze 和 Namoshek 的回答! 我昨晚找到了它的工作方式:

class Subscriber extends Model
{
    public function analytics()
    {
        return $this->hasMany('App\Models\Analytics');
    }

    public function subscriberCity()
    {
        return $this->belongsTo('Modules\Directories\Entities\City', 'city_id', 'id');
    }

}


class City extends Model
{
    public function subscriber()
    {
        return $this->hasMany('App\Models\Subscriber');
    }
}

而需要的结果可以这样得到:

$results = Analytics::where('survey_id', '4')
    ->with(['subscriber' => function($i){
            $i->with(['subscriberCity']);
        }])
    ->whereDate('created_at', '>=', $last_survey_date)
    ->orderBy('data')
    ->get();