X 市建了一个新的体育馆,每日人流量信息被记录在这三列信息中:序号 (id)、日期 (visit_date)、 人流量 (people)。
请编写一个查询语句,找出人流量的高峰期。高峰期时,至少连续三行记录中的人流量不少于100。
例如,表 stadium:
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
对于上面的示例数据,输出为:
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
提示:
每天只有一行记录,日期随着 id 的增加而增加。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/human-traffic-of-stadium
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
select stadium.*, id - cast(row_number() over(partition by people >= 100) as signed) rnk
from stadium
where people >= 100
{"headers": ["id", "visit_date", "people", "rnk"],
"values": [
[2, "2017-01-02", 109, 1],
[3, "2017-01-03", 150, 1],
[5, "2017-01-05", 145, 2],
[6, "2017-01-06", 1455, 2],
[7, "2017-01-07", 199, 2],
[8, "2017-01-08", 188, 2]]}
select id, visit_date, people,
count(*) over(partition by rnk) cnt
from
(
select stadium.*,
id - cast(row_number() over(partition by people >= 100) as signed) rnk
from stadium
where people >= 100
) t
{"headers": ["id", "visit_date", "people", "cnt"],
"values": [
[2, "2017-01-02", 109, 2],
[3, "2017-01-03", 150, 2],
[5, "2017-01-05", 145, 4],
[6, "2017-01-06", 1455, 4],
[7, "2017-01-07", 199, 4],
[8, "2017-01-08", 188, 4]]}
# Write your MySQL query statement below
select id, visit_date, people
from
(
select id, visit_date, people,
count(*) over(partition by rnk) cnt
from
(
select stadium.*,
id - cast(row_number() over(partition by people >= 100) as signed) rnk
from stadium
where people >= 100
) t
) t
where cnt >= 3
或者 3表连接
# Write your MySQL query statement below
select distinct a.*
from stadium a, stadium b, stadium c
where
(
(b.id = a.id+1 and c.id = b.id+1) or
(c.id = b.id+1 and a.id = c.id+1) or
(a.id = c.id+1 and b.id = a.id+1)
)
and a.people>=100
and b.people>=100
and c.people>=100
order by a.id
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