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` (

`emp_no` int(11) NOT NULL,  -- '员工编号'

`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`));

select *

from employees

order by hire_date desc

limit 0, 1;

2、查找入职员工时间排名倒数第三的员工所有信息，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

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`));

select *

from employees

order by hire_date desc

limit 1 offset 2;

3、查找各个部门当前(dept_manager.to_date='9999-01-01')领导当前(salaries.to_date='9999-01-01')薪水详情以及其对应部门编号dept_no

(注:输出结果以salaries.emp_no升序排序，并且请注意输出结果里面dept_no列是最后一列)

CREATE TABLE `salaries` (

`emp_no` int(11) NOT NULL, -- '员工编号',

`salary` int(11) NOT NULL,

`from_date` date NOT NULL,

`to_date` date NOT NULL,

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

CREATE TABLE `dept_manager` (

`dept_no` char(4) NOT NULL, -- '部门编号'

`emp_no` int(11) NOT NULL, --  '员工编号'

`to_date` date NOT NULL,

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

select s.emp_no, s.salary, s.from_date, s.to_date, d.dept_no

from salaries s

inner join dept_manager d on s.emp_no = d.emp_no

where d.to_date='9999-01-01' and s.to_date='9999-01-01'

order by s.emp_no;

4、查找所有已经分配部门的员工的last_name和first_name以及dept_no(请注意输出描述里各个列的前后顺序)

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`));

select e.last_name, e.first_name, d.dept_no

from employees e

inner join dept_emp d on d.emp_no = e.emp_no;

5、查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括暂时没有分配具体部门的员工(请注意输出描述里各个列的前后顺序)

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`));

select e.last_name, e.first_name, d.dept_no

from employees e

left join dept_emp d on d.emp_no = e.emp_no;

6、查找所有员工入职时候的薪水情况，给出emp_no以及salary，并按照emp_no进行逆序(请注意，一个员工可能有多次涨薪的情况)

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 `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`));

select e.emp_no, s.salary

from employees e

inner join salaries s on s.emp_no = e.emp_no

where e.hire_date = s.from_date

order by e.emp_no desc;

7、查找薪水变动超过15次的员工号emp_no以及其对应的变动次数t

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`));

select emp_no, count(emp_no) t

from salaries

group by emp_no

having t > 15;

8、找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示

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`));

select distinct(salary)

from salaries

where to_date='9999-01-01'

order by salary desc;

9、获取所有部门当前(dept_manager.to_date='9999-01-01')manager的当前(salaries.to_date='9999-01-01')薪水情况，给出dept_no, emp_no以及salary，输出结果按照dept_no升序排列(请注意，同一个人可能有多条薪水情况记录)

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 `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`));

select m.dept_no, m.emp_no, s.salary

from dept_manager m

inner join salaries s on s.emp_no = m.emp_no

where m.to_date='9999-01-01' and s.to_date='9999-01-01'

order by m.dept_no;

10、获取所有非manager的员工emp_no

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`));

如插入为:

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

INSERT INTO dept_manager VALUES('d002',10006,'1990-08-05','9999-01-01');

INSERT INTO dept_manager VALUES('d003',10005,'1989-09-12','9999-01-01');

INSERT INTO dept_manager VALUES('d004',10004,'1986-12-01','9999-01-01');

INSERT INTO dept_manager VALUES('d005',10010,'1996-11-24','2000-06-26');

INSERT INTO dept_manager VALUES('d006',10010,'2000-06-26','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');

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');

select e.emp_no

from employees e

left join dept_manager m on e.emp_no = m.emp_no

where m.dept_no is null;

11、获取所有员工当前的(dept_manager.to_date='9999-01-01')manager，如果员工是manager的话不显示(也就是如果当前的manager是自己的话结果不显示)。输出结果第一列给出当前员工的emp_no,第二列给出其manager对应的emp_no。

CREATE TABLE `dept_emp` (

`emp_no` int(11) NOT NULL, -- '所有的员工编号'

`dept_no` char(4) NOT NULL, -- '部门编号'

`from_date` date NOT NULL,

`to_date` date NOT NULL,

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

CREATE TABLE `dept_manager` (

`dept_no` char(4) NOT NULL, -- '部门编号'

`emp_no` int(11) NOT NULL, -- '经理编号'

`from_date` date NOT NULL,

`to_date` date NOT NULL,

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

如插入：

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

INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');

INSERT INTO dept_emp VALUES(10003,'d004','1995-12-03','9999-01-01');

INSERT INTO dept_emp VALUES(10004,'d004','1986-12-01','9999-01-01');

INSERT INTO dept_emp VALUES(10005,'d003','1989-09-12','9999-01-01');

