sna*_*hot 11 php mysql doctrine
我在MySQL中有一个查询:
SELECT * FROM (
SELECT COUNT(*) AS count, t.name AS name
FROM tag t
INNER JOIN video_has_tag v USING (idTag)
GROUP BY v.idTag
ORDER BY count DESC
LIMIT 10
) as tags ORDER BY name
Run Code Online (Sandbox Code Playgroud)
我想在学说中写这个.我怎么能这样做?我写:
Doctrine_Query::create()
->select('COUNT(t.idtag) as count, t.name')
->from('Tag t')
->innerJoin('t.VideoHasTag v')
->groupBy('v.idTag')
->orderBy('count DESC, t.name')
->limit(30)
->execute();
Run Code Online (Sandbox Code Playgroud)
但我不能把它放在"从"按名称排序.
这是一个答案:
$q = new Doctrine_RawSql();
$q->addComponent('t', 'Tag')
->select('{t.name}, {t.count}')
->from('(SELECT COUNT(*) as count, t.name,t.idtag
FROM Tag t
INNER JOIN Video_Has_Tag v USING(idTag)
GROUP BY v.idTag
ORDER BY count DESC
LIMIT 50) t')
->orderBy('name');
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
13660 次 |
| 最近记录: |