标签: eager-loading

我试图通过急切加载避免N + 1查询问题,但它无法正常工作.相关模型仍在单独加载.

以下是相关的ActiveRecords及其关系:

class Player < ActiveRecord::Base
  has_one :tableau
end

Class Tableau < ActiveRecord::Base
  belongs_to :player
  has_many :tableau_cards
  has_many :deck_cards, :through => :tableau_cards
end

Class TableauCard < ActiveRecord::Base
  belongs_to :tableau
  belongs_to :deck_card, :include => :card
end

class DeckCard < ActiveRecord::Base
  belongs_to :card
  has_many :tableaus, :through => :tableau_cards
end

class Card < ActiveRecord::Base
  has_many :deck_cards
end

class Turn < ActiveRecord::Base
  belongs_to :game
end

我正在使用的查询是在Player的这个方法中:

def tableau_contains(card_id)
  self.tableau.tableau_cards = TableauCard.find :all, :include => [ {:deck_card => (:card)}], :conditions => ['tableau_cards.tableau_id = …

有没有办法进行第二次查询并复制预先加载的功能？让我们说一个真实的基本例子.一个人有很多车,一辆车有很多零件..

通常情况下,你会说:

person.cars.includes(:parts)

我真的不明白这是做什么的......它只是在内存中加载其他对象吗？它们是否在这一点上与每辆车相关联,然后只是共享密钥？如果只是共享密钥..我可以在内存中加载这些"部件"并在不明确调用"包含"的情况下访问它们吗？例如..类似于:

cars = person.cars
cars_id_array << insert cars ids here
parts = Parts.where(:user_id => person_id, "car_id IN cars_id_array") (just an example)

在这种情况下(假设我做了写代码,我确定我没有..我可以去car1.parts,car2.parts,那些部分是否已经加载到内存中？

如果不是,那么渴望加载的是什么？

我正在尝试优化我们的一些查询.一个特别的挑战是试图加载多态模型.在这种情况下,UserView是多态的,并具有以下列:

user_id, user_viewable_id, user_viewable_type

当我尝试运行此查询时.

@user_views = UserView.select('user_views.* AS user_views, songs.* AS songs, users.* AS users')
  .joins('INNER JOIN users AS users ON user_views.user_id = users.id')
  .joins('INNER JOIN songs AS songs ON user_views.user_viewable_id = songs.id')
  .where(:user_viewable_type => 'Song').order('user_views.id DESC').limit(5)

它似乎并不急于加载查询.我正在使用名为MiniProfiler的gem ,它表示它实际上运行的是n + 1个查询,而不是一个.

以下AR查询返回此SQL:

SELECT user_views.* AS user_views, songs.* AS songs, users.* AS users FROM "user_views" INNER JOIN users AS users ON user_views.user_id = users.id INNER JOIN songs AS songs ON user_views.user_viewable_id = songs.id WHERE "user_views"."user_viewable_type" = …

我是laravel的新手,并创建一个基本的应用程序来控制关系.我实现了一对一的关系,并希望从基表和相关表中获取特定的列.我有两个表,即users和identity_cards.Relationship定义如下

class User extends Eloquent implements UserInterface, RemindableInterface {

    use UserTrait, RemindableTrait;

    /**
     * The database table used by the model.
     *
     * @var string
     */
     protected $table = 'users';

    /**
     * The attributes excluded from the model's JSON form.
     *
     * @var array
     */
    //protected $hidden = array('password', 'remember_token');

    protected $fillable = array('first_name','last_name','email','created_at','updated_at');

    public function identity_cards() 
    {
        return $this->hasOne('IdentityCard');
    }
}


class IdentityCard extends Eloquent {

    protected $table    = 'identity_cards';
    protected $fillable = array('issuance_date','expiry_date','issuance_location','user_id');
    public $timestamps  = false;

    public …

任何人都可以帮助我理解为什么以下代码工作

$x = $widget->objGallery->galleryItems()->with(array('captions' => function($query){ $query->where('locale', 'IT' );}))->get() ;

但是当我使用动态值时

$id='11';
$x = $widget->objGallery->galleryItems()->with(array('captions' => function($query){ $query->where('locale', $id );}))->get() ;

在说

方法Illuminate\View\View :: __ toString()不得抛出异常

在我的一个控制器中,我有以下内容:

return Lot::with(array('region', 'territory', 'manager')) -> get();

这很完美,并返回以下内容:

Array
(
[0] => stdClass Object
    (
        [id] => 1
        [region_id] => 3
        [territory_id] => 2
        [state_id] => 5
        [manager_id] => 2
        [lot_num] => 0
        [lot_type] => managed
        [name] => Some Name
        [address_1] => Some Address
        [address_2] => 
        [address_3] => 
        [city] => Some City
        [zip] => 00000
        [opened_at] => 
        [deleted_at] => 
        [created_at] => 2014-11-06 00:49:39
        [updated_at] => 2014-11-06 00:49:39
        [region] => stdClass Object
            (
                [id] => 3
                [name] => Corporate
                [deleted_at] => …

我有代码可以工作,但没有急切加载嵌套关系.

