【力扣 + 牛客 | SQL题 | 每日5题】牛客SQL热题255，253，256

1. 牛客SQL热题255：给出employees表中排名为奇数行的first_name

1.1 题目：

描述

对于employees表中，输出first_name排名(按first_name升序排序)为奇数的first_name
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,

PRIMARY KEY (`emp_no`));

如，输入为：

INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');

输出格式:

first

Georgi

Anneke

请你在不打乱原序列顺序的情况下，输出：按first_name排升序后，取奇数行的first_name。

如对以上示例数据的first_name排序后的序列为：Anneke、Bezalel、Georgi、Kyoichi。

则原序列中的Georgi排名为3，Anneke排名为1，所以按原序列顺序输出Georgi、Anneke。

示例1

输入：

drop table if exists  `employees` ; 
CREATE TABLE `employees` (
  `emp_no` int(11) NOT NULL,
  `birth_date` date NOT NULL,
  `first_name` varchar(14) NOT NULL,
  `last_name` varchar(16) NOT NULL,
  `gender` char(1) NOT NULL,
  `hire_date` date NOT NULL,
  PRIMARY KEY (`emp_no`));
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');
复制输出：

Georgi
Anneke

1.2 思路：

看注释。

1.3 题解：

with tep1 as (
    -- 先生成依据first_name排序后的临时表
    select first_name, rank() over (order by first_name) ranks
    from employees
), tep2 as (
    -- 筛选出来排名为奇数的姓名
    select first_name
    from tep1
    where ranks % 2 = 1
)
-- 输出当然是依据原表
select first_name first
from employees
where first_name in (select * from tep2)

2. 牛客SQL热题253：获取有奖金的员工相关信息。

2.1 题目：

描述

现有员工表employees如下：

emp_no	birth_date	first_name	last_name	gender	hire_date
10001	1953-09-02	Georgi	Facello	M	1986-06-26
10002	1964-06-02	Bezalel	Simmel	F	1985-11-21

有员工奖金表emp_bonus:

emp_no	recevied	btype
10001	2010-01-01	1
10002	2010-10-01	2

有薪水表salaries:

emp_no	salary	from_date	to_date
10001	60117	1986-06-26	1987-06-26
10001	62102	1987-06-26	1988-06-25
10001	66074	1988-06-25	1989-06-25
10001	66596	1989-06-25	1990-06-25
10001	66961	1990-06-25	1991-06-25
10001	71046	1991-06-25	1992-06-24
10001	74333	1992-06-24	1993-06-24
10001	75286	1993-06-24	1994-06-24
10001	75994	1994-06-24	1995-06-24
10001	76884	1995-06-24	1996-06-23
10001	80013	1996-06-23	1997-06-23
10001	81025	1997-06-23	1998-06-23
10001	81097	1998-06-23	1999-06-23
10001	84917	1999-06-23	2000-06-22
10001	85112	2000-06-22	2001-06-22
10001	85097	2001-06-22	2002-06-22
10001	88958	2002-06-22	9999-01-01
10002	72527	1996-08-03	1997-08-03
10002	72527	1997-08-03	1998-08-03
10002	72527	1998-08-03	1999-08-03
10002	72527	1999-08-03	2000-08-02
10002	72527	2000-08-02	2001-08-02
10002	72527	2001-08-02	9999-01-01

• 其中bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 to_date='9999-01-01'表示当前薪水。
• 请你给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。
• bonus结果保留一位小数，输出结果按emp_no升序排序。

以上数据集的输出结果如下:

emp_no	first_name	last_name	btype	salary	bonus
10001	Georgi	Facello	1	88958	8895.8000
10002	Bezalel	Simmel	2	72527	14505.4000

示例1

输入：

drop table if exists  `employees` ; 
drop table if exists  emp_bonus; 
drop table if exists  `salaries` ; 
CREATE TABLE `employees` (
  `emp_no` int(11) NOT NULL,
  `birth_date` date NOT NULL,
  `first_name` varchar(14) NOT NULL,
  `last_name` varchar(16) NOT NULL,
  `gender` char(1) NOT NULL,
  `hire_date` date NOT NULL,
  PRIMARY KEY (`emp_no`));
 create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
insert into emp_bonus values
(10001, '2010-01-01',1),
(10002, '2010-10-01',2);
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');

INSERT INTO salaries VALUES(10001,60117,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25');
INSERT INTO salaries VALUES(10001,66074,'1988-06-25','1989-06-25');
INSERT INTO salaries VALUES(10001,66596,'1989-06-25','1990-06-25');
INSERT INTO salaries VALUES(10001,66961,'1990-06-25','1991-06-25');
INSERT INTO salaries VALUES(10001,71046,'1991-06-25','1992-06-24');
INSERT INTO salaries VALUES(10001,74333,'1992-06-24','1993-06-24');
INSERT INTO salaries VALUES(10001,75286,'1993-06-24','1994-06-24');
INSERT INTO salaries VALUES(10001,75994,'1994-06-24','1995-06-24');
INSERT INTO salaries VALUES(10001,76884,'1995-06-24','1996-06-23');
INSERT INTO salaries VALUES(10001,80013,'1996-06-23','1997-06-23');
INSERT INTO salaries VALUES(10001,81025,'1997-06-23','1998-06-23');
INSERT INTO salaries VALUES(10001,81097,'1998-06-23','1999-06-23');
INSERT INTO salaries VALUES(10001,84917,'1999-06-23','2000-06-22');
INSERT INTO salaries VALUES(10001,85112,'2000-06-22','2001-06-22');
INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');
INSERT INTO salaries VALUES(10002,72527,'1997-08-03','1998-08-03');
INSERT INTO salaries VALUES(10002,72527,'1998-08-03','1999-08-03');
INSERT INTO salaries VALUES(10002,72527,'1999-08-03','2000-08-02');
INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
复制输出：

