笛卡尔积
将两表所有的数据一一对应,生成一张大表 不建议
select * from dep,emp; #两个表拼一起
select * from dep,emp where dep.id = emp.dep_id; #找到两表之间对应的关系记录
select * from dep,emp where dep.id = emp.dep_id and dep.name='技术'; #筛选部门名称为技术的大表中的记录
select emp.name from dep,emp where dep.id = emp.dep_id and dep.name='技术'; #拿到筛选后的记录的员工姓名字段数据连表查询
inner join 内连接
select * from 表名 inner join 表2 on dep.id=emp.dep_id;
第一步:连表
select * from dep inner join emp on dep.id=emp.dep_id;
第二步: 过滤
select * from dep inner join emp on dep.id=emp.dep_id where dep.name='技术';
第三步:找对应字段数据
select emp.name from dep inner join emp on dep.id=emp.dep_id where dep.name='技术';left join 左连接
(left join左边的表为主表,主表记录必须全部显示,辅表没办法对应上的,就通过null来补全)
select * from dep left join emp on dep.id=emp.dep_id;right join 右连接
select * from dep right join emp on dep.id=emp.dep_id;union 全连接
mysql> select * from dep left join emp on dep.id=emp.dep_id
-> union
-> select * from dep right join emp on dep.id=emp.dep_id;子查询
(一个查询结果集作为另一个查询的条件)
select name from emp where dep_id = (select id from dep where name = '技术');小练习1
书名 作者 出版社 价格 出版日期
倚天屠龙记 egon 北京工业地雷出版社 70 2019-7-1
九阳神功 alex 人民音乐不好听出版社 5 2018-7-4
九阴真经 yuan 北京工业地雷出版社 62 2017-7-12
九阴白骨爪 jinxin 人民音乐不好听出版社 40 2019–8-7
独孤九剑 alex 北京工业地雷出版社 12 2017-9-1
降龙十巴掌 egon 知识产权没有用出版社 20 2019-7-5
葵花宝典 yuan 知识产权没有用出版社 33 2019–8-2
0.建表book，并向表中插入数据
create table book(
书名 varchar(12),
作者 varchar(12),
出版社 varchar(12),
价格 int unsigned not null,
出版日期 date
);
insert into book values
('倚天屠龙记','egon','北京工业地雷出版社',70,'2019-7-1'),
('九阳神功','alex','人民音乐不好听出版社',5,'2018-7-4'),
('九阴真经','yuan','北京工业地雷出版社',62,'2017-7-12'),
('九阴白骨爪','jinxin','人民音乐不好听出版社',40,'2019–8-7'),
('独孤九剑','alex','北京工业地雷出版社',12,'2017-9-1'),
('降龙十巴掌','egon','知识产权没有用出版社',20,'2019-7-5'),
('葵花宝典','yuan','知识产权没有用出版社',33,'2019–8-2');
1.查询 egon 写的所有书和价格
select 书名,价格 from book where 作者='egon';
2.找出最贵的图书的价格
select 书名,价格 from book ;
3.求所有图书的均价
4.将所有图书按照出版日期排序
5.查询alex写的所有书的平均价格
6.查询人民音乐不好听出版社出版的所有图书
7.查询人民音乐出版社出版的alex写的所有图书和价格
8.找出出版图书均价最高的作者
9.找出最新出版的图书的作者和出版社
10.显示各出版社出版的所有图书
11.查找价格最高的图书，并将它的价格修改为50元
12.删除价格最低的那本书对应的数据
13.将所有alex写的书作业修改成alexsb
14.select year(publish_date) from book
自己研究上面sql语句中的year函数的功能，完成需求：
将所有2017年出版的图书从数据库中删除
15.有文件如下，使用python写代码将文件中的数据写入数据库
学python从开始到放弃|alex|人民大学出版社|50|2018-7-1
学mysql从开始到放弃|egon|机械工业出版社|60|2018-6-3
学html从开始到放弃|alex|机械工业出版社|20|2018-4-1
学css从开始到放弃|wusir|机械工业出版社|120|2018-5-2
学js从开始到放弃|wusir|机械工业出版社|100|2018-7-30
答案
#第一步手动创建表
# create table book(
# id int primary key auto_increment,
# book_name char(20) not null,
# author char(12) not null,
# press char(20) not null,
# price float(6,2),
# pub_date date
# );
# 写入数据
import pymysql
conn = pymysql.Connection(host='127.0.0.1', user='root', password="",
database='daycs')
cur = conn.cursor()#右边
with open('book.txt',encoding='utf-8') as f:
try:
for line in f:
line = line.strip()
if '\t' in line:
lst = line.split('\t')
elif '|' in line:
lst = line.split('|')
sql = 'insert into book(book_name,author,press,price,pub_date) values (%s,%s,%s,%s,%s);'
cur.execute(sql,lst)
except Exception:
conn.rollback()#操作不成功 回滚
conn.commit()
cur.close()
conn.close()
# select book_name,price from book where author = 'egon'
# 2.找出最贵的图书的价格
# select max(price) from book;
# select price,book_name from book order by price desc limit 1;
# 3.求所有图书的均价
# select avg(price) from book;
# 4.将所有图书按照出版日期排序
# select * from book order by pub_date;
# 5.查询alex写的所有书的平均价格
# select avg(price) from book where author = 'alex'
# 扩展: 求所有人自己出版的图书的平均价格
# select author,avg(price) from book group by author
# 扩展: 求所有人自己出版的图书的平均价格>30的所有人
# select author from book group by author having avg(price)>30
# 6.查询人民音乐不好听出版社出版的所有图书
# select * from book where press = '人民音乐不好听出版社';
# 7.