mysql 多表查询

一介绍

本节主题

多表连接查询

复合条件连接查询

子查询

准备表

company.employee
company.department

#建表
create table department(
id int,
name varchar(20) 
);

create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);

#插入数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营');

insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('liwenzhou','male',18,200),
('jingliyang','female',18,204)
;


#查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+

mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

mysql> select * from employee;
+----+------------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+------------+--------+------+--------+
| 1 | egon | male | 18 | 200 |
| 2 | alex | female | 48 | 201 |
| 3 | wupeiqi | male | 38 | 201 |
| 4 | yuanhao | female | 28 | 202 |
| 5 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
+----+------------+--------+------+--------+

二多表连接查询

#重点：外链接语法

SELECT 字段列表
    FROM 表1 INNER|LEFT|RIGHT JOIN 表2
    ON 表1.字段 = 表2.字段;

1 交叉连接：不适用任何匹配条件。生成笛卡尔积

mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name       | sex    | age  | dep_id | id   | name         |
+----+------------+--------+------+--------+------+--------------+
|  1 | egon       | male   |   18 |    200 |  200 | 技术         |
|  1 | egon       | male   |   18 |    200 |  201 | 人力资源     |
|  1 | egon       | male   |   18 |    200 |  202 | 销售         |
|  1 | egon       | male   |   18 |    200 |  203 | 运营         |
|  2 | alex       | female |   48 |    201 |  200 | 技术         |
|  2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|  2 | alex       | female |   48 |    201 |  202 | 销售         |
|  2 | alex       | female |   48 |    201 |  203 | 运营         |
|  3 | wupeiqi    | male   |   38 |    201 |  200 | 技术         |
|  3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|  3 | wupeiqi    | male   |   38 |    201 |  202 | 销售         |
|  3 | wupeiqi    | male   |   38 |    201 |  203 | 运营         |
|  4 | yuanhao    | female |   28 |    202 |  200 | 技术         |
|  4 | yuanhao    | female |   28 |    202 |  201 | 人力资源     |
|  4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|  4 | yuanhao    | female |   28 |    202 |  203 | 运营         |
|  5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
+----+------------+--------+------+--------+------+--------------+

2 内连接：只连接匹配的行

#找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了正确的结果
#department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
+----+-----------+------+--------+--------------+
| id | name      | age  | sex    | name         |
+----+-----------+------+--------+--------------+
|  1 | egon      |   18 | male   | 技术         |
|  2 | alex      |   48 | female | 人力资源     |
|  3 | wupeiqi   |   38 | male   | 人力资源     |
|  4 | yuanhao   |   28 | female | 销售         |
|  5 | liwenzhou |   18 | male   | 技术         |
+----+-----------+------+--------+--------------+

#上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;

3 外链接之左连接：优先显示左表全部记录

#以左表为准，即找出所有员工信息，当然包括没有部门的员工
#本质就是：在内连接的基础上增加左边有右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name       | depart_name  |
+----+------------+--------------+
|  1 | egon       | 技术         |
|  5 | liwenzhou  | 技术         |
|  2 | alex       | 人力资源     |
|  3 | wupeiqi    | 人力资源     |
|  4 | yuanhao    | 销售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+

4 外链接之右连接：优先显示右表全部记录

#以右表为准，即找出所有部门信息，包括没有员工的部门
#本质就是：在内连接的基础上增加右边有左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id   | name      | depart_name  |
+------+-----------+--------------+
|    1 | egon      | 技术         |
|    2 | alex      | 人力资源     |
|    3 | wupeiqi   | 人力资源     |
|    4 | yuanhao   | 销售         |
|    5 | liwenzhou | 技术         |
| NULL | NULL      | 运营         |
+------+-----------+--------------+

5 全外连接：显示左右两个表全部记录

#注意 union与union all的区别：union会去掉相同的纪录

全外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意：mysql不支持全外连接 full JOIN
#强调：mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|    2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|    3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|    4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|    6 | jingliyang | female |   18 |    204 | NULL | NULL         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+

三符合条件连接查询

#示例1：以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25,即找出公司所有部门中年龄大于25岁的员工
select employee.name,employee.age from employee,department
    where employee.dep_id = department.id
    and age > 25;

#示例2：以内连接的方式查询employee和department表，并且以age字段的升序方式显示
select employee.id,employee.name,employee.age,department.name from employee,department
    where employee.dep_id = department.id
    and age > 25
    order by age asc;

