【SQL】牛客网SQL训练Part1 简单难度

地址位置：

https://www.nowcoder.com/exam/oj?difficulty=2

查找入职员工时间排名倒数第三的员工所有信息

-- 准备脚本

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(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');

INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');

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');

INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');

INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');

INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');

INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');

INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');

-- 按默认的情况筛选

SELECT * FROM `employees` ORDER BY hire_date DESC LIMIT 2, 1

-- 如果时间存在多个同样的，去重处理后再查询

SELECT *

FROM employees

WHERE hire_date = (

    SELECT DISTINCT hire_date

    FROM employees

    ORDER BY hire_date DESC       -- 倒序

    LIMIT 1 OFFSET 2              -- 去掉排名倒数第一第二的时间，取倒数第三

);

查找所有已经分配部门的员工的last_name和first_name以及dept_no

drop table if exists  `dept_emp` ;

drop table if exists  `employees` ;

CREATE TABLE `dept_emp` (

`emp_no` int(11) NOT NULL,

`dept_no` char(4) NOT NULL,

`from_date` date NOT NULL,

`to_date` date NOT NULL,

PRIMARY KEY (`emp_no`,`dept_no`));

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 dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01');

INSERT INTO dept_emp VALUES(10002,'d002','1996-08-03','9999-01-01');

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(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');

INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');

-- 查询SQL

SELECT

    A.last_name,

    A.first_name,

    B.dept_no

FROM

    employees AS A

    LEFT JOIN dept_emp AS B ON A.emp_no = B.emp_no

WHERE B.DEPT_NO IS NOT NULL

查找薪水记录超过15条的员工号emp_no以及其对应的记录次数t

drop table if exists  `salaries` ;

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 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');

-- GROUP BY  + HAVING 筛选

SELECT `emp_no`, COUNT(`emp_no`) AS `t`

FROM `salaries`

GROUP BY `emp_no`

HAVING `t` > 15

找出所有员工当前薪水salary情况

drop table if exists  `salaries` ;

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 salaries VALUES(10001,72527,'2002-06-22','9999-01-01');

INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');

INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');

-- 对这个薪资进行去重即可

SELECT DISTINCT `salary` FROM `salaries` ORDER BY `salary` DESC

获取所有非manager的员工emp_no

drop table if exists  `dept_manager` ;

drop table if exists  `employees` ;

CREATE TABLE `dept_manager` (

`dept_no` char(4) NOT NULL,

`emp_no` int(11) NOT NULL,

`from_date` date NOT NULL,

`to_date` date NOT NULL,

PRIMARY KEY (`emp_no`,`dept_no`));

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 dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');

INSERT INTO dept_manager VALUES('d002',10003,'1990-08-05','9999-01-01');

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(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');

-- 按员工表作为主表，联表，筛选条件为 部门号为空的记录

SELECT

	A.emp_no

FROM

	`employees` AS A

	LEFT JOIN `dept_manager` AS B ON A.emp_no = B.emp_no

WHERE B.dept_no IS NULL

查找employees表emp_no与last_name的员工信息

请你查找

1、employees表所有emp_no为奇数，

2、且last_name不为Mary的员工信息，

3、并按照hire_date逆序排列，

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(10003,'1959-12-03','Bezalel','Mary','M','1986-08-28');

INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');

INSERT INTO employees VALUES(10005,'1953-11-07','Mary','Sluis','F','1990-01-22');

-- 按条件写SQL查询

SELECT *

FROM employees

WHERE `emp_no` MOD 2 != 0 AND `last_name` != 'Mary'

ORDER BY `hire_date` DESC

获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

drop table if exists  `salaries` ;

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 salaries VALUES(10001,88958,'2002-06-22','9999-01-01');

INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');

INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');

-- 按薪资倒序

SELECT `emp_no`, `salary`

FROM `salaries`

ORDER BY `salary` DESC LIMIT 1, 1

将employees表的所有员工的last_name和first_name拼接起来作为Name

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(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');

INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');

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');

INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');

INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');

INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');

INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');

INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');

-- CONCAT 合并函数

SELECT CONCAT(`last_name`, ' ', `first_name`) AS `Name`

FROM `employees`

批量插入数据

DROP TABLE if exists actor;

CREATE TABLE actor (

	actor_id  smallint(5)  NOT NULL PRIMARY KEY,

	first_name  varchar(45) NOT NULL,

	last_name  varchar(45) NOT NULL,

	last_update  DATETIME NOT NULL

);

-- 插入SQL

INSERT INTO `actor`VALUES

(1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),

(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33');

删除emp_no重复的记录，只保留最小的id对应的记录。

DROP TABLE if exists titles_test;

