力扣sql刷题(四)

本文主要介绍了力扣(LeetCode)中关于SQL的几道题目,包括项目员工的经验分析和销售数据的查询。内容涵盖找出每个项目中最富有经验的员工,以及总销售额最高的销售者等。同时,还探讨了游戏玩法分析中的玩家留存率计算方法。

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1076. 项目员工II
Create table If Not Exists Project (project_id int, employee_id int)
Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
Truncate table Project
insert into Project (project_id, employee_id) values ('1', '1')
insert into Project (project_id, employee_id) values ('1', '2')
insert into Project (project_id, employee_id) values ('1', '3')
insert into Project (project_id, employee_id) values ('2', '1')
insert into Project (project_id, employee_id) values ('2', '4')
Truncate table Employee
insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
insert into Employee (employee_id, name, experience_years) values ('3', 'John', '1')
insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

Table: Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
主键为 (project_id, employee_id)。
employee_id 是员工表 Employee 表的外键。

Table: Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+

主键是 employee_id。编写一个SQL查询,报告所有雇员最多的项目。

查询结果格式如下所示:

Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
第一个项目有3名员工,第二个项目有2名员工。

题解一

select
    project_id  
from project
group by project_id
having count(employee_id ) >= 
all(
    select count(employee_id) 
    from project group by project_id  
   )

题解二

select project_id from
(select 
    project_id, 
    dense_rank() over(order by count(employee_id) desc) as ranking
 	# 窗口函数适用范围广,在数据层面先构造再筛选
from project
group by project_id) t
where ranking = 1
1077. 项目员工 III
Create table If Not Exists Project (project_id int, employee_id int)
Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
Truncate table Project
insert into Project (project_id, employee_id) values ('1', '1')
insert into Project (project_id, employee_id) values ('1', '2')
insert into Project (project_id, employee_id) values ('1', '3')
insert into Project (project_id, employee_id) values ('2', '1')
insert into Project (project_id, employee_id) values ('2', '4')
Truncate table Employee
insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
insert into Employee (employee_id, name, experience_years) values ('3', 'John', '3')
insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

项目表 Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) 是这个表的主键
employee_id 是员工表 Employee 的外键

员工表 Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id 是这个表的主键

写 一个 SQL 查询语句,报告在每一个项目中经验最丰富的雇员是谁。如果出现经验年数相同的情况,请报告所有具有最大经验年数的员工。

查询结果格式在以下示例中:

Project 表:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee 表:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   
Case When语句可以用来根据给定的条件来判断性别。使用Case When语句时,我们可以在When后面写上条件,然后在Then后面写上对应的结果。比如,如果我们想把性别为1的记录标记为“男性”,性别为2的记录标记为“女性”,那么我们可以使用以下的Case When语句: ```sql Case When sex = 1 Then '男性' When sex = 2 Then '女性' End ``` 这段代码将根据不同的性别值返回相应的结果。注意,如果没有匹配到任何条件,那么Case When语句将返回Null。此外,如果我们想要处理一个范围的条件,可以使用Between关键字,比如: ```sql Case When age Between 18 and 30 Then '青年' When age Between 31 and 50 Then '中年' Else '其他' End ``` 这段代码将根据年龄的范围返回相应的结果。需要注意的是,每个条件都是按照顺序依次判断的,如果一个条件满足,那么后面的条件不会再被判断。另外,我们还可以在Case When语句的末尾添加一个Else子句,用于处理所有未被匹配到的情况。<span class="em">1</span><span class="em">2</span><span class="em">3</span> #### 引用[.reference_title] - *1* *2* [case when](https://blog.youkuaiyun.com/weixin_39864101/article/details/111852552)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"] - *3* [力扣学习SQL篇——1-5 修改(变更性别——使用判断if/case when)](https://blog.youkuaiyun.com/weixin_43987277/article/details/126320938)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"] [ .reference_list ]
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