leetcode_sql

组合两个表

表1: Person

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId 是上表主键
表2: Address

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId 是上表主键

编写一个 SQL 查询，满足条件：无论 person 是否有地址信息，都需要基于上述两表提供 person 的以下信息：

• 解:

  # Write your MySQL query statement below
  
  
  # select FirstName, LastName, City, State
  # from 
  # (
  # (select 
  # PersonId,FirstName,LastName 
  # from 
  # Person)a  left join 
  # (select
  # PersonId,City, State
  # from 
  # Address)b
  # on a.PersonId=b.PersonId
  # );
  
  select FirstName, LastName, City, State
  from Person left join Address
  on Person.PersonId = Address.PersonId
  ;
  

第二高的薪水

编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表，SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水，那么查询应返回 null。

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+

• 解:

  # Write your MySQL query statement below
  
  # SELECT
  #     IFNULL(
  #       (SELECT DISTINCT Salary
  #        FROM Employee
  #        ORDER BY Salary DESC
  #         LIMIT 1 OFFSET 1),
  #     NULL) as SecondHighestSalary
  
  select 
  (
      select DISTINCT Salary as SecondHighestSalary 
      from Employee 
      order by Salary desc
      limit 1 offset 1
  )as SecondHighestSalary
  

第N高的薪水

编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）。

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。

+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+

• 解:

  CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN
  
  DECLARE m INT;
  
  SET m = N - 1;
  
  RETURN ( # Write your MySQL query statement below.
  
  SELECT ifnull( ( SELECT DISTINCT salary FROM Employee ORDER BY salary DESC LIMIT 1 offset m ), NULL ) );
  
  END
  

分数排名

编写一个 SQL 查询来实现分数排名。

如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。

+----+-------+
| Id | Score |
+----+-------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
+----+-------+
例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：

+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+------+

• 解:

  /*
  窗口函数的区别
  rank()
  排名为相同时记为同一个排名, 并且参与总排序
  dense_rank() over (PARTITION BY xx ORDER BY xx [DESC])
  排名相同时记为同一个排名, 并且不参与总排序
  row_number() over (over (PARTITION BY xx ORDER BY xx [DESC]))
  排名相同时记为下一个排名
  窗口函数在hive sql中经常使用, 也可在 mysql 8.0 之上的版本中使用
  */
  
  select score as "Score",dense_rank() over (order by score desc) as "Rank" from scores;
  

标签：Salary,200,PersonId,薪水,sql,Employee,leetcode,select
来源： https://www.cnblogs.com/deepJL/p/15149536.html