子查询联接解决方案

Question

我有以下查询，当我运行它时给我错误，

SELECT events.id AS id, SUM(tickets_sold.quantity)
FROM (`events`)
JOIN `category` AS cat ON `events`.`category_id` = `cat`.`id` 
JOIN `category` AS sub_cat ON `events`.`subCategoryID` = `sub_cat`.`id` 
JOIN `events_custom_dates` AS events_date ON `events_date`.`event_id` = `events`.`id`
JOIN `my_promos` ON `events`.id = `my_promos`.`event_id`
LEFT JOIN `mycalendar` ON `mycalendar`.`event_id` = `my_promos`.`event_id` 
LEFT JOIN `promo_events_stats` ON `promo_events_stats`.`id` = `events`.`id` 

JOIN `tickets_sold` ON  `my_promos`.`link_code` = `tickets_sold`.`code` 
WHERE `my_promos`.`user_id` = '532' AND DATE(my_promos.date) >= '2013-11-01' AND DATE(my_promos.date) <= '2014-01-22' 
GROUP BY my_promos.event_id

它给我

id   sum(tickets.sold)
some id    `2`
some id    `4`
some id   `14`

但它应该给

some id    `2`
some id    `4`
some id    `7`

当我从上述查询中删除以下行时，它为我提供了正确的数据

JOIN `events_custom_dates` AS events_date ON `events_date`.`event_id` = `events`.`id`

所以请告诉我我现在应该怎么做。event_id是唯一的外键，可以使它与事件表连接。

Answer 1

这是您需要的解决方案

SELECT events.id AS id,A.sold_tickets
FROM (`events`)
JOIN `category` AS cat ON `events`.`category_id` = `cat`.`id` 
JOIN `category` AS sub_cat ON `events`.`subCategoryID` = `sub_cat`.`id` 
JOIN `events_custom_dates` AS events_date ON `events_date`.`event_id` = `events`.`id`
JOIN `my_promos` ON `events`.id = `my_promos`.`event_id`
LEFT JOIN `mycalendar` ON `mycalendar`.`event_id` = `my_promos`.`event_id` 
LEFT JOIN `promo_events_stats` ON `promo_events_stats`.`id` = `events`.`id` 

LEFT JOIN (SELECT my_promos.event_id, SUM(tickets_sold.quantity) AS sold_tickets FROM my_promos
JOIN tickets_sold ON tickets_sold.code = my_promos.link_code
WHERE my_promos.user_id = '532' AND DATE(my_promos.date) >= '2013-11-01' AND DATE(my_promos.date) <= '2014-01-22' 
GROUP BY my_promos.event_id) A ON A.event_id = events.id

WHERE `my_promos`.`user_id` = '532' AND DATE(my_promos.date) >= '2013-11-01' AND DATE(my_promos.date) <= '2014-01-22' 
GROUP BY my_promos.event_id

Answer 2

是否

select * 
from events
join events_custom_dates as events_date on events_date.event_id = events.id;

给出您期望的结果？ 我希望模式可能有问题，但是如果看不到它，那将是一团黑暗的事。