查询人民音乐出版社出版的alex写的所有图书和价格
# select * from book where press = '人民音乐不好听出版社' and author = 'alex';
# 8.找出出版图书均价最高的作者
# select author,avg(price) as avg_p from book group by author order by avg_p desc limit 1;
# 9.找出最新出版的图书的作者和出版社
# select author,press from book order by pub_date desc limit 1;
# 10.显示各出版社出版的所有图书
# select press,group_concat(book_name) from book group by press;
# 11.查找价格最高的图书，并将它的价格修改为50元
# select max(price) from book;
# update book set price = 50 where price = 70
# 12.删除价格最低的那本书对应的数据
# select min(price) from book;
# delete from book where price = 5;
# 13.将所有alex写的书作者修改成alexsb
# update book set author = 'alexsb' where author = 'alex';
# 14.select year(publish_date) from book
# 自己研究上面sql语句中的year函数的功能，完成需求：
# 将所有2017年出版的图书从数据库中删除
# delete from book where year(publish_date) = 2017;
小练习2
键表准备
数据导入
init.sql文件内容
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
查表
select * from class;#班级
+-----+--------------+
| cid | caption |
+-----+--------------+
| 1 | 三年二班 |
+-----+--------------+
select * from course;#课程
+-----+--------+------------+
| cid | cname | teacher_id |
+-----+--------+------------+
| 1 | 生物 | 1 |
+-----+--------+------------+
mysql> select * from score;#分数
+-----+------------+-----------+-----+
| sid | student_id | course_id | num |
+-----+------------+-----------+-----+
| 1 | 1 | 1 | 10 |
+-----+------------+-----------+-----+
mysql> select * from student;#学生
+-----+--------+----------+--------+
| sid | gender | class_id | sname |
+-----+--------+----------+--------+
| 1 | 男 | 1 | 理解 |
+-----+--------+----------+--------+
mysql> select * from teacher;#老师
+-----+-----------------+
| tid | tname |
+-----+-----------------+
| 1 | 张磊老师 |
+-----+-----------------+
1、查询男生、女生的人数；
select count(gender) from student group by gender;
2、查询姓“张”的学生名单；
select * from student where sname like '张%'；
3、课程平均分从高到低显示
select avg(num) from score group by course_id order by avg(num) desc;
4、查询有课程成绩小于60分的同学的学号、姓名；
1查询小于60的的学生id
select student_id from score where num<60;
2利用学生id在学生表查姓名
select sid,sname from student where sid in (select student_id from score where num<60);
5、查询至少有一门课与学号为1的同学所学课程相同的同学的学号和姓名；
1先查出学号1的所学课程
select course_id from score where student_id=1;
2查分数表 只要符合 学的课程在1号同学就行 获取学生id
select student_id from score where course_id in (select course_id from score where student_id=1) and student_id !=1 group by student_id;
3学生id获取姓名
select sid,sname from student where sid in (select student_id from score where course_id in (select course_id from score where student_id=1) and student_id !=1 group by student_id);
6、查询出只选修了一门课程的全部学生的学号和姓名；
1先查学生id
select student_id from score group by student_id order by count(course_id)=1;
2id查姓名
select sid,sname from student where sid in (select student_id from score group by student_id order by count(course_id)=1);
7、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),min(num) from score group by course_id;
8、查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名；
1先查出课程编号1的学生信息
select * from score where course_id=1;
2查出课程编号2的学生信息
select * from score where course_id=2;
3连表查询 左外连接 利用学生id相同 找出学生id
select t1.student_id from (select * from score where course_id=1) t2 inner join (select * from score where course_id=2) t1 on t1.student_id = t2.student_id where t1.num<t2.num;
4学生id 找姓名
select sid,sname from student where sid in (select t1.student_id from (select * from score where course_id=1) t2 inner join (select * from score where course_id=2) t1 on t1.student_id = t2.student_id where t1.num<t2.num);
9、查询“生物”课程比“物理”课程成绩高的所有学生的学号；
1先查生物id
select cid from course where cname='生物';
2查物流id
select cid from course where cname='物理';
3利用id 查生物成绩
select * from score where course_id =(select cid from course where cname='生物');
3利用id 查物理成绩
