数据库 OVER(PARITITON BY)函数和MySQL排序常用的方法_rank parititon by over

OVER(PARTITION BY)开窗函数

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       |
+------+------------+-----------+
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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       |
+------+------------+-----------+
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解题:

select id,visit_date,people from (
    select id,visit_date,people,count(b)over(partition by b) b2 from (
    	 select *,(id-row_number()over(order by id)) as b 
    	 from stadium where people >= 100
    	 ) t1 
) t2
where b2>=3;
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over(partition by)和group by 作用相同,但是group by 是聚合函数,会影响输出结果,而over(partition by)是开窗函数,会在每一行后面加上分析的结果

比如count(Id)over(partiton by),那么就会在每一行后面加一列数据,就是统计Id的数量值

------------------------------------------分割线--------------------------------------------------

排序

现有这么一个表,根据不同部门对薪水进行排序:

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
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现在需要对各部门每个人的工资进行排序
方法1-row_number()
select *,row_number()over(partition by DepartmentId order by Salary desc)as `Rank` from test;

+----+--------+--------+--------------+------+
| id | Name   | Salary | DepartmentId | Rank |
+----+--------+--------+--------------+------+
|  2 | Jim    |  90000 |            1 |    1 |
|  5 | Max    |  90000 |            1 |    2 |
|  1 | Joe    |  70000 |            1 |    3 |
|  3 | Henery |  80000 |            2 |    1 |
|  4 | Sam    |  60000 |            2 |    2 |
+----+--------+--------+--------------+------+
直接取排序后的行数,排名不重复
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方法2-rank()
select *,rank()over(partition by DepartmentId order by Salary desc)as `Rank` from test;

+----+--------+--------+--------------+------+
| id | Name   | Salary | DepartmentId | Rank |
+----+--------+--------+--------------+------+
|  2 | Jim    |  90000 |            1 |    1 |
|  5 | Max    |  90000 |            1 |    1 |
|  1 | Joe    |  70000 |            1 |    3 |
|  3 | Henery |  80000 |            2 |    1 |
|  4 | Sam    |  60000 |            2 |    2 |
+----+--------+--------+--------------+------+
相同薪水相同部门排名一样,且会空出名次
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方法3-dense_rank()
select *,dense_rank()over(partition by DepartmentId order by Salary desc)as `Rank` from test;

+----+--------+--------+--------------+------+
| id | Name   | Salary | DepartmentId | Rank |
+----+--------+--------+--------------+------+
|  2 | Jim    |  90000 |            1 |    1 |
|  5 | Max    |  90000 |            1 |    1 |
|  1 | Joe    |  70000 |            1 |    2 |
|  3 | Henery |  80000 |            2 |    1 |
|  4 | Sam    |  60000 |            2 |    2 |
+----+--------+--------+--------------+------+
相同薪水相同名次,不会空出名次
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以上这三种方法,基本可以满足工作中所有的排序需求了