目录
SQL228 批量插入数据
描述
SQL202 找出所有员工当前薪水salary情况
描述
示例1
SQL195 查找最晚入职员工的所有信息描述
示例1
SQL196 查找入职员工时间排名倒数第三的员工所有信息描述
SQL201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
描述
SQL228 批量插入数据
如果忘记了可以参考我这一篇文章
6.MySQL的增删改查-CSDN博客
批量插入数据_牛客题霸_牛客网 (nowcoder.com)
https://www.nowcoder.com/practice/51c12cea6a97468da149c04b7ecf362e?tpId=82&tqId=29802&tPage=2&rp=&ru=%2Fta%2Fsql&qru=%2Fta%2Fsql%2Fquestion-ranking
描述
对于表actor批量插入如下数据
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
| actor_id | first_name | last_name | last_update |
|---|---|---|---|
| 1 | PENELOPE | GUINESS | 2006-02-15 12:34:33 |
| 2 | NICK | WAHLBERG | 2006-02-15 12:34:33 |
输入：
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)
输出：
1|PENELOPE|GUINESS|2006-02-15 12:34:33
2|NICK|WAHLBERG|2006-02-15 12:34:33
插入语句语法：
insert into 表名 (字段名，，，，) values (要插入的数据字段对齐，，，)# INSERT INTO actor(actor_id,
# first_name,
# last_name,
# last_update)
# VALUES(1,'PENELOPE','GUINESS','2006-02-15 12:34:33'),
# (2,'NICK','WAHLBERG','2006-02-15 12:34:33');
insert into actor VALUES(1,"PENELOPE","GUINESS","2006-02-15 12:34:33"),
(2,"NICK","WAHLBERG","2006-02-15 12:34:33");SQL202 找出所有员工当前薪水salary情况
如果忘记了可以参考我这篇文章：
6.MySQL的增删改查-CSDN博客
找出所有员工当前薪水salary情况_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/ae51e6d057c94f6d891735a48d1c2397?tpId=82&tqId=29760&tPage=1&rp=&ru=%2Fta%2Fsql&qru=%2Fta%2Fsql%2Fquestion-ranking
描述
有一个薪水表，salaries简况如下:
| emp_no | salary | from_date | to_date |
| 10001 | 72527 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 2001-08-02 | 9999-01-01 |
| 10003 | 43311 | 2001-12-01 | 9999-01-01 |
请你找出所有员工具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示，以上例子输出如下:
| salary |
| 72527 |
| 43311 |
示例1
输入：
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');输出：
72527 43311
这是答案哦
Select distinct salary from salaries order by salary desc;SQL195 查找最晚入职员工的所有信息描述
有一个员工employees表简况如下:
| 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 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10004 | 1954-05-01 | Christian | Koblick | M | 1986-12-01 |
请你查找employees里最晚入职员工的所有信息，以上例子输出如下:
| emp_no | birth_date | first_name | last_name | gender | hire_date |
| 10004 | 1954-05-01 | Christian | Koblick | M | 1986-12-01 |
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(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');输出：
10008|1958-02-19|Saniya|Kalloufi|M|1994-09-15
答案在这哦
方法一：我们可以按照入职时间进行降序，入职最晚的就是时间最大的，
从 0 开始，筛选 n 条结果SELECT ... FROM table_name [WHERE ...] [ORDER BY ...] LIMIT n;
select * from employees order by hire_date desc limit 1;方法二：
-- 从 s 开始，筛选 n 条结果SELECT ... FROM table_name [WHERE ...] [ORDER BY ...] LIMIT s, n
select * from employees order by hire_date desc limit 1 offset 0;方法三：
我们可以使用子查询进行操作
select * from employees where hire_date = (select max(hire_date) from employees);SQL196 查找入职员工时间排名倒数第三的员工所有信息描述
查找入职员工时间排名倒数第三的员工所有信息_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/ec1ca44c62c14ceb990c3c40def1ec6c?tpId=82&tqId=29754&tPage=1&rp=&ru=%2Fta%2Fsql&qru=%2Fta%2Fsql%2Fquestion-ranking
有一个员工employees表简况如下:
| 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 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10004 | 1954-05-01 | Christian | Koblick | M | 1986-12-01 |
请你查找employees里入职员工时间排名倒数第三的员工所有信息，以上例子输出如下:
| emp_no | birth_date | first_name | last_name | gender | hire_date |
| 10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
注意：可能会存在同一个日期入职的员工，所以入职员工时间排名倒数第三的员工可能不止一个。
既然不止一个人那我们就选择distinct关键字进行去重
SELECT *
FROM employees
WHERE hire_date = (
SELECT DISTINCT hire_date
FROM employees
ORDER BY hire_date DESC
LIMIT 1 OFFSET 2
); SQL201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
描述
有一个薪水表，salaries简况如下:
| 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 |
| 10002 | 72527 | 1996-08-03 | 1997-08-03 |
请你查找薪水记录超过15条的员工号emp_no以及其对应的记录次数t，以上例子输出如下:
| emp_no | t |
| 10001 | 16 |
select
emp_no,
count(emp_no) as t
from
salaries
group by
emp_no
having
t > 15;