select * from score where course_id =(select cid from course where cname='物理');
4连表 找出学生id
select t1.student_id from (select * from score where course_id =(select cid from course where cname='生物')) t2 inner join (select * from score where course_id =(select cid from course where cname='物理')) t1 on t1.student_id = t2.student_id where t1.num>t2.num ;
5 id找姓名
select sid,sname from student where sid in (select t1.student_id from (select * from score where course_id =(select cid from course where cname='生物')) t2 inner join (select * from score where course_id =(select cid from course where cname='物理')) t1 on t1.student_id = t2.student_id where t1.num<t2.num);
10、查询平均成绩大于60分的同学的学号和平均成绩;
1学生分组
select student_id,avg(num) from score group by student_id order by avg(num) >60;
11、查询所有同学的学号、姓名、选课数、总成绩；
1学生分组
select student_id,avg(num),count(course_id) from score group by student_id;
2连表
select t1.student_id,t1.pjnum,t1.zcourse,t2.sname from (select student_id,avg(num) as pjnum,count(course_id) as zcourse from score group by student_id) as t1 inner join (select * from student) t2 on t1.student_id=t2.sid;
12、查询姓“李”的老师的个数；
1先查姓李的老师id 在计数
select count(tid) from teacher where tname like '李%';
13、查询没学过“张磊老师”课的同学的学号、姓名；
1找到张磊老师id
select tid from teacher where tname='张磊老师';
2查询张磊老师教什么课
select cid from course where teacher_id=(select tid from teacher where tname='张磊老师');
3找学过的
select student_id from score where course_id in (select cid from course where teacher_id=(select tid from teacher where tname='张磊老师'));
4逆向思维
select sid,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='张磊老师')));
14、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名；
1查询学过编号1的
select student_id from score where course_id=1;
2查询学过编号2的
select student_id from score where course_id=2;
3连表 找id
select t1.student_id from (select student_id from score where course_id=1) as t1 inner join (select student_id from score where course_id=2) as t2 on t1.student_id=t2.student_id;
4id找名字
select sid,sname from student where sid in (select t1.student_id from (select student_id from score where course_id=1) as t1 inner join (select student_id from score where course_id=2) as t2 on t1.student_id=t2.student_id);
15、查询学过“李平老师”所教的所有课的同学的学号、姓名；
select student_id from score where course_id in (select cid from course where teacher_id=(select tid from teacher where tname='张磊老师'));
id 找名字
select sid,sname from student where sid in (select student_id from score where course_id in (select cid from course where teacher_id=(select tid from teacher where tname='张磊老师')));
1、查询没有学全所有课的同学的学号、姓名；
1总课程数
select count(cid) from course;
2分组 找id
select student_id from score group by student_id having count(course_id)=(select count(cid) from course);
id 找名字
select sid,sname from student where sid in (select student_id from score group by student_id having count(course_id)=(select count(cid) from course));
删除学习“李平”老师课的SC(score)表记录；
# 先查出李平老师的id
select tid from teacher where tname = '李平老师';
# 查看李平老师所教授的课程
select cid from course where teacher_id = (select tid from teacher where tname = '李平老师’);
# 查看李平老师所教课程的成绩数据
select * from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老师'));
# 执行删除命令
delete from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老师'));
# 先查询2号同学学了哪些课程
select * from score where student_id =2;
# 找到学习了2号同学没学习课程的所有同学（找到所有和2号同学学习的课程不一样的同学）
select student_id from score where course_id not in (select course_id from score where student_id=2)
# 找到score表中所有的学生并且把 2号同学 以及（和2号同学学习的课程不一样的同学）排除出去
select student_id from score where student_id not in (select student_id from score where course_id not in (select course_id from score where student_id=2)) and student_id !=2
