問(wèn)題描述

gift_id 有100多種

gift_id,user_id 建立了索引

只需要找擁有某一gift_id的用戶(hù)的查詢(xún)?nèi)缦?，而且速度非常快select * from user_gift where gift_id = 1004302 group by user_id

怎么快速找到 同時(shí)擁有 gift_id 為1004302和1004004的用戶(hù)user_id呢 ？

問(wèn)題解答

查 gift_id 為1004302的用戶(hù)存list1 查 gift_id 為1004004的用戶(hù)存list2 兩個(gè)list取交集

select t.user_id, count(1) as c from table as twhere t.gift_id in(1004302, 1004004)group by t.user_idhaving count(1)>1

效率問(wèn)題, 沒(méi)數(shù)據(jù), 也測(cè)試不了

如果, (user_id, gift_id) 是有可能重復(fù)的, 那在計(jì)算同時(shí)擁有之前還得將 (user_id, gift_id) 去重.

select t.user_id, count(1) as c from (select user_id, gift_id from table group by user_id, gift_id) as twhere t.gift_id in(1004302, 1004004)group by t.user_idhaving count(1)>1