四子查询

#1：子查询是将一个查询语句嵌套在另一个查询语句中。
#2：内层查询语句的查询结果，可以为外层查询语句提供查询条件。
#3：子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4：还可以包含比较运算符：= 、 !=、> 、<等

1 带IN关键字的子查询

#查询employee表，但dep_id必须在department表中出现过
select * from employee
    where dep_id in
        (select id from department);

2 带比较运算符的子查询\

#比较运算符：=、!=、>、>=、<、<=、<>
#查询平均年龄在25岁以上的部门名
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);

#查看技术部员工姓名
select name from employee
    where dep_id in 
        (select id from department where name='技术');

#查看不足1人的部门名
select name from department
    where id in 
        (select dep_id from employee group by dep_id having count(id) <=1);

3 带EXISTS关键字的子查询

EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。
而是返回一个真假值。True或False
当返回True时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询

#department表中存在dept_id=203，Ture
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=200);
+----+------------+--------+------+--------+
| id | name       | sex    | age  | dep_id |
+----+------------+--------+------+--------+
|  1 | egon       | male   |   18 |    200 |
|  2 | alex       | female |   48 |    201 |
|  3 | wupeiqi    | male   |   38 |    201 |
|  4 | yuanhao    | female |   28 |    202 |
|  5 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | female |   18 |    204 |
+----+------------+--------+------+--------+

#department表中存在dept_id=205，False
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=204);
Empty set (0.00 sec)

五综合练习

init.sql文件内容

View Code

从init.sql文件中导入数据

#准备表、记录
mysql> create database db1;
mysql> use db1;
mysql> source /root/init.sql

！！！重中之重：练习之前务必搞清楚sql逻辑查询语句的执行顺序

链接：http://www.cnblogs.com/linhaifeng/articles/7372774.html

1、查询所有的课程的名称以及对应的任课老师姓名

2、查询学生表中男女生各有多少人

3、查询物理成绩等于100的学生的姓名

4、查询平均成绩大于八十分的同学的姓名和平均成绩

5、查询所有学生的学号，姓名，选课数，总成绩

6、 查询姓李老师的个数

7、 查询没有报李平老师课的学生姓名

8、 查询物理课程比生物课程高的学生的学号

9、 查询没有同时选修物理课程和体育课程的学生姓名

10、查询挂科超过两门(包括两门)的学生姓名和班级
、查询选修了所有课程的学生姓名

12、查询李平老师教的课程的所有成绩记录
 
13、查询全部学生都选修了的课程号和课程名

14、查询每门课程被选修的次数

15、查询之选修了一门课程的学生姓名和学号

16、查询所有学生考出的成绩并按从高到低排序（成绩去重）

17、查询平均成绩大于85的学生姓名和平均成绩

18、查询生物成绩不及格的学生姓名和对应生物分数

19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名

20、查询每门课程成绩最好的前两名学生姓名

21、查询不同课程但成绩相同的学号，课程号，成绩

22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；

23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；

24、任课最多的老师中学生单科成绩最高的学生姓名

1、查询所有的课程的名称以及对应的任课老师姓名
SELECT c.cname,t.tname FROM course c LEFT JOIN teacher t ON c.teacher_id=t.tid 

2、查询学生表中男女生各有多少人

SELECT gender,COUNT(sid) FROM student GROUP BY gender

3、查询物理成绩等于100的学生的姓名
SELECT stu.sname FROM student stu LEFT JOIN  score s ON s.student_id=stu.sid LEFT JOIN
course c ON c.cid=s.course_id WHERE s.num=100 AND c.cname="物理"

SELECT sname FROM student WHERE sid IN(SELECT student_id FROM score WHERE 
num=100 AND course_id =(SELECT cid FROM course WHERE cname="物理"))

4、查询平均成绩大于八十分的同学的姓名和平均成绩
SELECT s.sname, 平均成绩 FROM student s RIGHT JOIN 
(SELECT student_id ,AVG(num) 平均成绩 FROM score GROUP BY student_id HAVING AVG(num)>80 )AS sc
ON s.sid=sc.student_id

5、查询所有学生的学号，姓名，选课数，总成绩
SELECT s.sid 学号,s.sname 姓名,选课数,总成绩 FROM student s RIGHT JOIN  
(SELECT student_id,COUNT(course_id) 选课数,SUM(num) 总成绩 FROM score 
GROUP BY student_id)AS sc 
ON s.sid=sc.student_id

