MySQL练手题--体育馆的人流量(困难)

一、准备工作

Create table If Not Exists Stadium (id int, visit_date DATE NULL, people int);
Truncate table Stadium;
insert into Stadium (id, visit_date, people) values ('1', '2017-01-01', '10');
insert into Stadium (id, visit_date, people) values ('2', '2017-01-02', '109');
insert into Stadium (id, visit_date, people) values ('3', '2017-01-03', '150');
insert into Stadium (id, visit_date, people) values ('4', '2017-01-04', '99');
insert into Stadium (id, visit_date, people) values ('5', '2017-01-05', '145');
insert into Stadium (id, visit_date, people) values ('6', '2017-01-06', '1455');
insert into Stadium (id, visit_date, people) values ('7', '2017-01-07', '199');
insert into Stadium (id, visit_date, people) values ('8', '2017-01-09', '188');

# 编写解决方案找出每行的人数大于或等于 100 且 id 连续的三行或更多行记录。
# 返回按 visit_date 升序排列 的结果表。

输入：

输出：

二、分析

 三、实现

with t as (select * from stadium where people >=100    -- 每行人数大于等于100
), t1 as (select*,row_number() over (order by id) rn     -- 对大于等于100的id排序from t
), t2 as (select*,id-rn as 差值                          -- 对id和排序求差，差值相等的为连续from t1
),t3 as (select 差值,count(差值) cn from t2 group by 差值    -- 求差值相等且大于等于3的记录
)
select id,visit_date,people from t2 ,t3 where t2.差值=t3.差值 and cn>=3 order by id; 

四、总结

此题稍微微有一点难度，如果看不明白可以对题目进行分解分析，分析出一步之后再接下一步来查询，题目已经要求人数大于等于100且id连续三行，就可以先筛选出大于等于100 的id 然后排序求差 差值相等的均为id连续的 ，最后筛选差值总数大于等于3 的，便出来了最后结果；