python3 mysql 多表查询
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python3 mysql 多表查询
一、准备表
创建二张表：
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
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 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
/* 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
参考答案： 查询所有的课程的名称以及对应的任课老师姓名 1、SELECT cname,tname FROM course LEFT JOIN teacher on course.teacher_id=teacher.tid 2、查询“生物”课程比“物理”课程成绩高的所有学生的学号； 思路： 获取所有有生物课程的人（学号，成绩） - 临时表 获取所有有物理课程的人（学号，成绩） - 临时表 根据【学号】连接两个临时表： 学号 物理成绩 生物成绩 然后再进行筛选 select A.student_id,sw,ty from (select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = ‘生物‘) as A left join (select student_id,num as ty from score left join course on score.course_id = course.cid where course.cname = ‘体育‘) as B on A.student_id = B.student_id where sw > if(isnull(ty),0,ty); 3、查询平均成绩大于60分的同学的学号和平均成绩； 思路： 根据学生分组，使用avg获取平均值，通过having对avg进行筛选 select student_id,avg(num) from score group by student_id having avg(num) > 60 4、查询所有同学的学号、姓名、选课数、总成绩； select score.student_id,sum(score.num),count(score.student_id),student.sname from score left join student on score.student_id = student.sid group by score.student_id 5、查询姓“李”的老师的个数； select count(tid) from teacher where tname like ‘李%‘ select count(1) from (select tid from teacher where tname like ‘李%‘) as B 6、查询没学过“叶平”老师课的同学的学号、姓名； 思路： 先查到“李平老师”老师教的所有课ID 获取选过课的所有学生ID 学生表中筛选 select * from student where sid not in ( select DISTINCT student_id from score where score.course_id in ( select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘李平老师‘ ) ) 7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名； 思路： 先查到既选择001又选择002课程的所有同学 根据学生进行分组，如果学生数量等于2表示，两门均已选择 select student_id,sname from (select student_id,course_id from score where course_id = 1 or course_id = 2) as B left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名； 同上，只不过将001和002变成 in (叶平老师的所有课) 9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名； 同第1题 10、查询有课程成绩小于60分的同学的学号、姓名； select sid,sname from student where sid in ( select distinct student_id from score where num < 60 ) 11、查询没有学全所有课的同学的学号、姓名； 思路： 在分数表中根据学生进行分组，获取每一个学生选课数量 如果数量 == 总课程数量，表示已经选择了所有课程 select student_id,sname from score left join student on score.student_id = student.sid group by student_id HAVING count(course_id) = (select count(1) from course) 12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名； 思路： 获取 001 同学选择的所有课程 获取课程在其中的所有人以及所有课程 根据学生筛选，获取所有学生信息 再与学生表连接，获取姓名 select student_id,sname, count(course_id) from score left join student on score.student_id = student.sid where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id 13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名； 先找到和001的学过的所有人 然后个数 ＝ 001所有学科 ＝＝》 其他人可能选择的更多 select student_id,sname, count(course_id) from score left join student on score.student_id = student.sid where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1) 14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名； 个数相同 002学过的也学过 select student_id,sname from score left join student on score.student_id = student.sid where student_id in ( select student_id from score where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1) ) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1) 15、删除学习“叶平”老师课的score表记录； delete from score where course_id in ( select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = ‘叶平‘ ) 16、向SC表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩； 思路： 由于insert 支持 inset into tb1(xx,xx) select x1,x2 from tb2; 所有，获取所有没上过002课的所有人，获取002的平均成绩 insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2) from student where sid not in ( select student_id from score where course_id = 2 ) 17、按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分； select sc.student_id, (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy, (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl, (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty, count(sc.course_id), avg(sc.num) from score as sc group by student_id desc 18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分； select course_id, max(num) as max_num, min(num) as min_num from score group by course_id; 19、按各科平均成绩从低到高和及格率的百分数从高到低顺序； 思路：case when .. then select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc; 20、课程平均分从高到低显示（现实任课老师）； select avg(if(isnull(score.num),0,score.num)),teacher.tname from course left join score on course.cid = score.course_id left join teacher on course.teacher_id = teacher.tid group by score.course_id 21、查询各科成绩前三名的记录:(不考虑成绩并列情况) select score.sid,score.course_id,score.num,T.first_num,T.second_num from score 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 3,1) as second_num from score as s1 ) as T on score.sid =T.sid where score.num <= T.first_num and score.num >= T.second_num 22、查询每门课程被选修的学生数； select course_id, count(1) from score group by course_id; 23、查询出只选修了一门课程的全部学生的学号和姓名； select student.sid, student.sname, count(1) from score left join student on score.student_id = student.sid group by course_id having count(1) = 1 24、查询男生、女生的人数； select * from (select count(1) as man from student where gender=‘男‘) as A , (select count(1) as feman from student where gender=‘女‘) as B 25、查询姓“张”的学生名单； select sname from student where sname like ‘张%‘; 26、查询同名同姓学生名单，并统计同名人数； select sname,count(1) as count from student group by sname; 27、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列； select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg asc,course_id desc; 28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩； select student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id; 29、查询课程名称为“数学”，且分数低于60的学生姓名和分数； select student.sname,score.num from score left join course on score.course_id = course.cid left join student on score.student_id = student.sid where score.num < 60 and course.cname = ‘生物‘ 30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名； select * from score where score.student_id = 3 and score.num > 80 31、求选了课程的学生人数 select count(distinct student_id) from score select count(c) from ( select count(student_id) as c from score group by student_id) as A 32、查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩； select sname,num from score left join student on score.student_id = student.sid where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname=‘张磊老师‘) order by num desc limit 1; 33、查询各个课程及相应的选修人数； select course.cname,count(1) from score left join course on score.course_id = course.cid group by course_id; 34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩； select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id; 35、查询每门课程成绩最好的前两名； select score.sid,score.course_id,score.num,T.first_num,T.second_num from score 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 T on score.sid =T.sid where score.num <= T.first_num and score.num >= T.second_num 36、检索至少选修两门课程的学生学号； select student_id from score group by student_id having count(student_id) > 1 37、查询全部学生都选修的课程的课程号和课程名； select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student); 38、查询没学过“叶平”老师讲授的任一门课程的学生姓名； select student_id,student.sname from score left join student on score.student_id = student.sid where score.course_id not in ( select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘张磊老师‘ ) group by student_id 39、查询两门以上不及格课程的同学的学号及其平均成绩； select student_id,count(1) from score where num < 60 group by student_id having count(1) > 2 40、检索“004”课程分数小于60，按分数降序排列的同学学号； select student_id from score where num< 60 and course_id = 4 order by num desc; 41、删除“002”同学的“001”课程的成绩； delete from score where course_id = 1 and student_id = 2
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