mysql面试题(1)

mysql训练逻辑题(1)

首先看一下题目的是什么?

这个题目其实不难理解,看下这个题心里有个初步的思路通过concat将列连接起来,再通过group_concat将列基于用户号进行聚合

首先导入数据

create table b(
userid int,
cj varchar(50),
ft varchar(50)
);

insert into b values(1,'1001','1400');
insert into b values(2,'1002','1401');
insert into b values(1,'1002','1402');
insert into b values(1,'1001','1402');
insert into b values(2,'1003','1403');
insert into b values(2,'1004','1404');
insert into b values(3,'1003','1400');

简单看一下数据

场景重复的话,选择最早的时间

select userid,cj,min(ft) ft from b group by userid,cj;

  

你认真查看结果其实就是要每个userid的前两条记录,这让我们想起了排名函数

select concat(userid,'-',group_concat(cj separator'-'))as ppp from
(
select userid,cj,rank() over(partition by userid order by cj)as pm from
(
select userid,cj,min(ft) ft from b group by userid,cj order by userid,cj
)t
)t1
where t1.pm<=2 group by userid;
原文地址:https://www.cnblogs.com/shiji7/p/13674888.html