CREATE TABLE titles_test (

   id int(11) not null primary key,

   emp_no  int(11) NOT NULL,

   title  varchar(50) NOT NULL,

   from_date  date NOT NULL,

   to_date  date DEFAULT NULL);

insert into titles_test values

('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

-- 1、按员工编号分组Group筛选唯一记录，并按照最小主键值筛选，

-- 2、删除时进行取反获取重复记录

-- 3、删除表时查询不能再声明是这张表，需要设置别名

DELETE FROM `titles_test`

WHERE `id` NOT IN(

	SELECT * FROM (

		SELECT MIN(`id`) FROM `titles_test` GROUP BY emp_no

	) AS `tt`

);

将所有to_date为9999-01-01的全部更新为NULL

1、将所有to_date为9999-01-01的全部更新为NULL,

2、且 from_date更新为2001-01-01。

DROP TABLE IF EXISTS titles_test;

CREATE TABLE  titles_test (

   id int(11) not null primary key,

   emp_no  int(11) NOT NULL,

   title  varchar(50) NOT NULL,

   from_date  date NOT NULL,

   to_date  date DEFAULT NULL);

insert into  values

('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

-- 更新记录

UPDATE `titles_test`

SET `to_date` = NULL, `from_date` = '2001-01-01'

WHERE `to_date` = '9999-01-01'

将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005

其他数据保持不变，使用replace实现，直接使用update会报错。

drop table if exists titles_test;

CREATE TABLE titles_test (

   id int(11) not null primary key,

   emp_no  int(11) NOT NULL,

   title  varchar(50) NOT NULL,

   from_date  date NOT NULL,

   to_date  date DEFAULT NULL);

insert into titles_test values

('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

-- 1、主键重复警告风险 Duplicate entry '5' for key 'titles_test.PRIMARY'

UPDATE `titles_test`

SET `id` = 5, `emp_no` = 10005

WHERE `id` = 5 AND `emp_no` = 10001

-- 2、使用Replace函数更新？

UPDATE `titles_test`

SET `emp_no` = REPLACE(`emp_no`, 10001, 10005)

WHERE `id` = 5;

-- 3、ON DUPLICATE KEY UPDATE 主键冲突触发更新

INSERT INTO titles_test

VALUES(5, 10001 ,'Senior Engineer', '1986-06-26', '9999-01-01')

ON DUPLICATE KEY UPDATE emp_no = 10005;

-- 4、遇到主键冲突时，优先进行UPDATE操作

REPLACE INTO titles_test VALUES(5, 10005 ,'Senior Engineer', '1986-06-26', '9999-01-01') ;

将titles_test表名修改为titles_2017　

drop table if exists titles_test;

drop table if exists titles_2017;

CREATE TABLE titles_test (

   id int(11) not null primary key,

   emp_no  int(11) NOT NULL,

   title  varchar(50) NOT NULL,

   from_date  date NOT NULL,

   to_date  date DEFAULT NULL);

insert into  values

('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),

('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),

('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),

('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

-- ALTER TABLE 语法

ALTER TABLE `titles_test` RENAME TO `titles_2017`;

出现三次以上相同积分的情况

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);

-- 分组后筛选即可

SELECT `number`

FROM `grade`

GROUP BY `number`

HAVING COUNT(`number`) > 2

找到每个人的任务

drop table if exists person;

drop table if exists task;

CREATE TABLE `person` (

`id` int(4) NOT NULL,

`name` varchar(32) NOT NULL,

PRIMARY KEY (`id`));

CREATE TABLE `task` (

`id` int(4) NOT NULL,

`person_id` int(4) NOT NULL,

`content` varchar(32) NOT NULL,

PRIMARY KEY (`id`));

INSERT INTO person VALUES

(1,'fh'),

(2,'tm');

INSERT INTO task VALUES

(1,2,'tm works well'),

(2,2,'tm works well');

SELECT

	`p`.`id`,

	`p`.`name`,

	`t`.`content`

FROM

	`person` AS `p`

	LEFT JOIN `task` AS `t` ON `t`.person_id = `p`.`id`

每个人最近的登录日期

drop table if exists login;

drop table if exists user;

drop table if exists client;

CREATE TABLE login (

id int(4) NOT NULL,

user_id int(4) NOT NULL,

client_id int(4) NOT NULL,

date date NOT NULL,

PRIMARY KEY (id));

CREATE TABLE user (

id int(4) NOT NULL,

name varchar(32) NOT NULL,

PRIMARY KEY (id));

CREATE TABLE client (

id int(4) NOT NULL,

name varchar(32) NOT NULL,

PRIMARY KEY (id));

INSERT INTO login VALUES

(1,2,1,'2020-10-12'),

(2,3,2,'2020-10-12'),

(3,2,2,'2020-10-13'),

(4,3,2,'2020-10-13');

INSERT INTO user VALUES

(1,'tm'),

(2,'fh'),

(3,'wangchao');

