数据库 多表查询
阅读目录
一:准备表
1.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
#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 * 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 * 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
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 | +------+------------+--------+------+--------+------+--------------+ #注意 union与union all的区别：union会去掉相同的纪录
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,department.name dep_name from employee inner join department on employee.dep_id = department.id where employee.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.name,employee.age,department.name dep_name from employee inner join department on employee.dep_id = department.id where employee.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 * from department where id not 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: 练习
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;
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
1、查询所有的课程的名称以及对应的任课老师姓名 SELECT cname,tname from course inner 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(sid) 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 where sid in ( SELECT student_id from score where course_id = (SELECT cid from course 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 SELECT student.sname,t1.avg_num from student inner join (SELECT student_id,AVG(num) avg_num from score GROUP BY student_id HAVING avg(num) > 80) as t1 on student.sid = t1.student_id; 5、查询所有学生的学号，姓名，选课数，总成绩 SELECT student.sid,student.sname,t1.course_num,t1.total_num from student inner JOIN (SELECT student_id, count(course_id) course_num, sum(num) total_num FROM score GROUP BY student_id) as t1 on student.sid = t1.student_id; 6、 查询姓李老师的个数 SELECT COUNT(1) 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' ) ) ); 8、 查询物理课程比生物课程高的学生的学号 SELECT t1.student_id from (SELECT student_id,num from score where course_id = ( SELECT cid from course where cname = '物理' )) as t1 inner join (SELECT student_id,num from score where course_id = ( SELECT cid from course 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 SELECT sname from student where sid in ( SELECT student_id from score LEFT JOIN course on score.course_id = course.cid WHERE course.cname in ('物理','体育') GROUP BY student_id HAVING count(sid) < 2 ); 10、查询挂科超过两门(包括两门)的学生姓名和班级 SELECT sname,caption from student LEFT JOIN class on student.class_id = class.cid where student.sid in ( SELECT student_id from score 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 SELECT student_id from score GROUP BY student_id HAVING count(course_id) = ( SELECT count(1) from course ); 12、查询李平老师教的课程的所有成绩记录 SELECT * from score where course_id in ( SELECT cid from course inner JOIN teacher on course.teacher_id = teacher.tid 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 course_id from score group by course_id HAVING count(student_id) = ( SELECT count(1) from student ); 14、查询每门课程被选修的次数 SELECT course.cname,t1.count_student FROM course INNER JOIN ( SELECT course_id,count(student_id) count_student from score GROUP BY course_id ) as t1 ON course.cid = t1.course_id; 15、查询只选修了一门课程的学生姓名和学号 SELECT student_id from score GROUP BY student_id HAVING COUNT(course_id) = 1; 16、查询所有学生考出的总成绩并按从高到低排序（成绩去重） SELECT DISTINCT sum(num) sum_num from score group by student_id ORDER BY sum_num desc; 17、查询平均成绩大于85的学生姓名和平均成绩 SELECT student.sname,t1.avg_num from student inner join ( SELECT student_id,avg(num) avg_num from score GROUP BY student_id having avg(num) > 85 ) as 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 SELECT sname,t1.num from student INNER JOIN ( SELECT student_id,num from score LEFT JOIN course on score.course_id = course.cid where course.cname = '生物' and score.num < 60 ) as t1 on student.sid = t1.student_id; 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 ); #并对第一 #选修了李平老师课程的学生 select student_id from score where course_id in ( #李平老师教的课程cid select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id having avg(num) = 最大平均成绩 ; #找最大的平均成绩 select avg(num) from score where course_id in ( select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id order by avg(num) desc limit 1 #结合到一起 select student_id from score where course_id in ( #李平老师教的课程cid select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id having avg(num) = ( select avg(num) from score where course_id in ( select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id order by avg(num) desc limit 1 ) ; #最终结果 select sname from student where sid in ( select student_id from score where course_id in ( #李平老师教的课程cid select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id having avg(num) = ( select avg(num) from score where course_id in ( select cid from course where teacher_id in ( select tid from teacher where tname = '李平老师' ) ) group by student_id order by avg(num) desc limit 1 ) ); 20、查询每门课程成绩最好的前两名学生姓名 SELECT * from score ORDER BY course_id,num desc; #取得课程编号与第一高的成绩:course_id,first_num SELECT course_id,max(num) first_num from score GROUP BY course_id; #取得课程编号与第二高的成绩:course_id,second_num SELECT score.course_id,max(num) second_num from score LEFT JOIN ( SELECT course_id,max(num) first_num from score GROUP BY course_id ) as t1 on score.course_id = t1.course_id where score.num < t1.first_num GROUP BY score.course_id ; #链表得到一张新表，新表包含课程编号与这门课程前两名的成绩分数 select t1.course_id,t1.first_num,t2.second_num from (SELECT course_id,max(num) first_num from score GROUP BY course_id) as t1 inner join (SELECT score.course_id,max(num) second_num from score LEFT JOIN ( SELECT course_id,max(num) first_num from score GROUP BY course_id ) as t1 on score.course_id = t1.course_id where score.num < t1.first_num GROUP BY score.course_id) as t2 on t1.course_id = t2.course_id; #取前两名学生的编号 SELECT score.course_id,score.student_id from score LEFT JOIN ( select t1.course_id,t1.first_num,t2.second_num from (SELECT course_id,max(num) first_num from score GROUP BY course_id) as t1 inner join (SELECT score.course_id,max(num) second_num from score LEFT JOIN ( SELECT course_id,max(num) first_num from score GROUP BY course_id ) as t1 on score.course_id = t1.course_id where score.num < t1.first_num GROUP BY score.course_id) as t2 on t1.course_id = t2.course_id ) as t3 on score.course_id = t3.course_id where score.num >= t3.second_num and score.num <= t3.first_num ;
#最终结果 SELECT t4.course_id,student.sname from student inner join ( SELECT score.course_id,score.student_id from score LEFT JOIN ( select t1.course_id,t1.first_num,t2.second_num from (SELECT course_id,max(num) first_num from score GROUP BY course_id) as t1 inner join (SELECT score.course_id,max(num) second_num from score LEFT JOIN ( SELECT course_id,max(num) first_num from score GROUP BY course_id ) as t1 on score.course_id = t1.course_id where score.num < t1.first_num GROUP BY score.course_id) as t2 on t1.course_id = t2.course_id ) as t3 on score.course_id = t3.course_id where score.num >= t3.second_num and score.num <= t3.first_num ) as t4 on student.sid = t4.student_id ORDER BY t4.course_id ;
既然选择了远方，便是风雨兼程...
