MySQL之CRUD

insert into t_mysql_employees(first_name,last_name,email,phone_number,salary,commission_pct,manager_id,hiredate) values('li','si','29494247@qq.com','13874388318',3000,0.25,100,now());

3.查询数据

select * from t_mysql_employees where phone_number='13874388318'

4.修改数据

update t_mysql_employees set last_name='jiang' where phone_number='13874388318'

5.删除数据

delete from t_mysql_employees where phone_number='13874388318'

6.在oracle查询当前日期数据

select sysdate from dual;

7.在mysql里面查询当前日期数据

select now();

8.在mysql里面+号它只是一个运算符，它不做为拼接符合


select 'zhangsan'+1;    答案：1
select '100'+1;        答案：101

二、过滤查询案例

1、按条件表达式筛选

①、案例1：查询工资>12000的员工信息

	select * from t_mysql_employees where salary >12000;

②、案例2：查询部门编号不等于90号的员工名和部门编号


方式一：
    select last_name,department_id from t_mysql_employees where department_id <> 90;
方式二：
	select last_name,department_id from t_mysql_employees where not(department_id = 90)

2、按逻辑表达式筛选

①.案例1：查询工资在10000到20000之间的员工名、工资以及奖金


方式一：	
select last_name,salary,commission_pct from t_mysql_employees where salary between 10000 and 20000
 
方式二：
select last_name,salary,commission_pct from t_mysql_employees where salary>=10000 and salary<=20000

②.案例2：查询部门编号不是在90到110之间，或者工资高于15000的员工信息

	select * from t_mysql_employees where not(department_id between 90 and 110) or salary>15000;

3、模糊查询

①.like

案例1：查询员工名中包含字符a的员工信息

	select * from t_mysql_employees where last_name like '%a%';

案例2：查询员工名中第二个字符为e,第四个字符为a的员工名和工资

	select last_name,salary from t_mysql_employees where last_name like '_e_a%';

案例3：查询员工名中第二个字符为_的员工名

	select * from t_mysql_employees where last_name like '_$_%' escape '$';

②.between and

案例1：查询员工在100到120之间的员工信息

	select * from t_mysql_employees where employee_id between 100 and 120;

③.in

案例：查询员工的工种编号是IT_PROG、AD_VP、AF_PRES中的员工员工名和工种编号

	select last_name,job_id from t_mysql_employees where job_id in('IT_PROG','AD_VP','AD_PRES');

④.is null

案例1：查询没有奖金的员工名和奖金率

	select last_name,commission_pct from t_mysql_employees where commission_pct is null;

案例2：查询有奖金的员工名和奖金率

		select last_name,commission_pct from t_mysql_employees where commission_pct is not null;

⑤.安全等于

案例1：查询没有奖金的员工名和奖金率

				select last_name,commission_pct from t_mysql_employees where commission_pct <=> null;

案例2：查询工资为12000的员工信息

SELECT last_name,salary FROM t_mysql_employees WHERE salary <=> 12000;

它能等价于 is null 也能等价于 = 0.4


	select last_name,commission_pct from t_mysql_employees where commission_pct is null;
		
		select last_name,commission_pct from t_mysql_employees where commission_pct is 0.4;
		
			select last_name,commission_pct from t_mysql_employees where commission_pct = null;
			
				select last_name,commission_pct from t_mysql_employees where commission_pct <=> 0.4;

⑥.is null pk <=>

is null :仅仅可以判断null值，可读性较高，建议使用

<=>：即可以判断null值，又可以判断普通的数值，可读性较低

三、排序 order by 子句

1、按单个字段排序

①.案例：按员工表薪资排序

	select * from t_mysql_employees order by salary;

2、添加筛选条件再排序

①.案例：查询部门编号>=90的员工信息，并按员工编号降序

	select * from t_mysql_employees where department_id>=90 order by employee_id desc;

3、按表达式排序

①.案例：查询员信息，按年薪降序


	select em.salary*12*(1+IFNULL(commission_pct,0)),em.* from t_mysql_employees em order by em.salary*12*(1+IFNULL(commission_pct,0)) desc;
	
	select 1+'json';   答案：1
	select 1+null;     答案：null

4、按别名排序

①.案例1：查询总记录数，并且取一个别名n


		select count(1) as n from t_mysql_employees

②.案例2：查询员工信息，按年薪升序

	select em.salary*12*(1+IFNULL(commission_pct,0)),em.* from t_mysql_employees em order by em.salary*12*(1+IFNULL(commission_pct,0)) asc;

5、按函数排序

①.案例：查询员工名，并且按名字的长度降序


#先把名字拼接起来
	select first_name,last_name from t_mysql_employees;
 
	select last_name,CONCAT(first_name,last_name),LENGTH(CONCAT(first_name,last_name)) from t_mysql_employees order by LENGTH(CONCAT(first_name,last_name)) desc;

6、按多个字段排序

案例:查询员工信息，要求先按工资降序，再按employee_id升序

select * from t_mysql_employees order by salary desc,employee_id asc;

