• 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

    复制代码
    #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 |
    +----+------------+--------+------+--------+
    复制代码

    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

    复制代码
    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         |
    +----+------------+--------+------+--------+------+--------------+
    复制代码

    2 内连接:只连接匹配的行

    复制代码
    #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;
    复制代码

    3 外链接之左连接:优先显示左表全部记录

    复制代码
    #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         |
    +----+------------+--------------+
    复制代码

    4 外链接之右连接:优先显示右表全部记录

    复制代码
    #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         |
    +------+-----------+--------------+
    复制代码

    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

    复制代码
    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         |
    +------+------------+--------+------+--------+------+--------------+
    复制代码

    三 符合条件连接查询

    复制代码
    #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;
    复制代码

    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

    复制代码
    #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);
    复制代码

    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

    复制代码
    #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)
    复制代码

    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 
    

      

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  • 原文地址:https://www.cnblogs.com/Xanderzyl/p/10962446.html
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