当前位置:网站首页>1873. The special bonus calculation
1873. The special bonus calculation
2022-08-05 02:32:00 【just six z】
1873. 计算特殊奖金
前言
表: Employees
+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| employee_id | int |
| name | varchar |
| salary | int |
+-------------+---------+
employee_id 是这个表的主键.
此表的每一行给出了雇员id ,名字和薪水.
写出一个SQL 查询语句,计算每个雇员的奖金.如果一个雇员的id是奇数并且他的名字不是以’M’开头,那么他的奖金是他工资的100%,否则奖金为0.
Return the result table ordered by employee_id.
返回的结果集请按照employee_id排序.
查询结果格式如下面的例子所示.
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/calculate-special-bonus
著作权归领扣网络所有.商业转载请联系官方授权,非商业转载请注明出处.
示例1:
输入:
Employees 表:
+-------------+---------+--------+
| employee_id | name | salary |
+-------------+---------+--------+
| 2 | Meir | 3000 |
| 3 | Michael | 3800 |
| 7 | Addilyn | 7400 |
| 8 | Juan | 6100 |
| 9 | Kannon | 7700 |
+-------------+---------+--------+
输出:
+-------------+-------+
| employee_id | bonus |
+-------------+-------+
| 2 | 0 |
| 3 | 0 |
| 7 | 7400 |
| 8 | 0 |
| 9 | 7700 |
+-------------+-------+
解释:
因为雇员id是偶数,所以雇员id 是2和8的两个雇员得到的奖金是0.
雇员id为3的因为他的名字以'M'开头,所以,奖金是0.
其他的雇员得到了百分之百的奖金.
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/calculate-special-bonus
著作权归领扣网络所有.商业转载请联系官方授权,非商业转载请注明出处.
1、union + not like
select
employee_id, salary as bonus
from Employees
where employee_id % 2 != 0 and name not like 'M%'
union
select
employee_id, salary*0 as bonus
from Employees
where employee_id % 2 = 0 or name like 'M%'
order by employee_id;
2、if
select
employee_id,if(employee_id % 2 = 0 or name like 'M%', 0 , salary) as bonus
from
Employees
order by
employee_id ;
3、if + left + mod
select employee_id,
if(mod(employee_id,2) != 0 and left(name,1) != 'M',salary,0) bonus
from Employees
order by employee_id;
边栏推荐
- LPQ (local phase quantization) study notes
- C学生管理系统 指定位置插入学生节点
- 1484. 按日期分组销售产品
- [深入研究4G/5G/6G专题-51]: URLLC-16-《3GPP URLLC相关协议、规范、技术原理深度解读》-11-高可靠性技术-2-链路自适应增强(根据无线链路状态动态选择高可靠性MCS)
- Error: Not a signal or slot declaration
- [In-depth study of 4G/5G/6G topic-51]: URLLC-16-"3GPP URLLC related protocols, specifications, and technical principles in-depth interpretation"-11-High reliability technology-2-Link adaptive enhancem
- VSCode Change Default Terminal how to modify the Default Terminal VSCode
- DAY22: sqli-labs shooting range clearance wp (Less01~~Less20)
- Solve connect: The requested address is not valid in its context
- 力扣-二叉树的前序遍历、中序遍历、后序遍历
猜你喜欢
DAY23: Command Execution & Code Execution Vulnerability
VSCode Change Default Terminal how to modify the Default Terminal VSCode
[Decryption] Can the NFTs created by OpenSea for free appear in my wallet without being chained?
Flink 1.15.1 集群搭建(StandaloneSession)
J9数字货币论:web3的创作者经济是什么?
02 【开发服务器 资源模块】
使用SuperMap iDesktopX数据迁移工具迁移地图文档和符号
特殊矩阵的压缩存储
CPDA|运营人如何从负基础学会数据分析(SQL)
C language diary 9 3 kinds of statements of if
随机推荐
js中try...catch和finally的用法
DAY22:sqli-labs 靶场通关wp(Less01~~Less20)
22-07-31周总结
1527. 患某种疾病的患者
[深入研究4G/5G/6G专题-51]: URLLC-16-《3GPP URLLC相关协议、规范、技术原理深度解读》-11-高可靠性技术-2-链路自适应增强(根据无线链路状态动态选择高可靠性MCS)
转:查尔斯·汉迪:你是谁,比你做什么更重要
KingbaseES V8 GIS data migration solution (2. Introduction to the capabilities of Kingbase GIS)
【日常训练】1403. 非递增顺序的最小子序列
nodeJs--encapsulate routing
C语言实现简单猜数字游戏
STM32使用stm32cubemx LL库系列教程
[Decryption] Can the NFTs created by OpenSea for free appear in my wallet without being chained?
【genius_platform软件平台开发】第七十六讲:vs预处理器定义的牛逼写法!!!!(其他组牛逼conding人员告知这么配置来取消宏定义)
2022了你还不会『低代码』?数据科学也能玩转Low-Code啦!
lua learning
[ROS] (10) ROS Communication - Service Communication
Images using redis cache Linux master-slave synchronization server hard drive full of moved to the new directory which points to be modified
View handler stepping record
ARM Mailbox
OpenGL 工作原理