四、案例

1、表结构要求：

-- 1.学生表-t_student
-- sid 学生编号,sname 学生姓名,sage 学生年龄,ssex 学生性别

-- 2.教师表-t_teacher
-- tid 教师编号,tname 教师名称

-- 3.课程表-t_course
-- cid 课程编号,cname 课程名称,tid 教师名称

-- 4.成绩表-t_score
-- sid 学生编号,cid 课程编号,score 成绩

2、表数据

学生表


insert into t_student values('01' , '赵雷' , '1990-01-01' , '男');
insert into t_student values('02' , '钱电' , '1990-12-21' , '男');
insert into t_student values('03' , '孙风' , '1990-12-20' , '男');
insert into t_student values('04' , '李云' , '1990-12-06' , '男');
insert into t_student values('05' , '周梅' , '1991-12-01' , '女');
insert into t_student values('06' , '吴兰' , '1992-01-01' , '女');
insert into t_student values('07' , '郑竹' , '1989-01-01' , '女');
insert into t_student values('09' , '张三' , '2017-12-20' , '女');
insert into t_student values('10' , '李四' , '2017-12-25' , '女');
insert into t_student values('11' , '李四' , '2012-06-06' , '女');
insert into t_student values('12' , '赵六' , '2013-06-13' , '女');
insert into t_student values('13' , '孙七' , '2014-06-01' , '女');

教师表


insert into t_teacher values('01' , '张三');
insert into t_teacher values('02' , '李四');
insert into t_teacher values('03' , '王五');

课程表


insert into t_course values('01' , '语文' , '02');
insert into t_course values('02' , '数学' , '01');
insert into t_course values('03' , '英语' , '03');

成绩表


insert into t_score values('01' , '01' , 80);
insert into t_score values('01' , '02' , 90);
insert into t_score values('01' , '03' , 99);
insert into t_score values('02' , '01' , 70);
insert into t_score values('02' , '02' , 60);
insert into t_score values('02' , '03' , 80);
insert into t_score values('03' , '01' , 80);
insert into t_score values('03' , '02' , 80);
insert into t_score values('03' , '03' , 80);
insert into t_score values('04' , '01' , 50);
insert into t_score values('04' , '02' , 30);
insert into t_score values('04' , '03' , 20);
insert into t_score values('05' , '01' , 76);
insert into t_score values('05' , '02' , 87);
insert into t_score values('06' , '01' , 31);
insert into t_score values('06' , '03' , 34);
insert into t_score values('07' , '02' , 89);
insert into t_score values('07' , '03' , 98);

3.案例题目：

01）查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数


select * from t_student right join (select a.SId,a.score class1,b.score class2 from (
SELECT * from t_score WHERE t_score.cid='01'
) as a,(
SELECT * from t_score WHERE t_score.cid='02'
) as b where a.sid=b.sid and a.score>b.score
)r on t_student.sid=r.sid

思路：

02）查询同时存在" 01 "课程和" 02 "课程的情况


select * from 
(select * from t_score where t_score.cid='01')as a join (select * from t_score where t_score.cid='02') as b on a.sid=b.sid

03）查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

【既left join ，满足左边条件，右边可能为空】

【既right join ，满足右边条件，左边可能为空


select * from(
select * from t_score where t_score.cid='01')as a left join (select * from t_score where t_score.cid='02')as b on a.sid=b.sid

思路：

04）查询不存在" 01 "课程但存在" 02 "课程的情况

select * from t_score where t_score.sid not in (select sid from t_score where t_score.cid='01') and t_score.cid='02'

05）查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

select s.sid ,sname,AVG(score) from t_student s join t_score c on s.sid=c.sid GROUP BY s.sid HAVING AVG(score) >=60

06）查询在t_score表存在成绩的学生信息

select distinct s.* from t_student s join t_score c on s.sid=c.cid

07）查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

select s.sid,s.sname,count(c.cid),sum(c.score) from t_student s join t_score c on s.sid=c.sid group by s.sid

08）查询「李」姓老师的数量

select count(*) from t_teacher where tname like '李%'

09）查询学过「张三」老师授课的同学的信息

select distinct s.* from t_student s join t_score c on s.sid=c.sid join t_course o on c.cid=o.cid join t_teacher t on t.tid=o.tid where t.tname='张三'

10）查询没有学全所有课程的同学的信息


select * from t_student where sid not in(select a.sid from t_student a,t_score b,t_score c,t_score d where a.sid=b.sid and a.sid=c.sid and b.cid='01' and c.cid='02' and d.cid='03')

11）查询没学过"张三"老师讲授的任一门课程的学生姓名


 select s.sname from t_student s where sid not in (select sid from t_course c join t_teacher t on c.tid=t.tid join t_score on t_score.cid=c.cid where t.tname='张三')

12）查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select s.sid,s.sname,avg(score) from t_student as s,t_score where s.sid=t_score.sid and score<60 group by s.sid having count(score)>=2

13）检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select s.* from t_student s join t_score sc on s.sid=sc.sid where cid = 01 and score < 60 order by score desc

14）按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select s.*,score from t_score as s left join (select t_score.sid,avg(t_score.score) from t_score group by t_score.sid) r on s.sid=r.sid order by score desc

15）查询各科成绩最高分、最低分和平均分：
以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列


SELECT cid,MAX(score) AS '最高分', MIN(score)AS '最低分' ,AVG(score) AS '平均分' ,COUNT(*) AS '选修人数',
     SUM(CASE WHEN t_score.score >= 60 THEN 1 ELSE 0 END)/COUNT(*) AS '及格率',
     SUM(CASE WHEN t_score.score >= 70 AND t_score.score< 80 THEN 1 ELSE 0 END)/COUNT(*) AS '中等率',
     SUM(CASE WHEN t_score.score >= 80 AND t_score.score<90 THEN 1 ELSE 0 END)/COUNT(*) AS '优良率',
     SUM(CASE WHEN t_score.score >= 90 THEN 1 ELSE 0 END)/COUNT(*) AS '优秀率'
FROM t_score GROUP BY cid ORDER BY COUNT(*)DESC,t_score.cid ASC ;

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原文地址：https://blog.csdn.net/weixin_67465673/article/details/125573355