INSERT INTO client VALUES

(1,'pc'),

(2,'ios'),

(3,'anroid'),

(4,'h5');

SELECT * FROM login

-- 按用户ID进行分组，取日期最大值（即最近一次登录时间）

select user_id,MAX(date) as recent_login_date

from login group by user_id order by user_id;

考试分数

第1行表示用户id为1的用户选择了C++岗位并且考了11001分

。。。

第8行表示用户id为8的用户选择了JS岗位并且考了9999分

请你写一个sql语句查询各个岗位分数的平均数，并且按照分数降序排序，

结果保留小数点后面3位(3位之后四舍五入):

drop table if exists grade;

CREATE TABLE  grade(

`id` int(4) NOT NULL,

`job` varchar(32) NOT NULL,

`score` int(10) NOT NULL,

PRIMARY KEY (`id`));

INSERT INTO grade VALUES

(1,'C++',11001),

(2,'C++',10000),

(3,'C++',9000),

(4,'Java',12000),

(5,'Java',13000),

(6,'JS',12000),

(7,'JS',11000),

(8,'JS',9999);

-- 对job进行分组，然后取成绩平均值，并设置四舍五入

SELECT `job`, ROUND(AVG(`score`), 3) AS `AVG_SCORE`

FROM `grade`

GROUP BY `job`

ORDER BY `AVG_SCORE` DESC

课程订单分析：

请你写出一个sql语句查询

1、在2025-10-15以后

2、状态为购买成功的C++课程或者Java课程或者Python的订单，

3、并且按照order_info的id升序排序

drop table if exists order_info;

CREATE TABLE order_info (

id int(4) NOT NULL,

user_id int(11) NOT NULL,

product_name varchar(256) NOT NULL,

status varchar(32) NOT NULL,

client_id int(4) NOT NULL,

date date NOT NULL,

PRIMARY KEY (id));

INSERT INTO order_info VALUES

(1,557336,'C++','no_completed',1,'2025-10-10'),

(2,230173543,'Python','completed',2,'2025-10-12'),

(3,57,'JS','completed',3,'2025-10-23'),

(4,57,'C++','completed',3,'2025-10-23'),

(5,557336,'Java','completed',1,'2025-10-23'),

(6,557336,'Python','no_completed',1,'2025-10-24');

-- 按描述条件查询即可

SELECT *

FROM `order_info`

WHERE

	`date` > '2025-10-15'

	AND `product_name` IN ('C++', 'Java', 'Python')

	AND `status` = 'completed'

ORDER BY `id` ASC

简历分析：

写出SQL语句查询在2025年内投递简历的岗位和数量，并且按数量降序排序

drop table if exists resume_info;

CREATE TABLE resume_info (

id int(4) NOT NULL,

job varchar(64) NOT NULL,

date date NOT NULL,

num int(11) NOT NULL,

PRIMARY KEY (id));

INSERT INTO resume_info VALUES

(1,'C++','2025-01-02',53),

(2,'Python','2025-01-02',23),

(3,'Java','2025-01-02',12),

(4,'Java','2025-02-03',24),

(5,'C++','2025-02-03',23),

(6,'Python','2025-02-03',34),

(7,'Python','2025-03-04',54),

(8,'C++','2025-03-04',65),

(9,'Java','2025-03-04',92),

(10,'Java','2026-01-04',230);

-- 1、使用YEAR函数筛选25年内的数据

-- 2、再对job分组，求和NUM字段

SELECT `job`, SUM(`NUM`) AS `cnt`

FROM `resume_info`

WHERE YEAR(`date`) = 2025

GROUP BY `job`

ORDER BY `cnt` DESC

【SQL】牛客网SQL训练Part1 简单难度的更多相关文章

MySql面试题、知识汇总、牛客网SQL专题练习
点击名字直接跳转到链接: Linux运维必会的100道MySql面试题之(一) Linux运维必会的100道MySql面试题之(二) Linux运维必会的100道MySql面试题之(三) Linux运 ...
牛客网Sql
牛客网Sql: 1.查询最晚入职的员工信息 select * from employees where hire_date =(select max(hire_date) from employee ...
牛客网sql刷题解析-完结
查找最晚入职员工的所有信息解题步骤: 题目:查询最晚入职员工的所有信息目标:查询员工的所有信息筛选条件:最晚入职答案: SELECT *--查询所有信息就用* ...
牛客网sql练习
一建表语句 /* Navicat MySQL Data Transfer Source Server : test Source Server Version : 50717 Source Host ...
牛客网sql实战参考答案（mysql版）：16-21
16.统计出当前(titles.to_date='9999-01-01')各个title类型对应的员工当前(salaries.to_date='9999-01-01')薪水对应的平均工资.结果给出ti ...
牛客网sql实战参考答案（mysql版）：1-15
1.查找最晚入职员工的所有信息,为了减轻入门难度,目前所有的数据里员工入职的日期都不是同一天(sqlite里面的注释为--,mysql为comment) CREATE TABLE `employees ...
牛客网数据库SQL实战解析(51-61题)
牛客网SQL刷题地址: https://www.nowcoder.com/ta/sql?page=0 牛客网数据库SQL实战解析(01-10题): https://blog.csdn.net/u010 ...
牛客网数据库SQL实战解析(31-40题)
牛客网SQL刷题地址: https://www.nowcoder.com/ta/sql?page=0 牛客网数据库SQL实战解析(01-10题): https://blog.csdn.net/u010 ...
牛客网数据库SQL实战解析(21-30题)
牛客网SQL刷题地址: https://www.nowcoder.com/ta/sql?page=0 牛客网数据库SQL实战解析(01-10题): https://blog.csdn.net/u010 ...
牛客网数据库SQL实战解析(11-20题)
牛客网SQL刷题地址: https://www.nowcoder.com/ta/sql?page=0 牛客网数据库SQL实战解析(01-10题): https://blog.csdn.net/u010 ...

