文章目錄
1.批量插入數據
對於表actor批量插入如下數據
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 |
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');
- 批量插入數據
- 來源:牛客網(NOWCODER)
2.找出所有員工當前薪水salary情況
找出所有員工當前(to_date=‘9999-01-01’)具體的薪水salary情況,對於相同的薪水只顯示一次,並按照逆序顯示
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 salary from salaries where to_date='9999-01-01' group by salary order by salary desc;
select distinct salary from salaries where to_date = '9999-01-01' order by salary desc;
- 找出所有員工當前薪水salary情況
- 來源:牛客網(NOWCODER)
3.查找最晚入職員工的所有信息
查找最晚入職員工的所有信息,創建語句如下:
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 emp_no,birth_date,first_name,last_name,gender,hire_date from employees order by hire_date desc limit 1 offset 0;
-- select * from employees order by hire_date desc limit 1;
-- select * from employees order by hire_date desc limit 0,1;
-- select * from employees where hire_date in (select max(hire_date) from employees);
-- select * from employees where hire_date == (select max(hire_date) from employees);
- 查找最晚入職員工的所有信息
- 來源:牛客網(NOWCODER)
4.查找入職員工時間排名倒數第三的員工所有信息
查找入職員工時間排名倒數第三的員工所有信息,創建語句如下:
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 emp_no,birth_date,first_name,last_name,gender,hire_date from employees order by hire_date desc limit 1 offset 2;
-- select * from employees order by hire_date desc limit 1 offset 2;
select * from employees order by hire_date desc limit 2,1;
- 查找入職員工時間排名倒數第三的員工所有信息
- 來源:牛客網(NOWCODER)
5.查找薪水漲幅超過15次的員工號emp_no以及其對應的漲幅次數t
查找薪水漲幅超過15次的員工號emp_no以及其對應的漲幅次數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) as t from salaries group by emp_no having count(emp_no) > 15;
- 查找薪水漲幅超過15次的員工號emp_no以及其對應的漲幅次數t
- 來源:牛客網(NOWCODER)
6.獲取所有部門當前manager的當前薪水情況
獲取所有部門當前manager的當前薪水情況,給出dept_no, emp_no以及salary,當前表示to_date=‘9999-01-01’
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 TABLEsalaries
(
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
));
- 獲取所有部門當前manager的當前薪水情況
- 來源:牛客網(NOWCODER)
7.從titles表獲取按照title進行分組
從titles表獲取按照title進行分組,每組個數大於等於2,給出title以及對應的數目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);
select title,count(title) as t from titles group by title having t >= 2;
- 從titles表獲取按照title進行分組
- 來源:牛客網(NOWCODER)
8.查找重複的電子郵箱
編寫一個 SQL 查詢,查找 Person 表中所有重複的電子郵箱。
示例
±—±--------+
| Id | Email |
±—±--------+
| 1 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
±—±--------+
根據以上輸入,你的查詢應返回以下結果:
±--------+
| Email |
±--------+
| [email protected] |
±--------+
說明:所有電子郵箱都是小寫字母。
# Write your MySQL query statement below
# select Email from Person group by Email having count(Email) > 1;
select Email from person where Id in(select Id from person group by Email having count(Email) > 1);
- 查找重複的電子郵箱
- 來源:力扣(LeetCode)
9.第N高的薪水
編寫一個 SQL 查詢,獲取 Employee 表中第 n 高的薪水(Salary)
±—±-------+
| Id | Salary |
±—±-------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
±—±-------+
例如上述 Employee 表,n = 2 時,應返回第二高的薪水 200。如果不存在第 n 高的薪水,那麼查詢應返回 null。
±-----------------------+
| getNthHighestSalary(2) |
±-----------------------+
| 200 |
±-----------------------+
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
set N = N-1;
RETURN (
# Write your MySQL query statement below.
select distinct salary from Employee order by salary desc limit 1 offset N
);
END
- 第N高的薪水
- 來源:力扣(LeetCode)