数据库SQL实战题目详解（全61题）---（41-61）部分

题目来源：牛客网–《数据库SQL实战》
https://www.nowcoder.com/ta/sql?page=0
题目答案为博主自写已通过运行，题目难度近似于阶梯上升，可根据自身情况分部分作答。

其他题目链接：
1-20题链接：https://blog.csdn.net/weixin_41744624/article/details/104226505
21-40题链接：
https://blog.csdn.net/weixin_41744624/article/details/104412955
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另附可以使用的在线编程工具
sqlfiddle在线工具
网页版方便个人在闲暇时间的小题目练习
也可以从牛客网参考本答案直接作答
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41、

构造一个触发器audit_log，在向employees_test表中插入一条数据的时候，触发插入相关的数据到audit中。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
CREATE TABLE audit(
EMP_no INT NOT NULL,
NAME TEXT NOT NULL
);

CREATE TRIGGER audit_log AFTER INSERT ON employees_test
BEGIN
    INSERT INTO audit VALUES (NEW.ID, NEW.NAME);
END;

42、

删除emp_no重复的记录，只保留最小的id对应的记录。
CREATE TABLE IF NOT EXISTS 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’);

DELETE FROM titles_test WHERE id NOT IN 
(SELECT MIN(id) FROM titles_test GROUP BY emp_no) 

43、

将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。
CREATE TABLE IF NOT EXISTS 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’);

update titles_test
set to_date = null ,from_date='2001-01-01' where to_date='9999-01-01';

44、

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。
CREATE TABLE IF NOT EXISTS 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’);

UPDATE titles_test SET emp_no = REPLACE(emp_no,10001,10005) WHERE id = 5

45、

将titles_test表名修改为titles_2017。
CREATE TABLE IF NOT EXISTS 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’);

ALTER TABLE titles_test RENAME TO titles_2017

46、

在audit表上创建外键约束，其emp_no对应employees_test表的主键id。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);

CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL
);

DROP TABLE audit;
CREATE TABLE audit(
    EMP_no INT NOT NULL,
    create_date datetime NOT NULL,
    FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));

47、

存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
如何获取emp_v和employees有相同的数据？
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));
输出格式:
emp_no birth_date first_name last_name gender hire_date
10006 1953-04-20 Anneke Preusig F 1989-06-02
10007 1957-05-23 Tzvetan Zielinski F 1989-02-10
10008 1958-02-19 Saniya Kalloufi M 1994-09-15
10009 1952-04-19 Sumant Peac F 1985-02-18
10010 1963-06-01 Duangkaew Piveteau F 1989-08-24
10011 1953-11-07 Mary Sluis F 1990-01-22

SELECT em.* FROM employees AS em, emp_v AS ev WHERE em.emp_no = ev.emp_no

48、

将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));

update salaries 
set salary =salary*1.1
where emp_no in(select emp_no from emp_bonus)

49、

针对库中的所有表生成select count(*)对应的SQL语句
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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));
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));

输出格式:
cnts
select count() from employees;
select count() from departments;
select count() from dept_emp;
select count() from dept_manager;
select count() from salaries;
select count() from titles;
select count(*) from emp_bonus;

select "select count(*) from "||name||";" 
as cnts
from sqlite_master
where type='table'

50、

将employees表中的所有员工的last_name和first_name通过(’)连接起来。
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));

输出格式:
name
Facello’Georgi
Simmel’Bezalel
Bamford’Parto
Koblick’Chirstian
Maliniak’Kyoichi
Preusig’Anneke
Zielinski’Tzvetan
Kalloufi’Saniya
Peac’Sumant
Piveteau’Duangkaew
Sluis’Mary

select last_name||"'"||first_name from employees

51、

查找字符串’10,A,B’ 中逗号’,'出现的次数cnt。

SELECT (length("10,A,B")-length(replace("10,A,B",",","")))/length(",") AS cnt

52、

获取Employees中的first_name，查询按照first_name最后两个字母，按照升序进行排列
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));
输出格式：
first_name
Chirstian
Tzvetan
Bezalel
Duangkaew
Georgi
Kyoichi
Anneke
Sumant
Mary
Parto
Saniya

SELECT first_name FROM employees 
ORDER BY substr(first_name,length(first_name)-1) 

