MySQL体育馆的人流量

Posted 2021-09-06 willem_chen

mysql体育馆的人流量

• SQL架构
• 题目描述
• 题解
• 答A
• 答B
• 答C

SQL架构

Create table If Not Exists stadium (id int, visit_date DATE NULL, people int);

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');

mysql> SELECT * FROM `stadium` LIMIT 0, 1000;
+------+------------+--------+
| 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 | 2017-01-09 |    188 |
+------+------------+--------+
8 rows in set (0.00 sec)

题目描述

表：Stadium

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| visit_date    | date    |
| people        | int     |
+---------------+---------+

visit_date 是表的主键

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)

每天只有一行记录，日期随着 id 的增加而增加

编写一个 SQL 查询找出，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。

查询结果格式如下所示。

Stadium table:

+------+------------+-----------+
| 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    | 2017-01-09 | 188       |
+------+------------+-----------+

Result table:

+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-09 | 188       |
+------+------------+-----------+

id 为 5、6、7、8 的四行 id 连续，并且每行都有 >= 100 的人数记录。

请注意，即使第 7 行和第 8 行的 visit_date 不是连续的，输出也应当包含第 8 行，因为我们只需要考虑 id 连续的记录。

不输出 id 为 2 和 3 的行，因为至少需要三条 id 连续的记录。

题解

答A

解题思路

先用一个变量cnt记录连续不少于100人的情况，即当前一天及今天都超过99人时，cnt加1。

select id,visit_date,people,(@cnt:=IF(people>99,@cnt+1,0)) as cnt from stadium;
+------+------------+--------+------+
| id   | visit_date | people | cnt  |
+------+------------+--------+------+
|    1 | 2017-01-01 |     10 |    0 |
|    2 | 2017-01-02 |    109 |    1 |
|    3 | 2017-01-03 |    150 |    2 |
|    4 | 2017-01-04 |     99 |    0 |
|    5 | 2017-01-05 |    145 |    1 |
|    6 | 2017-01-06 |   1455 |    2 |
|    7 | 2017-01-07 |    199 |    3 |
|    8 | 2017-01-09 |    188 |    4 |
+------+------------+--------+------+
8 rows in set (0.00 sec)

此时，我们只要取出cnt大于 2 的数据及前 cnt-1 天的数据，由于id是连续的，可以取出当前id及前cnt-1个id的数据，存在超过3天连续的情况，对取出的结果去重。

select  distinct s.* from stadium s,
(
	select id,visit_date,people,(@cnt:=IF(people>99,@cnt+1,0)) as cnt from stadium
) c where c.cnt>2 and s.id between c.id-c.cnt+1 and c.id;

+------+------------+--------+
| id   | visit_date | people |
+------+------------+--------+
|    5 | 2017-01-05 |    145 |
|    6 | 2017-01-06 |   1455 |
|    7 | 2017-01-07 |    199 |
|    8 | 2017-01-09 |    188 |
+------+------------+--------+
4 rows in set (0.00 sec)

答B

SELECT distinct a.*
FROM stadium as a,stadium as b,stadium as c
where ((a.id = b.id-1 and b.id+1 = c.id) or
       (a.id-1 = b.id and a.id+1 = c.id) or
       (a.id-1 = c.id and c.id-1 = b.id))
  and (a.people>=100 and b.people>=100 and c.people>=100)
order by a.id;

+------+------------+--------+
| id   | visit_date | people |
+------+------------+--------+
|    5 | 2017-01-05 |    145 |
|    6 | 2017-01-06 |   1455 |
|    7 | 2017-01-07 |    199 |
|    8 | 2017-01-09 |    188 |
+------+------------+--------+
4 rows in set (0.00 sec)

答C

select distinct t1.*
from stadium t1, stadium t2, stadium t3
where t1.people >= 100 and t2.people >= 100 and t3.people >= 100
and
(
	(t1.id - t2.id = 1 and t1.id - t3.id = 2 and t2.id - t3.id =1)  -- t1, t2, t3
    or
    (t2.id - t1.id = 1 and t2.id - t3.id = 2 and t1.id - t3.id =1) -- t2, t1, t3
    or
    (t3.id - t2.id = 1 and t2.id - t1.id =1 and t3.id - t1.id = 2) -- t3, t2, t1
)
order by t1.id;

+------+------------+--------+
| id   | visit_date | people |
+------+------------+--------+
|    5 | 2017-01-05 |    145 |
|    6 | 2017-01-06 |   1455 |
|    7 | 2017-01-07 |    199 |
|    8 | 2017-01-09 |    188 |
+------+------------+--------+
4 rows in set (0.00 sec)