在cakephp的同一表上使用sql联合查找查询 [英] using sql union on same table in cakephp find query

问题描述

假设我有这样的查询：

 （SELECT * FROM user WHERE id< 5 order by id DESC限制1）
 UNION 
（选择*来自用户WHERE ID = 5）
 UNION 
（选择*来自用户WHERE ID> 5限制1）
  

如何将上述查询转换为 CakePHP find（'all'）查询？

谢谢

解决方案

使用Model :: find（'neighbors'）

将其重写为find（'all'）是有问题的，但是 find（'neighbors'）可能符合您的要求；

请参阅文档 find（'neighbors'）

这应该为您提供所需的数据，但是您可能需要稍微采用一些代码才能拥有它工作ng正确地使用返回数组的不同布局；

  $ data = $ this-> User-> find（'neighbors' ，array（'field'=> ‘id’，‘value’=> 5））； 
  

更新

刚刚看到您在CakePHP 1.3，上面发布的链接指向文档的2.x部分。虽然类似，但这是CakePHP 1.3的文档：

http://book.cakephp.org/1.3/en/The-Manual/Developing-with-CakePHP/Models.html#find-neighbors

let's say I have a query like this:

(SELECT * FROM user WHERE id < 5 order by id DESC LIMIT 1)
UNION
(SELECT * FROM user WHERE id = 5)
UNION
(SELECT * FROM user WHERE id > 5  LIMIT 1)

How can I translate the above query into a CakePHP find('all') query?

Thank you

解决方案

Using Model::find('neighbors')

Rewriting this to a find('all') will be problematic, however find('neighbors') may fit your requirements;

See the documentation find('neighbors')

This should give you the data you need, but you will probably have to adopt your code a bit to have it working correctly with the different 'layout' of the returned array;

 $data = $this->User->find('neighbors', array('field' => 'id', 'value' => 5));

update

Just saw you're on CakePHP 1.3, the link posted above points to the 2.x part of the documentation. Although similar, this is the documentation for CakePHP 1.3:

http://book.cakephp.org/1.3/en/The-Manual/Developing-with-CakePHP/Models.html#find-neighbors

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