hasMany条件显示不需要的记录cakephp 1.3
发布时间：2020-12-13 18:10:58 所属栏目：PHP教程 来源：网络整理
导读：我有问题表,存储了这么多问题.与question_topics相关的问题因此创建与问题有很多关系.现在它看起来像这样： $this-Question-bindModel( array( 'hasMany' = array( 'QuestionTopic'=array( 'className' = 'QuestionTopic','foreignKey' = 'question_id','dep
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我有问题表,存储了这么多问题.与question_topics相关的问题因此创建与问题有很多关系.现在它看起来像这样：
$this->Question->bindModel(
array(
'hasMany' => array(
'QuestionTopic'=>array(
'className' => 'QuestionTopic','foreignKey' => 'question_id','dependent' => true,'conditions' => array('QuestionTopic.areas_id' => 165),'type' => 'left'
)
)
)
);
print_r($this->Question->find('all'));
die;
当我看到结果时,它看起来像这样 Array
(
[0] => Array
(
[Question] => Array
(
[id] => 89
[user_id] => 1
[question_group_id] => 0
[question] => jQuery function here
[target_id] => 1
[type] => 1
[order] => 1
[description] => additional info here
[privacy_friend_id] =>
[channel_id] => 1
[status] => 0
[location] => Chandigarh,India
[regions] => 307
[select_country] => 381
[select_states] => 515
[created] => 2014-04-15 06:59:44
[modified] => 2014-04-15 06:59:44
)
[QuestionTopic] => Array
(
[0] => Array
(
[id] => 167
[areas_id] => 165
[question_id] => 89
)
)
)
[1] => Array
(
[Question] => Array
(
[id] => 90
[user_id] => 1
[question_group_id] => 0
[question] => Art
[target_id] => 2
[type] => 1
[order] => 1
[description] => addional infomation here
[privacy_friend_id] =>
[channel_id] => 1
[status] => 0
[location] => Chandigarh,India
[regions] => 307
[select_country] => 381
[select_states] => 515
[created] => 2014-04-15 07:52:17
[modified] => 2014-04-15 07:52:17
)
[QuestionTopic] => Array
(
)
)
)
我想要的第一个记录只有问题主题ID 167而不是第二个.我怎样才能做到这一点.
不要使用hasMany,使用有这样的一个
$this->Question->bindModel(
array(
'hasOne' => array(
'QuestionTopic'=>array(
'className' => 'QuestionTopic','type' => 'left'
)
)
)
);
print_r($this->Question->find('all',array('conditions'=>array('QuestionTopic.areas_id' => array('165')))));
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