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MySQL数据库学习【第八篇】多表查询

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一、准备

首先先准备表

员工表和部门表

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

create table employee1(
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 employee1(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:

技术分享图片

技术分享图片

 

二、多表连接查询

1.交叉连接:不适用任何匹配条件。生成笛卡尔积、

select * from employee1 ,department;

2.内连接: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。(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
select * from employee1,department where employee1.dep_id=department.id;

#上面用where表示的可以用下面的内连接表示,建议使用下面的那种方法
select * from employee1 inner join department on employee1.dep_id=department.id;

#也可以这样表示哈
select employee1.id,employee1.name,employee1.age,employee1.sex,department.name from 
employee1,department where employee1.dep_id=department.id;

 技术分享图片

注意:内连接的join可以忽略不写,但是还是加上看起来清楚点

技术分享图片

3.左连接:优先显示左表全部记录。

#左链接:在按照on的条件取到两张表共同部分的基础上,保留左表的记录
select * from employee1 left join department on department.id=employee1.dep_id;
select * from department left join employee1 on department.id=employee1.dep_id;

 技术分享图片

技术分享图片

4.右链接:优先显示右表全部记录。

#右链接:在按照on的条件取到两张表共同部分的基础上,保留右表的记录
select * from employee1 right join department on department.id=employee1.dep_id;
select * from department right join employee1 on department.id=employee1.dep_id;

技术分享图片

技术分享图片

5.全外连接:显示左右两个表的全部记录。

注意:mysql不支持全外连接 full join
强调:mysql可以使用union间接实现全外连接
select * from employee1 left join department on department.id=employee1.dep_id
union
select * from employee1 right join department on department.id=employee1.dep_id; 

技术分享图片

 

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 * from employee1 inner join department on employee1.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 * from employee1 inner join department on employee1.dep_id=department.id 
=

and age>25 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

小练习

#查询平均年龄在25岁以上的部门名
select 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 = (select id from department where name=‘技术‘);

#查看小于2人的部门名
select name from department where id in (
select dep_id from employee1 group by dep_id having count(id) < 2
)
union
select name from department where id not in (select distinct dep_id from employee1);

#提取空部门                              #有人的部门
select * from department where id not in (select distinct dep_id from employee1);


或者:
select name from department where id in
(
select dep_id from employee1 group by dep_id having count(id) < 2
union
select id from department where id not in (select distinct dep_id from employee1);
);

五、综合练习

一、SELECT语句关键字的定义顺序

#SELECT语句关键字的定义顺序

SELECT DISTINCT <select_list>
FROM <left_table>
<join_type> JOIN <right_table>
ON <join_condition>
WHERE <where_condition>
GROUP BY <group_by_list>
HAVING <having_condition>
ORDER BY <order_by_condition>
LIMIT <limit_number>

 

二 、SELECT语句关键字的执行顺序

(7)     SELECT 
(8)     DISTINCT <select_list>
(1)     FROM <left_table>
(3)     <join_type> JOIN <right_table>
(2)     ON <join_condition>
(4)     WHERE <where_condition>
(5)     GROUP BY <group_by_list>
(6)     HAVING <having_condition>
(9)     ORDER BY <order_by_condition>
(10)    LIMIT <limit_number>

init.sql文件内容

数据导入:

/*
Navicat Premium Data Transfer

Source Server : localhost
Source Server Type : MySQL
Source Server Version : 50624
Source Host : localhost
Source Database : sqlexam

Target Server Type : MySQL
Target Server Version : 50624
File Encoding : utf-8

Date: 10/21/2016 06:46:46 AM
*/

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
`cid` int(11) NOT NULL AUTO_INCREMENT,
`caption` varchar(32) NOT NULL,
PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES (‘1‘, ‘三年二班‘), (‘2‘, ‘三年三班‘), (‘3‘, ‘一年二班‘), (‘4‘, ‘二年九班‘);
COMMIT;

-- ----------------------------
-- Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
`cid` int(11) NOT NULL AUTO_INCREMENT,
`cname` varchar(32) NOT NULL,
`teacher_id` int(11) NOT NULL,
PRIMARY KEY (`cid`),
KEY `fk_course_teacher` (`teacher_id`),
CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES (‘1‘, ‘生物‘, ‘1‘), (‘2‘, ‘物理‘, ‘2‘), (‘3‘, ‘体育‘, ‘3‘), (‘4‘, ‘美术‘, ‘2‘);
COMMIT;