6、 查询姓李老师的个数
SELECT COUNT(tid) FROM  teacher WHERE tname LIKE "李%"


7、 查询没有报李平老师课的学生姓名
SELECT sname FROM student WHERE sid  NOT IN
(SELECT student_id FROM score WHERE course_id IN
(SELECT cid FROM course WHERE teacher_id =(SELECT tid FROM teacher WHERE tname="李平老师")))

SELECT sname FROM student WHERE sid NOT IN (
SELECT student_id FROM score s LEFT JOIN course c ON c.cid=s.course_id WHERE c.cid IN(
SELECT cid FROM course c INNER JOIN teacher t ON t.tid=c.teacher_id WHERE t.tname="李平老师"
)
)
8、查询物理课程比生物课程高的学生的学号
SELECT * FROM 
(SELECT  student_id 学生号, num 成绩,cname 课程 FROM score LEFT JOIN course ON course_id=score.course_id WHERE cname="物理")AS wl
 INNER JOIN
(SELECT  student_id 学生号, num 成绩,cname 课程 FROM score LEFT JOIN course ON course_id=score.course_id WHERE cname="生物")AS sw
 ON wl.成绩>sw.成绩
 
 SELECT t1.student_id FROM 
(SELECT student_id,num FROM score s INNER JOIN course c ON c.cid=s.course_id WHERE c.cname="物理")AS t1
JOIN
(SELECT student_id,num FROM score s INNER JOIN course c ON c.cid=s.course_id WHERE c.cname="生物")AS t2 ON
t1.student_id=t2.student_id WHERE t1.num>t2.num

9、查询没有同时选修物理课程和体育课程的学生姓名
SELECT sname FROM student WHERE sid NOT IN(
SELECT student_id FROM score WHERE course_id IN (SELECT cid FROM course
 WHERE cname IN ("物理","生物"))GROUP BY student_id  HAVING COUNT(course_id)=2
 )
10、查询挂科超过两门(包括两门)的学生姓名和班级
SELECT sname,cl.caption FROM student stu ,class cl WHERE stu.sid IN(
SELECT s.student_id FROM score s LEFT JOIN course c ON s.course_id = c.cid WHERE num<60 GROUP BY student_id HAVING COUNT(sid)>=2
)AND stu.class_id=cl.cid

SELECT sname,caption FROM student stu ,class cl WHERE stu.sid IN(
  SELECT student_id FROM score WHERE num<60 GROUP BY student_id HAVING COUNT(sid)>=2)
AND stu.class_id=cl.cid

SELECT t2.sname ,cl.caption FROM (
SELECT sname ,class_id FROM student stu INNER JOIN 
( SELECT student_id  FROM score WHERE num<60 GROUP BY student_id HAVING COUNT(course_id)>=2)
AS t1 ON stu.sid=t1.student_id 
)AS t2 INNER JOIN class cl  ON  cl.cid=t2.class_id

11 、查询选修了所有课程的学生姓名

SELECT s.sid, s.sname FROM student s WHERE s.sid IN(
SELECT s.student_id  FROM score s GROUP BY student_id HAVING COUNT(s.sid)=(SELECT COUNT(cid) FROM course)
)

12、查询李平老师教的课程的所有成绩记录
SELECT s.num ,c.cname FROM score s INNER JOIN course c  ON s.course_id=c.cid WHERE c.cid IN
(SELECT c.cid FROM course c  INNER JOIN  teacher t ON c.teacher_id=t.tid WHERE tname="李平老师")

SELECT s.num ,c.cname FROM score s LEFT JOIN  course c ON c.cid=s.course_id LEFT 
JOIN teacher t ON t.tid=c.teacher_id WHERE t.tname="李平老师"

13、查询全部学生都选修了的课程号和课程名
SELECT cid,cname FROM course WHERE cid IN(
SELECT course_id FROM score GROUP BY course_id  HAVING COUNT(sid)=(
SELECT COUNT(sid) FROM student
)
)

SELECT c.cid,c.cname FROM course c WHERE c.cid IN(
SELECT course_id FROM score GROUP BY course_id HAVING COUNT(sid)=(SELECT COUNT(sid) FROM student)
)
14、查询每门课程被选修的次数 

SELECT c.cname ,COUNT(sid) FROM course c LEFT JOIN score s ON  s.course_id =c.cid GROUP BY s.course_id

