MySQL体育馆的人流量

「爱情、让人受尽委屈。」 2023-10-06 18:59 25阅读 0赞

#### 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)