本文介绍了如何在 zend db 上构建嵌套选择的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这个查询有问题
SELECT * FROM(
SELECT `b`.*,`owner`.firstname,`owner`.lastname,`owner`.email,
(
SELECT COUNT(`ps`.profile_id) FROM `profile` AS `ps`
LEFT JOIN `xref_store_profile_brand` AS `xbp` ON `xbp`.profile_id = `ps`.profile_id
WHERE `xbp`.brand_id = b.brand_id AND ps.role = 'salesrep' AND `xbp`.store_id IS NULL
) AS `salesrepTotal`,
(
SELECT GROUP_CONCAT(`ms`.firstname) FROM `profile` AS `ps`
LEFT JOIN `xref_store_profile_brand` AS `xbp` ON `xbp`.profile_id = `ps`.profile_id
LEFT JOIN `member` AS `ms`ON `ms`.member_id = `ps`.member_id
WHERE `xbp`.brand_id = `b`.brand_id AND ps.role = 'salesrep' AND `xbp`.store_id IS NULL
) AS `salesrep`,
(
SELECT COUNT(`s`.store_id) FROM `store` AS `s`
LEFT JOIN `xref_store_profile_brand` AS `xbs` ON `xbs`.store_id = `s`.store_id
WHERE `xbs`.brand_id = `b`.brand_id AND `xbs`.brand_id IS NOT NULL
) AS `storeTotal`,
(
SELECT GROUP_CONCAT(`s`.name) FROM `store` AS `s`
LEFT JOIN `xref_store_profile_brand` AS `xbs` ON `xbs`.store_id = `s`.store_id
WHERE `xbs`.brand_id = `b`.brand_id AND `xbs`.brand_id IS NOT NULL
) AS `store`
FROM `brand` AS `b`
LEFT JOIN
(
SELECT `m`.firstname,`m`.lastname,`m`.email,`xspb`.brand_id FROM `member` AS `m`
LEFT JOIN `profile` as `p` ON `p`.member_id = `m`.member_id AND `p`.role = 'designer' AND `p`.isPrimary = 1
LEFT JOIN `xref_store_profile_brand` AS `xspb` ON `xspb`.profile_id = `p`.profile_id AND `xspb`.store_id IS NULL
) AS `owner` ON `owner`.brand_id =`b`.brand_id
GROUP BY `b`.brand_id
) AS `final`
如何将其转换为 Zend_Db_Select 对象?
how can i convert this in to Zend_Db_Select object?
主要问题是这部分
SELECT `b`.*,`owner`.firstname,`owner`.lastname,`owner`.email,
(
SELECT COUNT(`ps`.profile_id) FROM `profile` AS `ps`
LEFT JOIN `xref_store_profile_brand` AS `xbp` ON `xbp`.profile_id = `ps`.profile_id
WHERE `xbp`.brand_id = b.brand_id AND ps.role = 'salesrep' AND `xbp`.store_id IS NULL
) AS `salesrepTotal`,
推荐答案
您需要使用 Zend_Db_Expr 对象在您的查询和 选择 AS 的数组结构.
You need to use Zend_Db_Expr objects in your query and array structures for select AS.
以下是您正在寻找的解决方案:
below is the solution you are looking for:
<?php
$db = Zend_Db_Table::getDefaultAdapter();
// inner query
$sqlSalesRepTotal = $db->select()
->from(array('ps' => 'profile'))
->joinLeft(array('xbp' => 'xref_store_profile_brand'), 'xbp.profile_id = ps.profile_id')
->where('xbp.brand_id = b.brand_id')
->where('ps.role = ?', 'salesrep')
->where('xbp.store_id IS NULL');
// main query
$sql = $db->select()
->from(array('b' => 'brand'), array(
// NOTE: have to add parentesis around the expression
'salesrepTotal' => new Zend_Db_Expr("($sqlSalesRepTotal)")
))
->where('....')
->group('brand_id');
// debug
var_dump($db->fetchAll($sql));
这篇关于如何在 zend db 上构建嵌套选择的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!
本站部分内容来源互联网,如果有图片或者内容侵犯您的权益请联系我们删除!