让我们假设你有如下表:
CREATE TABLE `videos` (
`id` char(36) NOT NULL,
`name` varchar(50) NOT NULL,
PRIMARY KEY (`id`)
)
CREATE TABLE `votes` (
`id` char(36) NOT NULL,
`video_id` char(36) NOT NULL,
`user_id` char(36) NOT NULL,
PRIMARY KEY (`id`)
);
您试图编写的SQL如下所示:
SELECT
count(id),
video_id
FROM votes
GROUP BY video_id
ORDER BY count(id) DESC
LIMIT 25;
问题是,它不会返回视频名称。这可以通过JOIN或使用蛋糕递归功能轻松修复。鉴于上述情况,我会写:
$this->Vote->recursive = 1;
$vote_info = $this->Vote->find('all',
array(
'fields' => array('count(Vote.id)','Video.name'),
'group' => array('video_id'),
'order' => array('count(Vote.id) DESC'),
'limit' => 25,
)
);
当你把数据传回,你也将获得视频信息(由于被设置为1的递归选项),只要在模型中你的人际关系设置正确。
例如,数据看起来是这样的:
Array
(
[0] => Array
(
[0] => Array
(
[count(`Vote`.`id`)] => 3
)
[Video] => Array
(
[name] => Dumb and Dumber
)
)
[1] => Array
(
[0] => Array
(
[count(`Vote`.`id`)] => 1
)
[Video] => Array
(
[name] => Lord of the Rings
)
)
[2] => Array
(
[0] => Array
(
[count(`Vote`.`id`)] => 1
)
[Video] => Array
(
[name] => Just Do It
)
)
)
我想你可以从这里看到如何访问您的视图中的数据。
Hoppy Coding!
UPDATE: 如果你只需要视频ID,你可以这样做:
$this->Vote->recursive = 0;
$votes = $this->Vote->find('all',
array(
'fields' => array('count(Vote.id) as count', 'Vote.video_id'),
'order' => array('count(Vote.id) DESC'),
'group' => array('Vote.video_id'),
'limit' => 25,
)
);
这将返回:
Array
(
[0] => Array
(
[0] => Array
(
[count] => 3
)
[Vote] => Array
(
[video_id] => 4dac4de7-4f80-48c1-8a02-83b4dfa458e5
)
)
[1] => Array
(
[0] => Array
(
[count] => 1
)
[Vote] => Array
(
[video_id] => 4dac4ddc-712c-4795-8d9a-83b4dfa458e5
)
)
[2] => Array
(
[0] => Array
(
[count] => 1
)
[Vote] => Array
(
[video_id] => 4dac4df0-382c-4bb5-a5ee-83b4dfa458e5
)
)
)
你基本上需要在票COUNT(*)表? – Dunhamzzz 2011-04-18 12:53:32
你可以发布表格之间的关系吗?即我认为一个视频有很多票和一票投票属于一个视频。 – Belinda 2011-04-18 13:04:39
@Belinda绝对。投票属于视频,视频hasMany Vote。 – 2011-04-18 13:50:20