# 对剩余的和2号同学所选课程没有不同的同学所选课程数进行统计，如果和2号同学的课程数相同，就是选择了相同的课程
select student_id from score where student_id not in (
select student_id from score where course_id not in (select course_id from score where student_id=2)
) and student_id !=2
group by student_id
having count(course_id)= (select count(course_id) from score where student_id=2);
6、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 课程ID,max(num) 最高分,min(num) 最低分 from score group by course_id;
7、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；
# 方法1:
# 先求平均成绩
select course_id,avg(num) from score group by course_id;
# 解决计算各科及格率的问题
所有及格的人/所有人数
select t1.course_id,t1.count1/t2.count2 from
(select course_id,count(course_id) count1 from score where num>60 group by course_id) t1
left join
(select course_id,count(course_id) count2 from score group by course_id) t2
on t1.course_id = t2.course_id;
# 根据上述内容进行表的拼接
select t_out1.course_id,t_out1.avgnum, t_out2.pass_per from
(select course_id,avg(num) avgnum from score group by course_id ) t_out1
left join
(select t1.course_id,t1.count1/t2.count2 pass_per from
(select course_id,count(course_id) count1 from score where num>60 group by course_id) t1
left join
(select course_id,count(course_id) count2 from score group by course_id) t2
on t1.course_id = t2.course_id) t_out2
on t_out1.course_id = t_out2.course_id
# 加上排序
select t_out1.course_id,t_out1.avgnum, t_out2.pass_per from (select course_id,avg(num) avgnum from score group by course_id ) t_out1 left join (select t1.course_id,t1.count1/t2.count2 pass_per from (select course_id,count(course_id) count1 from score where num>60 group by course_id) t1 left join (select course_id,count(course_id) count2 from score group by course_id) t2 on t1.course_id = t2.course_id) t_out2 on t_out1.course_id = t_out2.course_id order by avgnum ,pass_per desc;
# 方法2
# 使用case when直接计算合格率
select
sum(case when num>60 then 1 else 0 end)/count(course_id)
from score group by course_id
# 加上课程id和平均值
select course_id,avg(num),
sum(case when num>60 then 1 else 0 end)/count(course_id)
from score group by course_id
# 加上排序
select course_id,avg(num) avgnum,
sum(case when num>60 then 1 else 0 end)/count(course_id) pass_per
from score group by course_id
order by avgnum ,pass_per desc;
9、查询每门课程被选修的学生数；
select course_id,count(course_id) from score group by course_id;
10、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;
11、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;
12、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;
13、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 = '数学'
14、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
15、求选了课程的学生人数
select sid,sname from student where sid not in (select student_id from score group by student_id);
16、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；
# 先找到“杨艳”老师的教师id
select tid from teacher where tname = '杨艳';
# 再找到杨艳老师教的所有课程
select cid from course where teacher_id in (select tid from teacher where tname = '杨艳');
# 再找到杨艳老师教的所有课程的最高分
select max(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'));
# 再找到杨艳老师教的所有课程的最高分对应的学生
select distinct student_id,num from score
where num = (select max(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')))
and course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'));
# 找到学生的姓名
select student.sname,t1.num from(
select distinct student_id,num from score
where num = (select max(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')))
and course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'))
) t1
left join
student
on
t1.student_id = student.sid;
17、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;
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;
9、查询每门课程成绩最好的前两名；
先查询每条数据对应学科成绩的第一名和第二名，这里必须要保留所有的s1，以便后续进行连表查询
