数据库SQL实战练习(上)
题目来源 牛客网
1.查找最晚入职员工的所有信息
题目描述
查找最晚入职员工的所有信息
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;2.查找入职员工时间排名倒数第三的员工所有信息
题目描述
查找入职员工时间排名倒数第三的员工所有信息
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 2, 1;limit用法
SELECT * FROM tablename LIMIT 0,1
即取出第一条记录。
SELECT * FROM tablename LIMIT 1,1
第二条记录
SELECT * FROM tablename LIMIT 10,20
从第11条到31条（共计20条）3.查找各个部门当前领导当前薪水详情以及其对应部门编号
题目描述
查找各个部门当前(to_date=’9999-01-01’)领导当前薪水详情以及其对应部门编号dept_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 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 salaries.*, dept_manager.dept_no
FROM salaries INNER JOIN dept_manager
ON dept_manager.emp_no = salaries.emp_no
WHERE dept_manager.to_date = '9999-01-01'
AND salaries.to_date ='9999-01-01';4.查找所有已经分配部门的员工的last_name和first_name
题目描述
查找所有已经分配部门的员工的last_name和first_name
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 employees.last_name,employees.first_name, dept_emp.dept_no
FROM employees INNER JOIN dept_emp
ON employees.emp_no = dept_emp.emp_no;5.查找所有员工入职时候的薪水情况
题目描述
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
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 employees.emp_no,salaries.salary
FROM employees,salaries
WHERE employees.emp_no = salaries.emp_no
AND employees.hire_date = salaries.from_date
ORDER BY employees.emp_no DESC;6.查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
题目描述
查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数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) AS t
FROM salaries
GROUP BY emp_no
HAVING t>15;用COUNT()统计出现的次数，因为COUNT()是聚合函数所以要和GROUP BY配合使用对属于一组的数据起作用，GROUP BY 后跟处聚合函数以外的量
就是以xxx为分类标志统计xxx
WHERE 和HAVING：
WHERE语句在GROUP BY之前；SQL会在分组前计算WHERE语句
HAVING语句在GROUP BY之后；SQL会在分组后计算HAVING语句
7.找出所有员工当前具体的薪水情况
题目描述
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;SELECT DISTINCT可以去除重复值
8.获取所有部门当前manager的当前薪水情况
题目描述
获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示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 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 dept_manager.dept_no,dept_manager.emp_no,salaries.salary
FROM dept_manager INNER JOIN salaries
ON dept_manager.emp_no = salaries.emp_no
WHERE dept_manager.to_date = '9999-01-01'
AND salaries.to_date = '9999-01-01';9.获取所有非manager的员工emp_no
题目描述
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));
SELECT emp_no
FROM employees
WHERE emp_no NOT IN
(SELECT emp_no FROM dept_manager);使用NOT IN选出在A但不在B的记录
10.获取所有员工当前的manager
题目描述
获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date=’9999-01-01’。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。
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 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));
SELECT dept_emp.emp_no, dept_manager.emp_no
FROM dept_emp INNER JOIN dept_manager
ON dept_emp.dept_no = dept_manager.dept_no
WHERE dept_emp.emp_no <> dept_manager.emp_no
AND dept_manager.to_date = '9999-01-01';11.获取所有部门中当前员工薪水最高的相关信息
题目描述
获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
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));
SELECT dept_emp.dept_no, salaries.emp_no, MAX(salaries.salary) AS salary
FROM dept_emp INNER JOIN salaries
on dept_emp.emp_no = salaries.emp_no
WHERE dept_emp.to_date='9999-01-01' AND salaries.to_date='9999-01-01'
GROUP BY dept_emp.dept_no, salaries.emp_no;GROUP BY 后应该必须包含SELECT字段中除聚合函数以外的所有字段，但标准答案里却是不包含emp_no的？
12.从titles表获取按照title进行分组
题目描述
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);
SELECT title, COUNT(emp_no) AS t
FROM titles
GROUP BY title
HAVING COUNT(emp_no)>=2;13.从titles表获取按照title进行分组,注意对于重复的emp_no进行忽略。
题目描述
