原则 - 子查询 [英] Doctrine - subquery in from

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问题描述

我在MySQL中有一个查询:

  SELECT * FROM(
SELECT COUNT(*)AS计数,t.name AS名称
FROM标签t
INNER JOIN video_has_tag v USING(idTag)
GROUP BY v.idTag
ORDER BY count DESC
LIMIT 10
)作为标签ORDER BY名称

我想写这个教义。我该怎么办?我写道:

  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();

但我不能把它放在从按名称排序。

解决方案

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');


I have a query in MySQL:

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

and I want to write this in doctrine. How I can do that? I wrote:

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();

But I can't put it in "from" to sort by name.

解决方案

This is a answer:

$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');

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