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嗨,我有麻煩與此查詢如何建立嵌套選擇在Zend的分貝
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`
我如何轉換這對Zend_Db_Select對象的對象?
釷的主要問題是 這部分
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`,
可能重複http://stackoverflow.com/questions/1718888/zend-db-table-subquery ) – Ankur