-- ----------------------------
-- Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
`sid` int(11) NOT NULL AUTO_INCREMENT,
`student_id` int(11) NOT NULL,
`course_id` int(11) NOT NULL,
`num` int(11) NOT NULL,
PRIMARY KEY (`sid`),
KEY `fk_score_student` (`student_id`),
KEY `fk_score_course` (`course_id`),
CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES (‘1‘, ‘1‘, ‘1‘, ‘10‘), (‘2‘, ‘1‘, ‘2‘, ‘9‘), (‘5‘, ‘1‘, ‘4‘, ‘66‘), (‘6‘, ‘2‘, ‘1‘, ‘8‘), (‘8‘, ‘2‘, ‘3‘, ‘68‘), (‘9‘, ‘2‘, ‘4‘, ‘99‘), (‘10‘, ‘3‘, ‘1‘, ‘77‘), (‘11‘, ‘3‘, ‘2‘, ‘66‘), (‘12‘, ‘3‘, ‘3‘, ‘87‘), (‘13‘, ‘3‘, ‘4‘, ‘99‘), (‘14‘, ‘4‘, ‘1‘, ‘79‘), (‘15‘, ‘4‘, ‘2‘, ‘11‘), (‘16‘, ‘4‘, ‘3‘, ‘67‘), (‘17‘, ‘4‘, ‘4‘, ‘100‘), (‘18‘, ‘5‘, ‘1‘, ‘79‘), (‘19‘, ‘5‘, ‘2‘, ‘11‘), (‘20‘, ‘5‘, ‘3‘, ‘67‘), (‘21‘, ‘5‘, ‘4‘, ‘100‘), (‘22‘, ‘6‘, ‘1‘, ‘9‘), (‘23‘, ‘6‘, ‘2‘, ‘100‘), (‘24‘, ‘6‘, ‘3‘, ‘67‘), (‘25‘, ‘6‘, ‘4‘, ‘100‘), (‘26‘, ‘7‘, ‘1‘, ‘9‘), (‘27‘, ‘7‘, ‘2‘, ‘100‘), (‘28‘, ‘7‘, ‘3‘, ‘67‘), (‘29‘, ‘7‘, ‘4‘, ‘88‘), (‘30‘, ‘8‘, ‘1‘, ‘9‘), (‘31‘, ‘8‘, ‘2‘, ‘100‘), (‘32‘, ‘8‘, ‘3‘, ‘67‘), (‘33‘, ‘8‘, ‘4‘, ‘88‘), (‘34‘, ‘9‘, ‘1‘, ‘91‘), (‘35‘, ‘9‘, ‘2‘, ‘88‘), (‘36‘, ‘9‘, ‘3‘, ‘67‘), (‘37‘, ‘9‘, ‘4‘, ‘22‘), (‘38‘, ‘10‘, ‘1‘, ‘90‘), (‘39‘, ‘10‘, ‘2‘, ‘77‘), (‘40‘, ‘10‘, ‘3‘, ‘43‘), (‘41‘, ‘10‘, ‘4‘, ‘87‘), (‘42‘, ‘11‘, ‘1‘, ‘90‘), (‘43‘, ‘11‘, ‘2‘, ‘77‘), (‘44‘, ‘11‘, ‘3‘, ‘43‘), (‘45‘, ‘11‘, ‘4‘, ‘87‘), (‘46‘, ‘12‘, ‘1‘, ‘90‘), (‘47‘, ‘12‘, ‘2‘, ‘77‘), (‘48‘, ‘12‘, ‘3‘, ‘43‘), (‘49‘, ‘12‘, ‘4‘, ‘87‘), (‘52‘, ‘13‘, ‘3‘, ‘87‘);
COMMIT;

-- ----------------------------
-- Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
`sid` int(11) NOT NULL AUTO_INCREMENT,
`gender` char(1) NOT NULL,
`class_id` int(11) NOT NULL,
`sname` varchar(32) NOT NULL,
PRIMARY KEY (`sid`),
KEY `fk_class` (`class_id`),
CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES (‘1‘, ‘男‘, ‘1‘, ‘理解‘), (‘2‘, ‘女‘, ‘1‘, ‘钢蛋‘), (‘3‘, ‘男‘, ‘1‘, ‘张三‘), (‘4‘, ‘男‘, ‘1‘, ‘张一‘), (‘5‘, ‘女‘, ‘1‘, ‘张二‘), (‘6‘, ‘男‘, ‘1‘, ‘张四‘), (‘7‘, ‘女‘, ‘2‘, ‘铁锤‘), (‘8‘, ‘男‘, ‘2‘, ‘李三‘), (‘9‘, ‘男‘, ‘2‘, ‘李一‘), (‘10‘, ‘女‘, ‘2‘, ‘李二‘), (‘11‘, ‘男‘, ‘2‘, ‘李四‘), (‘12‘, ‘女‘, ‘3‘, ‘如花‘), (‘13‘, ‘男‘, ‘3‘, ‘刘三‘), (‘14‘, ‘男‘, ‘3‘, ‘刘一‘), (‘15‘, ‘女‘, ‘3‘, ‘刘二‘), (‘16‘, ‘男‘, ‘3‘, ‘刘四‘);
COMMIT;