INSERT INTO dept_emp VALUES(10006,'d002','1990-08-05','9999-01-01');

INSERT INTO dept_emp VALUES(10007,'d005','1989-02-10','9999-01-01');

INSERT INTO dept_emp VALUES(10008,'d005','1998-03-11','2000-07-31');

INSERT INTO dept_emp VALUES(10009,'d006','1985-02-18','9999-01-01');

INSERT INTO dept_emp VALUES(10010,'d005','1996-11-24','2000-06-26');

INSERT INTO dept_emp VALUES(10010,'d006','2000-06-26','9999-01-01');

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

INSERT INTO dept_manager VALUES('d002',10006,'1990-08-05','9999-01-01');

INSERT INTO dept_manager VALUES('d003',10005,'1989-09-12','9999-01-01');

INSERT INTO dept_manager VALUES('d004',10004,'1986-12-01','9999-01-01');

INSERT INTO dept_manager VALUES('d005',10010,'1996-11-24','2000-06-26');

INSERT INTO dept_manager VALUES('d006',10010,'2000-06-26','9999-01-01');

select e.emp_no, m.emp_no manager_no

from dept_emp e, dept_manager m

where e.emp_no != m.emp_no

and e.dept_no = m.dept_no

and m.to_date = '9999-01-01';

12、获取所有部门中当前(dept_emp.to_date = '9999-01-01')员工当前(salaries.to_date='9999-01-01')薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary，按照部门编号升序排列。

【这题做错过】

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 `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`));

如插入：

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

INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');

INSERT INTO dept_emp VALUES(10003,'d001','1996-08-03','1997-08-03');

INSERT INTO salaries VALUES(10001,90000,'1986-06-26','1987-06-26');

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,'2000-08-02','2001-08-02');

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

INSERT INTO salaries VALUES(10003,90000,'1996-08-03','1997-08-03');

参考别人的答案

select uni.dept_no, uni.emp_no, max_salary.salary

from

    (select d.dept_no, s.emp_no, s.salary

     from dept_emp d join salaries s

     on d.emp_no = s.emp_no

     and d.to_date = '9999-01-01'

     and s.to_date = '9999-01-01'

    ) as uni, /* 部门编号,员工编号,当前薪水 */

    (select d.dept_no, max(s.salary) as salary

     from dept_emp d join salaries s

     on d.emp_no = s.emp_no

     and d.to_date = '9999-01-01'

     and s.to_date = '9999-01-01'

     group by d.dept_no

    ) as max_salary /* 部门编号,当前最高薪水 */

where uni.salary = max_salary.salary

and uni.dept_no = max_salary.dept_no

order by uni.dept_no;

13、从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。

CREATE TABLE IF NOT EXISTS "titles" (

`emp_no` int(11) NOT NULL,

`title` varchar(50) NOT NULL,

`from_date` date NOT NULL,

`to_date` date DEFAULT NULL);

如插入：

INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01');

INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01');

INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01');

INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01');

INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01');

INSERT INTO titles VALUES(10005,'Senior Staff','1996-09-12','9999-01-01');

INSERT INTO titles VALUES(10005,'Staff','1989-09-12','1996-09-12');

INSERT INTO titles VALUES(10006,'Senior Engineer','1990-08-05','9999-01-01');

INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01');

INSERT INTO titles VALUES(10007,'Staff','1989-02-10','1996-02-11');

INSERT INTO titles VALUES(10008,'Assistant Engineer','1998-03-11','2000-07-31');

INSERT INTO titles VALUES(10009,'Assistant Engineer','1985-02-18','1990-02-18');

INSERT INTO titles VALUES(10009,'Engineer','1990-02-18','1995-02-18');

INSERT INTO titles VALUES(10009,'Senior Engineer','1995-02-18','9999-01-01');

INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');

INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');

select title, count(title) t

from titles

group by title

having t >= 2;

14、从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。

注意对于重复的emp_no进行忽略(即emp_no重复的title不计算，title对应的数目t不增加)。

CREATE TABLE IF NOT EXISTS `titles` (

`emp_no` int(11) NOT NULL,

`title` varchar(50) NOT NULL,

`from_date` date NOT NULL,

`to_date` date DEFAULT NULL);

如插入：

INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01');

INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01');

INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01');

INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01');

INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01');

INSERT INTO titles VALUES(10005,'Senior Staff','1996-09-12','9999-01-01');

INSERT INTO titles VALUES(10005,'Staff','1989-09-12','1996-09-12');

INSERT INTO titles VALUES(10006,'Senior Engineer','1990-08-05','9999-01-01');

INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01');

INSERT INTO titles VALUES(10007,'Staff','1989-02-10','1996-02-11');