10001|Georgi|Facello|1|88958|8895.8
10002|Bezalel|Simmel|2|72527|14505.4

2.2 思路：

简单的多表连接题。

2.3 题解：

with tep1 as (
    -- 先得到当前薪水的记录
    select emp_no, salary
    from salaries
    where to_date = '9999-01-01'
), tep2 as (
    -- 多表连接
    select t1.emp_no, btype, salary, first_name, last_name
    from employees t1
    join emp_bonus t2
    on t1.emp_no = t2.emp_no
    join tep1 t3
    on t1.emp_no = t3.emp_no
)
-- case when完事
select emp_no, first_name, last_name, btype, salary,

case btype when 1 then round(salary*0.1, 4)
when 2 then round(salary*0.2, 4)
else round(salary*0.3, 4)
end bonus
from tep2

3. 牛客SQL热题256：出现三次以上相同积分的情况

3.1 题目：

描述

在牛客刷题的小伙伴们都有着牛客积分，积分(grade)表简化可以如下:

id	number
1 2 3 4 5	111 333 111 111 333

id为用户主键id，number代表积分情况，让你写一个sql查询，积分表里面出现三次以及三次以上的积分，查询结果如下:

111

注意：若有多个符合条件的number，则按number升序排序输出。

示例1

输入：

drop table if exists grade;
CREATE TABLE `grade` (
`id` int(4) NOT NULL,
`number` int(4) NOT NULL,
PRIMARY KEY (`id`));

INSERT INTO grade VALUES
(1,111),
(2,333),
(3,111),
(4,111),
(5,333);
复制输出：

111

3.2 思路：

分组过滤。

3.3 题解：

select number
from grade
group by number
having count(*) >= 3

4. 力扣SQL1294：不同国家的天气类型

4.1 题目：

表：Countries

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| country_id    | int     |
| country_name  | varchar |
+---------------+---------+
country_id 是这张表的主键(具有唯一值的列)。
该表的每行有 country_id 和 country_name 两列。

表：Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| country_id    | int     |
| weather_state | varchar |
| day           | date    |
+---------------+---------+
(country_id, day) 是该表的复合主键(具有唯一值的列)。
该表的每一行记录了某个国家某一天的天气情况。

编写解决方案找到表中每个国家在 2019 年 11 月的天气类型。

天气类型的定义如下：

• 当 weather_state 的平均值小于或等于 15 返回 Cold，
• 当 weather_state 的平均值大于或等于 25 返回 Hot，
• 否则返回 Warm。

以 任意顺序 返回你的查询结果。

返回结果格式如下所示：

示例 1：

输入：
Countries table:
+------------+--------------+
| country_id | country_name |
+------------+--------------+
| 2          | USA          |
| 3          | Australia    |
| 7          | Peru         |
| 5          | China        |
| 8          | Morocco      |
| 9          | Spain        |
+------------+--------------+
Weather table:
+------------+---------------+------------+
| country_id | weather_state | day        |
+------------+---------------+------------+
| 2          | 15            | 2019-11-01 |
| 2          | 12            | 2019-10-28 |
| 2          | 12            | 2019-10-27 |
| 3          | -2            | 2019-11-10 |
| 3          | 0             | 2019-11-11 |
| 3          | 3             | 2019-11-12 |
| 5          | 16            | 2019-11-07 |
| 5          | 18            | 2019-11-09 |
| 5          | 21            | 2019-11-23 |
| 7          | 25            | 2019-11-28 |
| 7          | 22            | 2019-12-01 |
| 7          | 20            | 2019-12-02 |
| 8          | 25            | 2019-11-05 |
| 8          | 27            | 2019-11-15 |
| 8          | 31            | 2019-11-25 |
| 9          | 7             | 2019-10-23 |
| 9          | 3             | 2019-12-23 |
+------------+---------------+------------+
输出：
+--------------+--------------+
| country_name | weather_type |
+--------------+--------------+
| USA          | Cold         |
| Austraila    | Cold         |
| Peru         | Hot          |
| China        | Warm         |
| Morocco      | Hot          |
+--------------+--------------+
解释：
USA 11 月的平均 weather_state 为 (15) / 1 = 15 所以天气类型为 Cold。
Australia 11 月的平均 weather_state 为 (-2 + 0 + 3) / 3 = 0.333 所以天气类型为 Cold。
Peru 11 月的平均 weather_state 为 (25) / 1 = 25 所以天气类型为 Hot。
China 11 月的平均 weather_state 为 (16 + 18 + 21) / 3 = 18.333 所以天气类型为 Warm。
Morocco 11 月的平均 weather_state 为 (25 + 27 + 31) / 3 = 27.667 所以天气类型为 Hot。
我们并不知道 Spain 在 11 月的 weather_state 情况所以无需将他包含在结果中。

4.2 思路：

easy题。

4.3 题解：

with tep1 as (
    select country_id , avg(weather_state) avg_wea 
    from Weather 
    where substring(day, 1, 7) = '2019-11'
    group by country_id 
)
select country_name , 

case when avg_wea <= 15 then 'Cold'
when avg_wea >= 25 then 'Hot'
else 'Warm'
end weather_type 
from tep1 t1 
join Countries t2 
on t1.country_id = t2.country_id