所以我有这个查询:
select u.*, count(mi.media_id) as media_count
from user as u
left join media_items as mi on mi.user_id = u.user_id
where u.is_enabled = 1 and
group by u.user_type
having media_count > 0;
我正在尝试将其转换为 Zend_Db_Select。到目前为止,除了媒体的数量,我什么都有。这是我尝试过的:
$select->from(array('u' => 'user'), array('*', new Zend_Db_Expr('COUNT(mi.media_id) AS media_count'))
->joinLeft('mi' => 'media_items'), 'mi.user_id = u.user_id')
->where('u.is_enabled = 1')
->group('u.user_type')
->having('media_count > 0');
这让我得到了错误:
"Mysqli prepare error: Unknown column 'media_count' in 'having clause'"
如何以 Zend 方式创建此语句?
另外,我输出了这个创建的查询并在 MySQLWorkBench 上运行它,它运行得很好。
编辑我刚刚尝试过:
->columns('media_count' => new Zend_Db_Expr('COUNT(mi.media_id)'))
同样的错误。