INSERT INTO titles VALUES(10008,'Assistant Engineer','1998-03-11','2000-07-31');

INSERT INTO titles VALUES(10009,'Assistant Engineer','1985-02-18','1990-02-18');

INSERT INTO titles VALUES(10009,'Engineer','1990-02-18','1995-02-18');

INSERT INTO titles VALUES(10009,'Senior Engineer','1995-02-18','9999-01-01');

INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');

INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');

select title, count(distinct emp_no) t

from titles

group by title

having count(*) >= 2;

只需在count中加入distinct '指定列'就可以在计算行数时去掉指定列中重复的值：在count函数计数时去掉重复的emp_no

15、查找employees表所有emp_no为奇数，且last_name不为Mary(注意大小写)的员工信息，并按照hire_date逆序排列

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`));

如插入：

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');

select *

from employees

where emp_no % 2 = 1

and last_name != "Mary"

order by hire_date desc;

牛客网sql实战参考答案（mysql版）：1-15的更多相关文章

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 ...
MySQL：怒刷牛客网“sql实战”
MySQL:怒刷牛客网"sql实战" 在对MySQL有一定了解后,抽空刷了一下牛客网上的数据库SQL 实战,在此做一点小小的记录 SQL1 查找最晚入职员工的所有信息 sele ...
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-数据库SQL实战
查找最晚入职员工的所有信息 CREATE TABLE `employees` ( `emp_no` ) NOT NULL, `birth_date` date NOT NULL, `first_nam ...
牛客网sql练习
一建表语句 /* Navicat MySQL Data Transfer Source Server : test Source Server Version : 50717 Source Host ...
牛客网数据库SQL实战解析(1-10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(41-50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

hdu4784 不错的搜索（买卖盐，要求整钱最多）
题意: 给你一个有向图,每个节点上都有一个盐价,然后给你k个空间,么个空间上节点与节点的距离不变,但盐价不同,对于每一个节点,有三种操作,卖一袋盐,买一袋盐 ,不交易,每一个节点可以跳掉( ...
使用SSH端口做端口转发以及反向隧道
目录 SSH做本地端口转发 SSH做反向隧道(远程端口转发) 用autossh建立稳定隧道 SSH开启端口转发需要修改 /etc/ssh/sshd_config配置文件,将 GatewayPorts修 ...
使用同步或异步的方式完成 I/O 访问和操作（Windows核心编程）
0x01 Windows 中对文件的底层操作 Windows 为了方便开发人员操作 I/O 设备(这些设备包括套接字.管道.文件.串口.目录等),对这些设备的差异进行了隐藏,所以开发人员在使用这些设备 ...
SqlServer数据库主从同步
分发/订阅模式实现SqlServer主从同步在文章开始之前,我们先了解一下几个关键的概念: 分发服务器分发服务器是负责存储在同步过程中所用复制信息的服务器.可以比喻成报刊发行商. 分发数据库分发数据 ...
mysql-.frm,.myd,myi备份如何导入mysql
.frm..myd..myi文件,也就是说是MySQL的原始数据文件,这三个文件分别是: .frm 表结构文件 .myd 表数据文件 .myi 表索引文件方法,如下: 新建一个数据库在my.ini ...
五、postman公共函数及newman运行与生成测试报告
一.公共函数 postman中定义公共函数如下 1.每次断言的时候都需要重写或者复制之前的断言代码,可以通过如下方法定义断言的公共函数,以后每次断言的时候只需要调用公共函数即可进行断言设置公共函数对 ...
mysql.data.entityframeworkcore 已弃用
转官网有方案: https://dev.mysql.com/doc/connector-net/en/connector-net-entityframework-core.html General R ...
SQLFlow使用中的注意事项--设置篇
SQLFlow 是用于追溯数据血缘关系的工具,它自诞生以来以帮助成千上万的工程师即用户解决了困扰许久的数据血缘梳理工作. 数据库中视图(View)的数据来自表(Table)或其他视图,视图中字段(Co ...
uboot1：启动流程和移植框架
目录 0 环境 1 移植框架 3 执行流程 3.0 链接地址 3.1 start.S, 入口 3.2 __main 3.3 board_init_f()和init_sequence_f[] 3.4 r ...
QFNU-ACM 2019.5.23组队赛 2019山东省赛复现
A.Calandar 题意:一年12个月,一个月30天,5天一周,已知某天的年月日星期数,求所给年月日的星期数是多少思路:直接进行计算,其实每个月每年都是等长度的就使得计算的时候忽略年月,可以直接进 ...

牛客网sql实战参考答案（mysql版）：1-15

牛客网sql实战参考答案（mysql版）：1-15的更多相关文章

随机推荐

热门专题