CakePHP：几个连接,belongsTo和hasMany关系在两个查询中完成

是)我有的

我现在有以下设置：

邮寄hasMany附件

它工作正常,页面生成2个查询：

SELECT *,`Post`.`id` 
FROM `posts` AS `Post` 
WHERE 1 = 1 
ORDER BY `Post`.`created` DESC

SELECT 
    `Attachment`.`id`,`Attachment`.`post_id`,`Attachment`.`created` 
FROM 
    `attachments` AS `Attachment` 
WHERE 
    `Attachment`.`post_id` IN (1,2,3,...,n)

我想要的是

我想将关系扩展如下：

邮政有很多附件;每个附件属于所有类型

我不太热衷于让CakePHP遵循它.
基本上,我需要的是：

SELECT *,`Attachment`.`created`,`Type`.`title`,`Type`.`icon` 
FROM 
    `attachments` AS `Attachment` 
LEFT JOIN 
    `types` AS `Type` 
    ON (`Attachment`.`type_id`=`Type`.`id`) 
WHERE 
    `Attachment`.`post_id` IN (1,n)

请注意添加了LEFT JOIN类型.

所以我在第二个查询中得到了相应的类型数据.我知道我可以循环获取数据或使用 – > query()调用,但我希望它尽可能有效和灵活.

问题

我尝试了Containable,Model Unbinding trick(and this one)但没有成功.我尝试了不同的选项组合,我相信我甚至删除了连接.这就是我的PostsController现在的样子.

class PostsController extends AppController {
    public function index() {

        $this->Post->unbindModel(array('hasMany' => array('Attachment')));
        $this->Post->Attachment->unbindModel(array('belongsTo' => array('Type')));

        $this->Post->bindModel(array(
            'hasMany' => array(
                'Attachment' => array(
                    'className'  => 'Attachment',// when uncommented,throws the "Unknown column Post.id" SQLSTATE error
                //  'conditions' => array('Post.id' => 'Attachment.post_id'),'foreignKey' => false,),));
        $this->Post->Attachment->bindModel(array(
            'belongsTo' => array(
                'Filetype' => array(
                    'className'  => 'Filetype',//  'conditions' => array('Type.id' => 'Attachment.type_id'),));
        $all = $this->Post->find('all',array(
            'joins' => array(
                array(
                    'table' => 'users','prefix' => '','alias' => 'User','type' => 'INNER','conditions' => array(
                        'User.id = Post.user_id',)
                ),'contain' => array('Attachment','Type'),'conditions' => array(),'fields' => array('*'),'order' => 'Post.created ASC'
        ));
        var_dump($all);exit;
    }
}

但它只是在循环中每次迭代运行一个额外的查询并获取所有附件：

SELECT `Attachment`.`id`,... 
FROM `attachments` AS `Attachment` 
WHERE 1 = 1

当我取消注释这个关联的条件时,它会抛出SQLSTATE“Column Post.id not found error” – 我想因为这里没有Post表加入.

我需要帮助设置它.

请帮忙！谢谢

UPDATE

我按如下方式更改了控制器.请注意,没有bindModel / unbindModel代码,关系是在models类中设置的(在这种情况下是正确的吗？).

class PostsController extends AppController {
    public function index() {
        $options = array(
            'contain' => array(
                'Post','Type'
            ),'order' => 'Post.created DESC','conditions' => array(
            //  'Post.title LIKE' => 'my post'
            )
        );

    //  The following throws "Fatal error: Call to a member function find() on a non-object"
    //  $posts = $this->Attachment->find('all',$options); 

    //  So I had to use $this->Post->Attachment instead of $this->Attachment
        $posts = $this->Post->Attachment->find('all',$options);
        $this->set(compact('posts'));
    }   
}

这是附件模型：

class Attachment extends AppModel {
    public $belongsTo = array(
        'Type' => array(
            'className' => 'Type','foreignKey' => 'type_id','Post' => array(
            'className' => 'Post','foreignKey' => 'post_id',);
}

上面的代码运行此查询：

SELECT 
    `Attachment`.`id`,`Attachment`.`type_id`,`Type`.`id`,`Post`.`id`,`Post`.`text`,`Post`.`created` 
FROM 
    `attachments` AS `Attachment` 
    LEFT JOIN `types` AS `Type` ON (`Attachment`.`type_id` = `Type`.`id`) 
    LEFT JOIN `posts` AS `Post` ON (`Attachment`.`post_id` = `Post`.`id`) 
WHERE 
    1 = 1 
ORDER BY 
    `Post`.`created` ASC

一切都是关于这里的附件.我的意思是帖子已加入附件,因此如果帖子没有附件,则不会返回.这可能是因为调用是Attachment-> find()所以它来自附件的观点.我猜它应该是：

// ...
FROM 
    `posts` AS `Post`
    LEFT JOIN `attachments` AS `Attachment`  ON (`Attachment`.`post_id` = `Post`.`id`) 
    LEFT JOIN `types` AS `Type` ON (`Attachment`.`type_id` = `Type`.`id`) 
// ...

但它不会起作用,是吗？您会看到有帖子,附件和类型,但它们确实具有不同的关系类型.最初,我发布了CakePHP运行的两个单独的查询 – 必须有这样的理由.

UPDATE2

我仍然相信在初始设置中将第二个查询更改为附件模型(请参阅“我想要的内容”部分).所以我将获得附件类型以及附件本身.我的意思是在那种情况下LEFT JOINING类型表到附件不会打破任何数据库关系逻辑,是吗？
我只想确保一个复杂但单一的find()调用无法做到这一点.