Question

我有以下关联

用户拥有多个域名
域名有很多访问者
域名包含多个网页浏览量

以下是查询

    SELECT
    pv.name,pv.domain_id,pv.counter as page_visits, v.ip, v.counter as visits
FROM (
    SELECT  `domains`.`name` ,  `pageviews`.`domain_id` , COUNT(  `pageviews`.`domain_id` ) as counter
    FROM  `pageviews` 
    LEFT JOIN  `domains` ON (  `domains`.`id` =  `pageviews`.`domain_id` 
    AND  `domains`.`user_id` =  '129' ) 
    WHERE  `pageviews`.`user_id` =  '129'
    GROUP BY  `pageviews`.`domain_id` 
    ORDER BY COUNT(  `pageviews`.`domain_id` ) DESC 
) as pv
LEFT JOIN (
    SELECT  `domains`.`name` ,  `visitors`.`ip` ,  `visitors`.`domain_id` , COUNT(  `visitors`.`domain_id` ) as counter
    FROM  `visitors` 
    LEFT JOIN  `domains` ON (  `domains`.`id` =  `visitors`.`domain_id` 
    AND  `domains`.`user_id` =  '129' ) 
    WHERE  `visitors`.`user_id` =  '129'
    GROUP BY  `visitors`.`domain_id` 
    ORDER BY COUNT(  `visitors`.`domain_id` ) DESC
) as v
    ON pv.domain_id = v.domain_id

我想转换为cakephp查找paginate的查询，因为cakephp不允许$ this-＆gt; paginate（“mysql query here”）;

Answer 1

我建议您使用此工具。

http://dogmatic69.com/sql-to-cakephp-find-converter

这将为您创建此查询的自定义查找方法，允许您在模型中实现它。

Answer 2

您只需执行

查询即可

$this->ModelName->query('your_sql_query');

这将为上面的复杂查询工作

将简单查询转换为cakephp查找控制器中的paginate查询

2 个答案: