MySQL体育馆的人流量

待我称王封你为后i 2023-10-06 18:46 22阅读 0赞

#### MySQL体育馆的人流量 ####

*  SQL架构
 *  表：Stadium
 *   *  编写一个 SQL 查询以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。
     *  正确答案
 *  题解
 *   *  第一步：查询人流量超过 100 的记录，然后将结果与其自身的临时表连接。
 *  答案2

## SQL架构 ##

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')

![在这里插入图片描述][ff7ee28387934172aa86fdbf81965771.png]

## 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 查询以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。

### 正确答案 ###

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;

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaWd1YW5nMTAy_size_16_color_FFFFFF_t_70]

## 题解 ##

方法：使用 JOIN 和 WHERE 子句【通过】

思路

在表 stadium 中查询人流量超过 100 的记录，将查询结果与其自身的临时表连接，再使用 WHERE 子句获得满足条件的记录。

算法

### 第一步：查询人流量超过 100 的记录，然后将结果与其自身的临时表连接。 ###

select distinct t1.*
    from stadium t1, stadium t2, stadium t3
    where t1.people >= 100 and t2.people >= 100 and t3.people >= 100;

共有 6 天人流量超过 100 人，笛卡尔积 后有 216（666） 条记录。  
前 3 列来自表 t1，中间 3 列来自表 t2，最后 3 列来自表 t3。

mysql> select distinct t1.* from stadium t1, stadium t2, stadium t3 where t1.people >= 100 and t2.people >= 100 and t3.people >= 100;
    +------+------------+--------+
    | id   | visit_date | people |
    +------+------------+--------+
    |    2 | 2017-01-02 |    109 |
    |    3 | 2017-01-03 |    150 |
    |    5 | 2017-01-05 |    145 |
    |    6 | 2017-01-06 |   1455 |
    |    7 | 2017-01-07 |    199 |
    |    8 | 2017-01-09 |    188 |
    +------+------------+--------+
    6 rows in set (0.01 sec)

表 t1，t2 和 t3 相同，需要考虑添加哪些条件能够得到想要的结果。以 t1 为例，它有可能是高峰期的第 1 天，第 2 天，或第 3 天。

t1 是高峰期第 1 天：(t1.id - t2.id = 1 and t1.id - t3.id = 2 and t2.id - t3.id =1) – t1, t2, t3  
t1 是高峰期第 2 天：(t2.id - t1.id = 1 and t2.id - t3.id = 2 and t1.id - t3.id =1) – t2, t1, t3  
t1 是高峰期第 3 天：(t3.id - t2.id = 1 and t2.id - t1.id =1 and t3.id - t1.id = 2) – t3, t2, t1

select 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
    );

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

可以看到查询结果中存在重复的记录，再使用 DISTINCT 去重。

mysql> 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)

## 答案2 ##

select DISTINCT t1.* from stadium t1, stadium t2, stadium t3 
    where 
    t1.people >= 100 and 
    t2.people >= 100 and 
    t3.people >= 100 and ( 
    (t2.id-t1.id=1 and t3.id-t2.id=1 ) or 
    (t1.id-t2.id=1 and t3.id-t1.id=1 ) or 
    (t1.id-t2.id=1 and t2.id-t3.id=1 ) ) 
    ORDER BY t1. visit_date;
    
    +------+------------+--------+
    | 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)

[ff7ee28387934172aa86fdbf81965771.png]: https://img-blog.csdnimg.cn/ff7ee28387934172aa86fdbf81965771.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaWd1YW5nMTAy_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/0d6dfb40da374ba19566f0452bb448d6.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaWd1YW5nMTAy,size_16,color_FFFFFF,t_70