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。
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(DISTINCT emp_no) AS t
FROM titles
GROUP BY title
HAVING t >= 2;SQL COUNT() 语法 SQL COUNT(column_name) 语法 COUNT(column_name)
函数返回指定列的值的数目（NULL 不计入）： SELECT COUNT(column_name) FROM table_name SQL
COUNT() 语法 COUNT() 函数返回表中的记录数： SELECT COUNT(*) FROM table_name SQL
COUNT(DISTINCT column_name) 语法 COUNT(DISTINCT column_name)
函数返回指定列的不同值的数目： SELECT COUNT(DISTINCT column_name) FROM table_name
14. 查找employees表
题目描述
查找employees表所有emp_no为奇数，且last_name不为Mary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));
SELECT * FROM employees
WHERE emp_no%2 = 1
AND last_name != 'Mary'
ORDER BY hire_date desc;15. 统计出当前各个title类型对应的员工当前薪水对应的平均工资
题目描述
统计出当前各个title类型对应的员工当前薪水对应的平均工资。结果给出title以及平均工资avg。
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));
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 titles.title, AVG(salaries.salary) AS avg
FROM titles INNER JOIN salaries
ON titles.emp_no = salaries.emp_no
WHERE titles.to_date = '9999-01-01'
AND salaries.to_date = '9999-01-01'
GROUP BY titles.title;16. 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
题目描述
获取当前（to_date=’9999-01-01’）薪水第二多的员工的emp_no以及其对应的薪水salary
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, salary
FROM salaries
ORDER BY salary DESC
LIMIT 1,117.查找当前薪水排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
题目描述
查找当前薪水(to_date=’9999-01-01’)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
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 salaries.emp_no,MAX(salaries.salary),employees.last_name,employees.first_name
FROM salaries INNER JOIN employees
ON salaries.emp_no = employees.emp_no
WHERE salaries.to_date = '9999-01-01'
AND salaries.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = '9999-01-01');18.查找所有员工的last_name和first_name以及对应的dept_name
题目描述
查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工
CREATE TABLE departments (
dept_no char(4) NOT NULL,
dept_name varchar(40) NOT NULL,
PRIMARY KEY (dept_no));
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 employees.last_name,employees.first_name,departments.dept_name
FROM (employees LEFT JOIN dept_emp ON employees.emp_no = dept_emp.emp_no)
LEFT JOIN departments ON dept_emp.dept_no = departments.dept_no;19.查找员工编号emp_now为10001其自入职以来的薪水salary涨幅值
题目描述
查找员工编号emp_now为10001其自入职以来的薪水salary涨幅值growth
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 (
(SELECT salary FROM salaries WHERE emp_no = '10001' ORDER BY to_date DESC LIMIT 1)
-(SELECT salary FROM salaries WHERE emp_no = '10001' ORDER BY to_date ASC LIMIT 1)
) AS growth;20.查找所有员工自入职以来的薪水涨幅情况
题目描述
查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_noy以及其对应的薪水涨幅growth，并按照growth进行升序
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));
(1) 选出当前工资
SELECT employees.emp_no, salaries.salary
FROM employees LEFT JOIN salaries
ON employees.emp_no = salaries.emp_no
WHERE salaries.to_date = '9999-01-01';
(2) 选出入职工资
SELECT employees.emp_no,salaries.salary
FROM employees LEFT JOIN salaries
ON employees.emp_no = salaries.emp_no
WHERE employees.hire_date = salaries.from_date;
(3) 查找所有员工自入职以来的薪水涨幅
SELECT Current.emp_no, (Current.salary - Start.salary) AS growth
FROM
(SELECT employees.emp_no, salaries.salary
FROM employees LEFT JOIN salaries
ON employees.emp_no = salaries.emp_no
WHERE salaries.to_date = '9999-01-01') AS Current
INNER JOIN
(SELECT employees.emp_no,salaries.salary
FROM employees LEFT JOIN salaries
ON employees.emp_no = salaries.emp_no
WHERE employees.hire_date = salaries.from_date) AS Start
ON Current.emp_no = Start.emp_no
ORDER BY growth ASC;