$projects = Project::with('organization')
        ->leftJoin('stages', 'stages.project_id', '=', 'projects.id')
        ->leftJoin('activities', 'activities.stage_id', '=', 'stages.id')
        ->leftJoin('tasks', 'tasks.activity_id', '=', 'activities.id')
        ->select('projects.*',  DB::raw('SUM(IF(tasks.status = 4, score, 0)) AS score'), 
                                DB::raw('SUM(tasks.score) AS total_score'))
        ->groupBy('projects.id')
        ->get();

我想通过急切加载嵌套关系来做到这一点,如果我没有这些自定义选择(total_score和得分),我会做

$projects = Project::with('stages.activities.tasks');

但问题出现在那些自定义选择(得分和total_score)上.我尝试了类似的东西,但没有奏效

$projects = Project::with(['stages', 'activities', 'tasks' => function($q) { 
        $q->select( DB::raw('SUM(IF(tasks.status = 4, score, 0)) AS score'), 
                    DB::raw('SUM(tasks.score) AS total_score')); 
    }])->get();

我正在使用Rails 4.1和我的模型:

Client has_many TicketLists
TicketList has many projects

现在,我正在尝试在客户端模型中使用eager loading,例如:

class Client < ActiveRecord::Base
  def ticket_lists_with_project_id(project_id)
    ticket_lists.includes(:projects).where("projects.id = ?", project_id)
  end  
end  

当我这样做时:

Client.find(2).ticket_lists_with_project_id(1)
  Client Load (1.7ms)  SELECT  "clients".* FROM "clients"  WHERE "clients"."is_destroyed" = 'f' AND "clients"."is_closed" = 'f' AND "clients"."id" = $1 LIMIT 1  [["id", 2]]
PG::UndefinedTable: ERROR:  missing FROM-clause entry for table "projects"
LINE 1: ...d" = 'f' AND "ticket_lists"."client_id" = $1 AND (projects.i...
                                                             ^
: SELECT "ticket_lists".* FROM "ticket_lists"  WHERE "ticket_lists"."is_destroyed" = 'f' AND "ticket_lists"."client_id" = …

我有四个模型Tehsil，Ilr，Patwar和Villages。他们的联系是

Tehsil-> 1：m-> Ilr-> 1：m-> Patwar-> 1：m->村庄

我想对所有四个模型应用订单。

查询：

var tehsilQuery = {
    include: [{
        model: Ilr,
        as: 'GirdawariKanoongo',
        include: [{
            model: Patwar,
            as: 'GirdawariPatwar',
            include: [{
                model: Villages,
                as: 'GirdawariVillages',
            }]
        }]
    }],
    order: [
        ['tehsil_name', 'ASC'],
        [ {model: Ilr, as: 'GirdawariKanoongo'}, 'kanoongo_name', 'ASC'],
        [ {model: Patwar, as: 'GirdawariPatwar'}, 'patwar_area', 'ASC'],
        [ {model: Villages, as: 'GirdawariVillages'}, 'village_name', 'ASC'],
    ]
};
return Tehsils.findAll(tehsilQuery);

[Error: 'girdawari_patwar' in order / group clause is not valid association]

如果我从中删除Patwar和Villages（经纬度两个模型），则order by工作正常 …

我知道我可以渴望或懒惰地在Laravel中建立关系。我也知道关系对象基本上是伪装的查询对象，并且调用会$user->load('teams')执行该查询并将其添加到$ user对象中（即使我不知道该怎么精确）。

假设我有一位User从评论中获得积分的人。为了在数据库调用中获取用户的分数，我将执行以下操作：

SELECT `user_id`, sum(`points`) AS `total_points` FROM `user_comments` where `id` = ?

如果我想一次全部加载它们，我将运行以下查询：

SELECT `users`.*, sum(`user_comments`.`points`) AS `total_points`
FROM `users` LEFT JOIN `user_points` ON `users`.`id` = `user_comments`.`user_id`
GROUP BY `users`.`id`

我想使用Laravel完成这些任务。我喜欢写这样的事情

User::where('age', '>', 40)->with('total_points')->get();

并像这样访问值

$user->total_points

同时加载用户及其总分。我希望可能会有很多很多记录，并且不想每次有用户时都总是调用数据库联接查询。如果我只需要它们的总分值，我也不想不为每个用户自己加载所有注释。

有没有一种方法可以使用关系生成器在Laravel中完成此任务？