SQL面试题学习与总结
数据库MySQL经典面试题之SQL语句
1.需要数据库表1.学生表
Student(SID,Sname,Sage,Ssex) --SID 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
2.课程表
Course(CID,Cname,TID) --CID --课程编号,Cname 课程名称,TID 教师编号
3.教师表
Teacher(TID,Tname) --TID 教师编号,Tname 教师姓名
4.成绩表
SC(SID,CID,score) --SID 学生编号,CID 课程编号,score 分数
添加测试数据
1.学生表
create table Student(SID varchar(10),Sname nvarchar(10),Sage datetime,Ssex
nvarchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
2.课程表
create table Course(CID varchar(10),Cname nvarchar(10),TID varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
3.教师表
create table Teacher(TID varchar(10),Tname nvarchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
4.成绩表
create table SC(SID varchar(10),CID varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
练习与总结
题目一：
1、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60
order by a.SID
2、该SQL涉及的函数
（1）cast（表达式 as 数据类型）
举例：
SELECT CAST('9.6' AS DECIMAL(18,2))：结果保留两位小数
SELECT CAST('9.6' AS DECIMAL)：结果进行四舍五入
（2）having
having是group by分组后的筛选条件，分组后的数据内再筛选
在说where和having之前，需要了解group by，在了解group by之前，需要了解聚合函数
（3）group by
聚合函数，作用在多条记录上，sum，count，max，avg
（4）order by
将结果进行顺序、倒序排列
3、由简入繁编写SQL
--按照学生来查询三门课程的总成绩
SELECT SID, SUM(score) FROM sc GROUP BY SID;
--按照学生来查询三门课程的总成绩,并且仅仅显示总分大于90分的人
SELECT SID, SUM(score) FROM sc GROUP BY SID HAVING SUM(score) > 90;
--按照学生来查询三门课程的总成绩,并且仅仅显示总分大于90分的人,使用CAST AS函数进行转换
SELECT SID, SUM(score) FROM sc GROUP BY SID HAVING CAST(SUM(score) AS DECIMAL) > 90;
--按照学生来查询三门课程的总成绩,并且仅仅显示总分大于90分的人,使用CAST AS函数进行转换，分数顺序排列
SELECT SID, SUM(score) FROM sc GROUP BY SID HAVING CAST(SUM(score) AS DECIMAL) > 90 ORDER BY SUM(score);