随机推荐

阿里云ECS后台CPU占用100%，top却找不到
上周公司阿里云服务器后台报警,CPU占用瞬间飙升到100%: 首先想到使用top命令查询CPU占用详情: 发现进程占用CPU都比较低,在CPU占用一栏发现只有ni的占用比较高. 先了解一下CPU相关监 ...
vs2019 调试服务器代码
前言&使用背景: 又是我,一个不是在解决bug就是在解决bug路上的黄金程序员.今天给大家介绍一种方法是vs自带的调试工具.通常我们会遇见一种情况,那就是本地调试没有一点问题! 可一但发布到服 ...
[SWPUCTF 2021 新生赛]easy_sql
这道题呢就是很简单的sql注入,我们直接用sqlmap来跑. 首先我们打开页面可以看见提示,参数为wllm **然后我们启动虚拟机,输入sqlmap的命令:sqlmap -u "url地址/ ...
Java第二次Blog
7-4~6题目集前言这些题目主要用到对象与类的处理继承与多态的使用: 继承和多态是面向对象编程中相互关联的两个概念.继承为多态提供了基础,而多态则通过继承实现了代码的灵活性和可扩展性. 1.字符串 ...
什么是spring框架？
spring是一个开放源代码的设计层面框架,它解决的是业务逻辑层和其他各层的松耦合问题,是一个分层的javaEE一站式轻量级开源框架
spring.jackson 相差8小时,restful接收Date参数处理
spring.jackson 相差8小时,restful接收Date参数处理前端提交字符串到后台映射日期类型的话,加上@DateTimeFormat(pattern = "yyyy-M ...
洛谷 P1226 快速幂
题目链接:快速幂思路简单快速幂模板.a ^ 17 = (a ^ 2) ^ 8 * a,此时pow()中的y就可以视为17 -> 8(y >>= 1),pow()中的x就是底数a ...
Playbook条件语句
目录 Playbook条件语句 1. when的基本使用 1.1 when的基本示例 1.2比较运算符 1.3 比较运算符示例 1.4 逻辑运算符 1.5 逻辑运算符示例 2. 条件判断与block ...
获取ImageView的触摸点所对应的UIImage的坐标
获取ImageView的触摸点所对应的UIImage的坐标功能描述实现前分析注意事项代码求打赏功能描述在imageview上触摸图片,求对应UIImage的触摸点. 实现前分析从ima ...
Android程序获取鸿蒙手机设备信息（是否鸿蒙手机、版本号、小版本号等）
1.效果图鸿蒙手机 --> 关于手机的截图: Android程序获取鸿蒙手机设备信息的截图: 2.实现本案例DEMO的实现主要借鉴了网上现有的资料: https://blog.csdn.ne ...

【SQL】 牛客网SQL训练Part1 简单难度

查找入职员工时间排名倒数第三的员工所有信息

查找所有已经分配部门的员工的last_name和first_name以及dept_no

查找薪水记录超过15条的员工号emp_no以及其对应的记录次数t

找出所有员工当前薪水salary情况

获取所有非manager的员工emp_no

查找employees表emp_no与last_name的员工信息

获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

将employees表的所有员工的last_name和first_name拼接起来作为Name

批量插入数据

删除emp_no重复的记录，只保留最小的id对应的记录。

将所有to_date为9999-01-01的全部更新为NULL

将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005

将titles_test表名修改为titles_2017

出现三次以上相同积分的情况

找到每个人的任务

每个人最近的登录日期

考试分数

课程订单分析：

简历分析：

【SQL】 牛客网SQL训练Part1 简单难度的更多相关文章

随机推荐

热门专题

【SQL】牛客网SQL训练Part1 简单难度

将titles_test表名修改为titles_2017　

【SQL】牛客网SQL训练Part1 简单难度的更多相关文章