53、

按照dept_no进行汇总，属于同一个部门的emp_no按照逗号进行连接，结果给出dept_no以及连接出的结果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));
输出格式:
dept_no employees
d001 10001,10002
d002 10006
d003 10005
d004 10003,10004
d005 10007,10008,10010
d006 10009,10010

SELECT dept_no, group_concat(emp_no) AS employees
FROM dept_emp GROUP BY dept_no

54、

查找排除当前最大、最小salary之后的员工的平均工资avg_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));
输出格式:
avg_salary
69462.5555555556

SELECT AVG(salary) AS avg_salary FROM salaries 
WHERE to_date = '9999-01-01' 
AND salary NOT IN (SELECT MAX(salary) FROM salaries)
AND salary NOT IN (SELECT MIN(salary) FROM salaries)

55、

分页查询employees表，每5行一页，返回第2页的数据
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 limit 5,5

56、

获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和received ，没有分配具体的员工不显示
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));
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));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
输出格式:
e.emp_no dept_no btype received
10001 d001 1 2010-01-01
10002 d001 2 2010-10-01
10003 d004 3 2011-12-03
10004 d004 1 2010-01-01
10005 d003
10006 d002
10007 d005
10008 d005
10009 d006
10010 d005
10010 d006

SELECT em.emp_no, de.dept_no, eb.btype, eb.recevied
FROM employees AS em INNER JOIN dept_emp AS de
ON em.emp_no = de.emp_no
LEFT JOIN emp_bonus AS eb 
ON de.emp_no = eb.emp_no

57、

使用含有关键字exists查找未分配具体部门的员工的所有信息。
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));
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));
输出格式:
emp_no birth_date first_name last_name gender hire_date
10011 1953-11-07 Mary Sluis F 1990-01-22

select * from employees 
where not exists(
select emp_no from dept_emp  WHERE emp_no = employees.emp_no)

58、

存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
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));
获取employees中的行数据，且这些行也存在于emp_v中。注意不能使用intersect关键字。
输出格式:
emp_no birth_date first_name last_name gender hire_date
10006 1953-04-20 Anneke Preusig F 1989-06-02
10007 1957-05-23 Tzvetan Zielinski F 1989-02-10
10008 1958-02-19 Saniya Kalloufi M 1994-09-15
10009 1952-04-19 Sumant Peac F 1985-02-18
10010 1963-06-01 Duangkaew Piveteau F 1989-08-24
10011 1953-11-07 Mary Sluis F 1990-01-22

SELECT em.* FROM employees AS em, emp_v AS ev WHERE em.emp_no = ev.emp_no

59、

获取有奖金的员工相关信息。
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));
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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));
给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date=‘9999-01-01’
输出格式:
emp_no first_name last_name btype salary bonus
10001 Georgi Facello 1 88958 8895.8
10002 Bezalel Simmel 2 72527 14505.4
10003 Parto Bamford 3 43311 12993.3
10004 Chirstian Koblick 1 74057 7405.7

SELECT e.emp_no, e.first_name, e.last_name, b.btype, s.salary, 
(CASE b.btype 
 WHEN 1 THEN s.salary * 0.1
 WHEN 2 THEN s.salary * 0.2
 ELSE s.salary * 0.3 END) AS bonus
FROM employees AS e INNER JOIN emp_bonus AS b ON e.emp_no = b.emp_no
INNER JOIN salaries AS s ON e.emp_no = s.emp_no AND s.to_date = '9999-01-01'

60、

按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。。
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));
输出格式:
emp_no salary running_total
10001 88958 88958
10002 72527 161485
10003 43311 204796
10004 74057 278853
10005 94692 373545
10006 43311 416856
10007 88070 504926
10009 95409 600335
10010 94409 694744
10011 25828 720572

SELECT s1.emp_no, s1.salary, 
(SELECT SUM(s2.salary) FROM salaries AS s2 
 WHERE s2.emp_no <= s1.emp_no AND s2.to_date = '9999-01-01') AS running_total 
FROM salaries AS s1 WHERE s1.to_date = '9999-01-01' ORDER BY s1.emp_no

61、

对于employees表中，给出奇数行的first_name
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));
输出格式:
first_name
Georgi
Chirstian
Anneke
Tzvetan
Saniya
Mary

SELECT e1.first_name FROM 
  (SELECT e2.first_name, 
    (SELECT COUNT(*) FROM employees AS e3 
     WHERE e3.first_name <= e2.first_name) 
   AS rowid FROM employees AS e2) AS e1
WHERE e1.rowid % 2 = 1