Question

我试图通过卡纳克民阵(mouth!)的关系,把我的头接在两头上。

I m working on a picture matching website.

Pictures are associated to other pictures via a match (the join model).
Each match has two pictures and stores the current rating and the total number of votes.
When viewing a picture, all of its related images from either direction should be available (via its matches).

我先通过与加入模式的关系来界定 has。

表格中的图象是:

id | picture_id | partner_id | rating | total_votes

我加入示范协会就是这样:

class PictureMatch extends AppModel {

...

    public $belongsTo = array(
         Picture  => array(
             className  =>  Picture ,
             foreignKey  =>  picture_id ,
             conditions  =>   ,
             fields  =>   ,
             order  =>   
        ),
         Partner  => array(
             className  =>  Picture ,
             foreignKey  =>  partner_id ,
             conditions  =>   ,
             fields  =>   ,
             order  =>   
        )
    );
}

Each picture needs to be able to access its related pictures from either direction, but this is where my grasp is slipping. It looks like I need to save both sides of the relationship but this destroys the extra data stored in the join model - with two db entries, voting could vary depending on the direction.

Can anyone shed any light on the best way to do this in CakePHP? I m rather confused.
Is it possible to create the inverse relationships on the fly?

Answer 1

你可以就“爱生模式”提出现实:Bbind Model()非常有用,这预示着你要控制逆向关系,或者说是你希望的飞跃。

http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html

此外,你还利用可维持的行为,可以建立无限链,使您的相关日期提前结束。

包含(照片:PictureMatch.Partner.PictureMatch.Picture....)

基本上,只要每一链条都与下一个链条有某种关系,就可以通过你的所有模式来解释其更简单的例子(请在其中不考虑逻辑)

Circle belongsTo Square Square belongsTo Triangle

So Triangle与环圈(直接)没有关系,但广场在环圈之间是友好的。

Circle->find( all , array( ... , contain => array( Square.Triangle ));

or to have more fun lets get circle by circle with loop around

Circle->find( all , array( ... , contain => array( Square.Trinagle.Square.Circle ));

因此,当然,这些例子是毫无用处的,没有任何方案逻辑,但我希望你理解,你可以 lo住多少关系。

Answer 2

I am not sure if this is the best solution but if you did this:

public $belongsTo = array(
     Picture1  => array(
         className  =>  Picture ,
         foreignKey  =>  picture_id ,
    ),
     Picture2  => array(
         className  =>  Picture ,
         foreignKey  =>  picture_id ,
    ),
     Partner  => array(
         className  =>  Partner ,
         foreignKey  =>  partner_id ,
    ),
);

然后,在你进行查询时,你才查询<代码>($this-gt;data[图片] = var ́ ́ ́ «this->data[] 页: 1 页: 1

Answer 3

我假定这一点已经放弃,但很容易再解决——它必须与阶段有关。

销毁合并模型中储存的额外数据

这意味着你们的储蓄正在运行一个<代码>deleteAll,并插入关于匹配的记录......相反,你需要找到并更新这一记录。

可以用几种方式做到这一点,但最容易的是,在你发出呼吁之前,看着这种呼吁,并将主要关键内容列入相应的记录数据。基本上不将其作为现金自动取款机予以节省,而且只有在你试图找到现有对应记录的主要钥匙(id)并更新数据以予以节省时,才将其作为 has予以节省。

友情链接