select sid,course_id,
(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
按照sid连表，把学生的成绩和对应的第一名、第二名成绩连起来
select
* from score t1
left join
(
select sid,course_id,
(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
) t2
on t1.sid = t2.sid
判断如果学生的成绩是第一名、第二名的成绩，那么就符合条件，显示学生的id、学科和成绩
select
t1.sid,t1.student_id,t1.course_id,t1.num from score t1
left join
(
select sid,course_id,
(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
) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num;
20、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;
21、查询全部学生都选修的课程的课程号和课程名；
# 先查看一共有多少学生
select count(sid) from student;
# 查看哪一门课选秀的学生个数和学生的总个数相等
select course_id from score group by course_id having count(student_id) = (select count(sid) from student);
2、查询没学过“叶平”老师讲授的任一门课程的学生姓名；
# 先查看要查找老师的id
select tid from teacher where tname = '李平老师';
# 查看该老师交了哪些课程
select cid from course where teacher_id in (select tid from teacher where tname = '李平老师')
# 看看有多少学生学习了该老师的课程
select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'));
# 把不在上表中的学生姓名查出来
select sname from student where sid not in (select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师')));
23、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,avg(num) from score where num<60 group by student_id having count(num)>=2;
24、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;
25、删除“002”同学的“001”课程的成绩；
delete from score where course_id = 1 and student_id = 2
5、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；
# 查看每个学生的数学成绩
select student_id,num from score where course_id = (select cid from course where cname = '数学');
# 查看每个学生的语文成绩
select student_id,num from score where course_id = (select cid from course where cname = '语文');
# 查看每个学生的英语成绩
select student_id,num from score where course_id = (select cid from course where cname = '英语');
# 查看每个学生的平均成绩
select student_id,avg(num),count(num) from score group by student_id;
# 将上面的几张表拼接起来,为了生成所有学生的信息，用student表作为左连接的第一张表
select sid 学生ID,t2.num 语文,t1.num 数学, t3.num 英语,t4.count_course 有效课程数,t4.avg_num 有效平均分 from student
left join (select student_id,num from score where course_id = (select cid from course where cname = '数学')) t1
on student.sid = t1.student_id
left join (select student_id,num from score where course_id = (select cid from course where cname = '语文')) t2
on student.sid = t2.student_id
left join (select student_id,num from score where course_id = (select cid from course where cname = '英语')) t3
on student.sid = t3.student_id
left join (select student_id,avg(num) avg_num,count(num) count_course from score group by student_id) t4
on student.sid = t4.student_id
4、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
# 先找寻上过2号课程的同学
select student_id from score where course_id = 2;
# 再找到没上过2号课程的所有同学
select * from student where sid not in (select student_id from score where course_id = 2);
#计算出学习2号课程的同学的平均成绩
select avg(num) from score where course_id = 2 group by course_id;
# 用笛卡尔积将上述两个表拼起来
select * from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) from score where course_id = 2 group by course_id) t2;
# 向SC表中插入记录
insert into score (course_id,student_id,num) select 2,t1.sid,t2.avg_num from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) avg_num from score where course_id = 2 group by course_id) t2;
8、查询各科成绩前三名的记录:(不考虑成绩并列情况)
select
t1.sid,t1.student_id,t1.course_id,t1.num from score t1
left join
(
select sid,course_id,
(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,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 2, 1) as third_num
from score as s1
) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num or t1.num = t2.third_num;
原文地址：https://www.cnblogs.com/saoqiang/p/12388393.html
时间： 2024-10-09 14:46:53