现有一个数据表，需要获取同一个用户，number 累计达到大于等于 10 的那一条数据，也就是第 6 行和第 10 行。
大佬们这个 sql 怎么写。
本贴最后更新于 1799 天前，其中的信息可能已经时移世异
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someone48938 • • 1 • 1 评论 via macOS
我在本地模拟了一下，借助窗口函数可以轻松解决。（注意，MySQL 8.0 以后的版本才支持这一特性）
select distinct r.user_id, first_value(r.number) over (partition by r.user_id) as number from ( select c.user_id, c.number, sum(c.number) over (partition by c.user_id rows unbounded preceding) AS f from ( select 1 as user_id, 1 as number union all select 1 as user_id, 2 as number union all select 1 as user_id, 2 as number union all select 1 as user_id, 4 as number union all select 1 as user_id, 7 as number union all select 1 as user_id, 2 as number union all select 2 as user_id, 1 as number union all select 2 as user_id, 8 as number union all select 2 as user_id, 3 as number union all select 2 as user_id, 6 as number ) c group by c.user_id, c.number ) r where r.f >= 10 group by r.user_id, r.number;🤔 group by c.user_id, c.number 为什么 c.numberPeterChu • -
PeterChu •
id user_id number 1 1 1 2 1 2 3 1 2 4 1 4 5 1 7 6 1 2 7 2 1 8 2 3 9 2 8 10 2 6 SELECT * FROM ( SELECT c.id, c.user_id, c.number, sum( c.number ) over ( PARTITION BY c.user_id rows unbounded preceding ) AS f FROM user001 c ) r WHERE r.f >= 10 GROUP BY r.user_id ORDER BY r.id;结果：
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yang17762622 •作者
这个项目是一个外包，在之前的开发基础上加一下功能，然后他们那边有个 review 代码的，然后每次提交他都会讲代码风格必须不一致，然后又把我们做的功能又改一遍
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