-- ----------------------------
-- Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
`tid` int(11) NOT NULL AUTO_INCREMENT,
`tname` varchar(32) NOT NULL,
PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES (‘1‘, ‘张磊老师‘), (‘2‘, ‘李平老师‘), (‘3‘, ‘刘海燕老师‘), (‘4‘, ‘朱云海老师‘), (‘5‘, ‘李杰老师‘);
COMMIT;

SET FOREIGN_KEY_CHECKS = 1;

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

#题目

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 

 答案

1、查询所有的课程的名称以及对应的任课老师姓名
select cname 课程名称,tname 老师姓名 from course  left join teacher on course.teacher_id = teacher.tid;

2、查询学生表中男女生各有多少人
select gender ,count(gender) from student group by gender;

3、查询物理成绩等于100的学生的姓名
select sname from student inner join score on student.sid = score.student_id
join course on score.course_id = course.cid
where cname = 物理 and num = 100;

4、查询平均成绩大于八十分的同学的姓名和平均成绩
(1):
select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_id 
group by student_id having avg(num)>80;
(2):
select sname,avg(num) from student inner join score on student.sid = score.student_id 
group by student_id having avg(num)>80

5、查询所有学生的学号,姓名,选课数,总成绩
select student.sid 学号,sname 姓名,count(course_id) 选课数,sum(num) 总成绩 from student left join score on student.sid = score.student_id
group by student.sid ;
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 姓名 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
)));

8、 查询物理课程比生物课程高的学生的学号
select t1.student_id  学号 from 
(select student_id ,num from score inner join course on score.course_id=course.cid  where cname=物理 )as t1
inner join 
(select student_id , num from score inner join course on score.course_id=course.cid  where 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
(1):
select sname from student inner join score on student.sid = score.student_id
join course on course.cid=score.course_id and cname in (物理,体育) 
group by student_id having count(course_id)!=2;

(2):
select sname from student join score  on student.sid = score.student_id
join course on course.cid=score.course_id where cname =物理 or cname= 体育
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 姓名,caption 班级 from student inner join score on student.sid = score.course_id
join class on class.cid = score.course_id
where num<60 group by student_id having count(course_id)>=2;

11、查询选修了所有课程的学生姓名
(1)
select sname 姓名 ,所有的课程数 from student inner join 
(select student_id,count(course_id) 所有的课程数 from score group by student_id having count(course_id) = (
select count(cid) from course)) as t1
on t1.student_id = student.sid;

(2)
select sname,count(course_id) from student inner join score on student.sid = score.student_id
group by student_id having count(course_id)=(select count(cid) from course);

12、查询李平老师教的课程的所有成绩记录
(1):
select num from score inner join course on course.cid=score.course_id
join teacher on teacher.tid=course.teacher_id
where tname = 李平老师;

(2):
select num from score where course_id in(select cid from course where teacher_id=
(select tid from teacher where 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 课程号,cname 课程名 from course
select ;
14、查询每门课程被选修的次数 
(1)
select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cid
group by course_id;

(2):也可以按照name分组
select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cid
group by cname;

15、查询之选修了一门课程的学生姓名和学号
select sname 姓名,student_id 学号 from student inner join score on student.sid = score.student_id
group by student_id having count(course_id)=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)
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
(1):
select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_id
group by student_id having avg(num)>85;

(2):
select student.sname,avg_num from student inner join
(select student_id,avg(num) as avg_num from score group by student_id having avg(num) > 85
) t1
on student.sid=t1.student_id;


18、查询生物成绩不及格的学生姓名和对应生物分数
(1):
select student.sname ,num 生物成绩 from student inner join score on student.sid = score.student_id
join course on course.cid=score.course_id
where cname=生物 and num<60;

(2):
select student.sname,t1.num from student inner join
(
select student_id,num from score
where course_id=(select cid from course where cname=生物) and num < 60
) t1
on t1.student_id=student.sid
;

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 sname from student where sid=(
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)
) group by student_id order by avg(num) desc limit 1
)

 

MySQL数据库学习【第八篇】多表查询

标签:mil   tar   文件内容   structure   value   commit   必须   记录   host   

原文地址:https://www.cnblogs.com/xiaohema/p/8454928.html

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