SQL力扣练习(六)

目录

1. 部门工资前三高的所有员工(185)

题解一(dense_rank()窗口函数)

题解二(自定义函数)

 2.删除重复的电子邮箱(196)

题解一

题解二(官方解析)

3.上升的温度(197)

解法一(DATEDIFF())

解法二(TIMESTAMPDIFF())

解法三(ADDDATE()力扣)


1. 部门工资前三高的所有员工(185)

表: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
Id是该表的主键列。
departmentId是Department表中ID的外键。
该表的每一行都表示员工的ID、姓名和工资。它还包含了他们部门的ID。

表: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
Id是该表的主键列。
该表的每一行表示部门ID和部门名。

公司的主管们感兴趣的是公司每个部门中谁赚的钱最多。一个部门的 高收入者 是指一个员工的工资在该部门的 不同 工资中 排名前三 。

编写一个SQL查询,找出每个部门中 收入高的员工 。

以 任意顺序 返回结果表。

题解一(dense_rank()窗口函数)

dense_rank() 排序规则 1 2 3 3  4

rank() 1 2  3 3 5

partition by 根据某字段分区

# Write your MySQL query statement below
select  Department, Employee, Salary 
from (select d.name Department,e.name Employee,e.salary Salary ,dense_rank() over(partition by d.name order by e.salary desc) no from employee e join department d
on e.departmentId=d.id) temp
where temp.no<=3

题解二(自定义函数)

到现在,我对这个自定义函数的用法才稍微熟练了一点,这自定义函数的出现,可以让sql变的更加灵活。

CASE 
         WHEN @pre = DepartmentId THEN @rank:= @rank + 1
         WHEN @pre := DepartmentId THEN @rank:= 1
END AS 'RANK'

这个意思是,如果@pre等于当前的DepartmentId则@rank加一,否则更新@pre的值且@rank=1。

这个@rank的值命名为RANK字段。用这个进行排序的关键就是from 的表需要有序。

SELECT dep.Name Department, emp.Name Employee, emp.Salary
FROM (## 自定义变量RANK, 查找出 每个部门工资前三的排名
        SELECT te.DepartmentId, te.Salary,
               CASE 
                    WHEN @pre = DepartmentId THEN @rank:= @rank + 1
                    WHEN @pre := DepartmentId THEN @rank:= 1
               END AS 'RANK'
        FROM (SELECT @pre:=null, @rank:=0)tt,
             (## (部门,薪水)去重,根据 部门(升),薪水(降) 排序
                 SELECT DepartmentId,Salary
                 FROM Employee
                 GROUP BY DepartmentId,Salary
                 ORDER BY DepartmentId,Salary DESC
             )te
       )t
INNER JOIN Department dep ON t.DepartmentId = dep.Id
INNER JOIN Employee emp ON t.DepartmentId = emp.DepartmentId and t.Salary = emp.Salary and t.RANK <= 3
ORDER BY t.DepartmentId, t.Salary DESC ## t 结果集已有序,根据该集合排序

 2.删除重复的电子邮箱(196)

表: Person

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| email       | varchar |
+-------------+---------+
id是该表的主键列。
该表的每一行包含一封电子邮件。电子邮件将不包含大写字母。

编写一个 SQL 删除语句来 删除 所有重复的电子邮件,只保留一个id最小的唯一电子邮件。

以 任意顺序 返回结果表。 (注意: 仅需要写删除语句,将自动对剩余结果进行查询)

题解一

首先反向思维,找出不满足的,根据分组可以很简单找到

DELETE from Person
WHERE id NOT IN #不在满足条件内的肯定就是不满足的,直接删除
(
    SELECT ID #先把满足条件的找出来
    From
    (
     SELECT  MIN(id) as ID
     From Person
     Group by Email
    )t
)

题解二(官方解析)

# Write your MySQL query statement below
DELETE p1 FROM Person p1,
    Person p2
WHERE
    p1.Email = p2.Email AND p1.Id > p2.Id

3.上升的温度(197)

表: Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id 是这个表的主键
该表包含特定日期的温度信息

编写一个 SQL 查询,来查找与之前(昨天的)日期相比温度更高的所有日期的 id 。

返回结果 不要求顺序 。

示例 1:

输入:
Weather 表:
+----+------------+-------------+
| id | recordDate | Temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10          |
| 2  | 2015-01-02 | 25          |
| 3  | 2015-01-03 | 20          |
| 4  | 2015-01-04 | 30          |
+----+------------+-------------+
输出:
+----+
| id |
+----+
| 2  |
| 4  |
+----+
解释:
2015-01-02 的温度比前一天高(10 -> 25)
2015-01-04 的温度比前一天高(20 -> 30)

解法一(DATEDIFF())

DATEDIFF 函数,可以计算两者的日期差

DATEDIFF('2007-12-31','2007-12-30');   # 1
DATEDIFF('2010-12-30','2010-12-31');   # -1

# Write your MySQL query statement below  

select a.ID
from weather as a cross join weather as b 
     on datediff(a.recordDate, b.recordDate) = 1
where a.temperature > b.temperature;

解法二(TIMESTAMPDIFF())

TIMESTAMPDIFF能干什么,可以计算相差天数、小时、分钟和秒,相比于datediff函数要灵活很多。格式是时间小的前,时间大的放在后面。 计算相差天数

select a.ID
from weather as a cross join weather as b 
     on timestampdiff(day,a.recordDate, b.recordDate) = -1
where a.temperature > b.temperature;

解法三(ADDDATE()力扣)

SQL力扣练习(六)_第1张图片

select a.id 
    from weather a join weather b 
    on (a.recorddate = adddate(b.recorddate,INTERVAL 1 day))
where a.temperature > b.temperature

 

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