我正试图在分类列表中找到用户的杂货.关联是Category hasMany Item hasMany User through Grocery.我正在使用可包含行为,并没有过滤掉所有其他Grocery.它基本上返回每个项目.
我的控制器功能:
function showlist() {
$categories = $this->Category->find('all', array(
'contain' => array(
'Item' => array(
'Grocery' => array(
'conditions' => array(
'Grocery.user_id =' => $this->Auth->user('id')
)
)
)
)
));
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返回的数组:
Array
(
[0] => Array
(
[Category] => Array
(
[id] => 10
[parent_id] =>
[name] => Dairy
[lft] => 1
[rght] => 6
)
[Item] => Array
(
)
)
[1] => Array
(
[Category] => Array
(
[id] => 11
[parent_id] …Run Code Online (Sandbox Code Playgroud) 我有下列协会
后>伯 - >次级
$results = $this->Post->find('all', array(
'conditions' => array(
'Post.post_id =' => 2,
'Primary.secondary_id !=' => null
),
'contain' => array(
'Primary' => array(
'Secondary' => array(
'conditions' => array('Secondary.short_code =' => 'code')
)
)
)
));
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返回此.
Array
(
[0] => Array
(
[Post] => Array
(
[id] => 2
[created] => 2012-10-29 09:48:29
[modified] => 2012-10-29 09:48:29
)
[Primary] => Array
(
[id] => 3
[secondary_id] => 6
[Secondary] => Array
(
[id] => 6
[short_code] …Run Code Online (Sandbox Code Playgroud) 我有以下型号.
我需要找到6个最新的预告片Industry.每部电影都不需要预告片或可以有多个[0-n].
结果必须包含至少一个预告片的电影数组.
$this->Industry->find('all', array(
'contain' => array(
'Movie' => array(
'Trailer' => array(
'limit' => 1
),
'order' => 'Movie.release DESC',
'limit' => 6
)
),
'order' => 'Industry.name ASC'
));
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