Cakephp检索HABTM的条件

Question

我正在使用cakephp,并希望显示属于类别'X'的所有提交,我有4个具有HABTM关系的表.

用户 - >(hasMany) - >提交< - >(hasAndBelongsToMany)< - >类别

但我想使用$ this-> paginate()这样做,并且对于每个提交,我想显示发布提交的用户.

用户表

Id   |      Name         
-----+-------------------
1    |   User 1      
2    |   User 2     

提交表

Id   |      Name         |   User_id
-----+-------------------+--------------
1    |   Submission 1    |      1    
2    |   Submission 2    |      2    

分类表

Id   |      Name 
-----+-------------------
1    |   Category 1        
2    |   Category 2        

SubmissionCategory表

Id   |   Submission_id   |   Category_id 
-----+-------------------+-------------------
1    |         1         |        1     
2    |         1         |        2  
3    |         2         |        1      

我真的很难创建一个可以做到这一点的分页,我开始认为它不可能,除非我错过了什么.

如果我没有使用cakephp这是我想要的查询

SELECT 
     * 
FROM 
     submissions_categories, 
     submissions, 
     users
WHERE 
     submissions_categories.category_id = 8 
          AND 
     submissions_categories.submission_id = submissions.id 
          AND 
     submissions.user_id = users.id

Answer 1

我发现CakePHP中的HABTM关系非常难以处理.您将需要使用$ paginate变量手动设置连接.然后,您可以将可选的条件数组传递给paginate()函数.例:

<?php
class SubmissionsController extends AppController {

 var $name = 'Submissions';
 var $helpers = array('Html', 'Form');
 var $paginate = array('joins' => array(
     array( 
               'table' => 'submissions_categories', 
               'alias' => 'SubmissionsCategory', 
               'type' => 'inner',  
               'conditions'=> array('SubmissionsCategory.submission_id = Submission.id') 
           ), 
           array( 
               'table' => 'categories', 
               'alias' => 'Category', 
               'type' => 'inner',  
               'conditions'=> array( 
                   'Category.id = SubmissionsCategory.category_id'
               ) 
           )));

 function index() {
  $this->Submission->recursion = 1;
  $this->set('submissions', $this->paginate(array('Category.id'=>1)));
 }
}

?>