15、查询只选修了一门课程的学生姓名和学号
SELECT sid,sname FROM student WHERE sid IN(SELECT student_id FROM score GROUP BY student_id HAVING COUNT(sid)=1)

16、查询所有学生考出的成绩并按从高到低排序（成绩去重）
查询每个考生对应的成绩
SELECT stu.sname,s.num FROM student stu LEFT JOIN score s ON s.student_id=stu.sid ORDER BY s.num DESC
查询所有学生考出的成绩并按从高到低排序（成绩去重）
SELECT DISTINCT num FROM score ORDER BY num DESC


17查询平均成绩大于85的学生姓名和平均成绩

SELECT stu.sname 姓名,AVG(s.num )平均成绩 FROM student stu INNER JOIN score s ON 
stu.sid =s.student_id GROUP BY s.student_id HAVING AVG(num)>85 

18、查询生物成绩不及格的学生姓名和对应生物分数

SELECT stu.sname 姓名,s.num 生物成绩 FROM score s 
LEFT JOIN student stu  ON stu.sid=s.student_id
LEFT JOIN course c ON s.course_id=c.cid 
WHERE c.cname="生物" AND s.num<60

19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)
平均成绩最高的学生姓名

SELECT stu.sname FROM score s LEFT JOIN student stu ON s.student_id=stu.sid
LEFT JOIN course c ON c.cid=s.course_id WHERE c.cid IN (SELECT c.cid FROM course c LEFT JOIN
teacher t ON c.teacher_id =t.tid WHERE t.tname="李平老师"
) GROUP BY student_id ORDER BY AVG(s.num) DESC LIMIT 1

 
20、查询每门课程成绩最好的前两名学生姓名

 SELECT s.sid,s.course_id,s.num,ss.first_num,ss.second_num FROM score s LEFT JOIN
    (
    SELECT
        sid,
        (SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 0,1) AS first_num,
        (SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 1,1) AS second_num
    FROM
        score AS s1
    ) AS ss
    ON s.sid =ss.sid
    WHERE s.num <= ss.first_num AND s.num >= ss.second_num
    

21、查询不同课程但成绩相同的学号，课程号，成绩

SELECT  s1.student_id,  s1.course_id, s1.num, s2.student_id , s2.course_id,s2.num FROM score AS s1, 
score AS s2 WHERE s1.num = s2.num AND s1.course_id != s2.course_id AND s1.student_id!=
s2.student_id;
 

22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；

SELECT stu.sname 姓名 FROM student stu WHERE stu.sid NOT IN (
SELECT student_id FROM score WHERE course_id IN(
SELECT cid FROM course WHERE teacher_id=(
SELECT tid FROM teacher WHERE tname="李平老师"
)
)
)



23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；
SELECT DISTINCT stu.sid, stu.sname FROM student stu LEFT JOIN score s ON  s.student_id=stu.sid WHERE s.course_id IN(
SELECT course_id FROM score WHERE student_id =1)

SELECT stu.sid,stu.sname  FROM student stu WHERE stu.sid IN(
SELECT student_id FROM score WHERE course_id IN (SELECT course_id FROM score WHERE student_id =1)
)

24、任课最多的老师中学生单科成绩最高的学生姓名
SELECT stu.sname FROM score s LEFT JOIN  student stu ON stu.sid=s.student_id
LEFT JOIN course c ON s.course_id =c.cid WHERE c.teacher_id =(
SELECT teacher_id FROM course  GROUP BY teacher_id ORDER BY COUNT(cid) DESC LIMIT 1)
GROUP BY s.student_id  ORDER BY AVG(s.num) DESC LIMIT 2

SELECT stu.sname FROM student stu LEFT JOIN score s ON s.student_id =stu.sid
LEFT JOIN course c ON c.cid=s.course_id WHERE cid IN( SELECT cid FROM course WHERE teacher_id =(SELECT teacher_id FROM course 
GROUP BY teacher_id ORDER BY COUNT(cid) DESC LIMIT 1))GROUP BY s.student_id  ORDER BY AVG(s.num) DESC LIMIT 2

posted on 2019-06-02 11:56 Zander-zhao 阅读(219) 评论(0) 收藏举报

mysql 多表查询

一 介绍

二 多表连接查询

三 符合条件连接查询

四 子查询

五 综合练习

一介绍

二多表连接查询

三符合条件连接查询

四子查询

五综合练习