作为一个SQL小白,在一个月的努力后终于把这50道练习题做完了,分享一下我的答案给大家,说不定能给你提供一些思路。
我使用的是在线SQL,个人认为十分方便,使用的版本是“Ms SQL Server 2017”。
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(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('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2017-12-30' , '女');
insert into Student values('12' , '赵六' , '2017-01-01' , '女');
insert into Student values('13' , '孙七' , '2018-01-01' , '女');
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');
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
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);
select t1.sid, t1.score as class1,t2.score as class2 from
(select sid,cid,score from sc where cid = '01') t1,
(select sid,cid,score from sc where cid = '02') t2
where t1.sid = t2.sid and t1.score > t2.score
Select sc.sid from sc
Where sc.cid = '01' and sc.sid in (select sid from sc where cid = '02')
Select * from
(select * from sc where sc.cid = ‘01’) t1
Left join (select * from sc where sc.cid = ‘02’) t2
On t1.sid = t2.sid
Select * from sc
Where sc.cid = ‘02’ and sc.sid not in (select sid from sc where cid = ‘01’)
Select sc.sid, st.sname, avg(sc.score) avg_score from sc
Join student st
On sc.sid = st.sid
Group by sc.sid
Having avg_score >= 60
Select sc.*, st.* from sc
Left join student st
On sc.sid = st.sid
Select st.sid, st.sname, count(sc.cid), sum(sc.score) from student st
Left join sc
On st.sid = sc.sid
Group by st.sid
Select st.* from student t
Where st.sid in (select sid from sc)
Select count(tid) from teacher
Where tname like’李%’
Select sc.sid, st.* from sc
Left join student st
On sc.sid = st.sid
Where sc.cid in (Select c.cid from course c
Join teacher t
On c.tid = t.tid
Where t.tname = ‘张三’)
Select st.* from student t
Where st.sid not in
(select sid from sc group by sid having count(cid) = (select count(cid) from course))
Select distinct sc.sid, st.* from sc
Join student st
On sc.sid = st.sid
Where sc.cid in (Select cid from sc where sid = ‘01’)
Select sc.sid, st.* from sc
Join student st
On sc.sid = st.sid
where sc.cid in (Select cid from sc where sid = '01') and sc.sid != '01'
group by sc.sid
having count(sc.cid) = (Select count(cid) from sc where sid = '01')
select st.sname from student st
where st.sname not in
(select st.sname from student st,sc,course c,teacher t
where st.sid = sc.sid and sc.cid = c.cid and c.tid = t.tid and t.tname = '张三')
select sc.sid,st.sname,avg(sc.score) from sc
join student st
on sc.sid = st.sid
group by sc.sid
having sum(case when sc.score < 60 then 1 else 0 end)>=2
select sc.sid,st.sname from sc
join student st
on sc.sid = st.sid
where sc.cid = '01' and sc.score <60
order by sc.score desc
select sid,avg(score) over(partition by sid) as avg_score,cid,score from sc
order by avg_score desc
select sc.sid,t1.avg_score,sc.cid,sc.score from sc
left join (select sid,avg(score) as avg_score,cid,score from sc group by sid) t1
on sc.sid = t1.sid
order by 2 desc
select cid,max(score),avg(score) from sc
group by cid
select sc.cid,c.cname,max(sc.score),min(sc.score),avg(sc.score),
sum(case when sc.score>=60 then 1 else 0 end)*1.0/count(sc.score) 及格,
sum(case when sc.score between 70 and 80 then 1 else 0 end)*1.0/count(sc.score) 中等,
sum(case when sc.score between 80 and 90 then 1 else 0 end)*1.0/count(sc.score) 优良,
sum(case when sc.score>=90 then 1 else 0 end)*1.0/count(sc.score) 优秀
from sc
join course c
on sc.cid = c.cid
group by sc.cid
select cid, rank() over (partition by cid order by score desc) rank,score from sc
select cid, dense_rank() over (partition by cid order by score desc) rank,score from sc
select cid, count(sid) from sc
group by cid
order by 2 desc,1
select sum(score) sum_score, rank() over (order by sum(score) desc) rank from sc
group by sid
查询学生的总成绩,并进行排名,总分重复时不保留名次空缺
统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[60-0] 及所占百分比
select cid,
sum(case when score between 85 and 100 then 1 else 0 end)*1.0/count(sid) '85-100',
sum(case when score between 70 and 85 then 1 else 0 end)*1.0/count(sid) '70-85',
sum(case when score between 60 and 70 then 1 else 0 end)*1.0/count(sid) '60-70',
sum(case when score between 85 and 100 then 1 else 0 end)*1.0/count(sid) '0-60'
from sc group by cid
select * from (select cid, sid,score, row_number() over (partition by cid order by score desc) as rank from sc) t
where t.rank <= 3
select cid, count(sid) from sc
group by cid
select sc.sid from sc
group by sid
having count(distinct cid)=2
select count(distinct t1.sid) 男,count(distinct t2.sid) 女 from
(select sid from student where ssex = '男') t1,
(select sid from student where ssex = '女') t2
select * from student
where sname like '%风%'
select sname, count(*) from student
group by sname
having count(*)>1;
select * from
(select t1.* from student t1, student t2
where t1.sname = t2.sname and t1.sid != t2.sid group by t1.sid) a
select * from student
where sage between '1990-01-01' and '1991-01-01'
select distinct cid, avg(score) over (partition by cid) avg_score from sc
order by 2 desc,1
select * from
(select distinct sid , avg(score) over (partition by sid) avg_score from sc) t1
where avg_score >= 85
select st.sname,sc.*,c.cname from sc
join course c
on sc.cid = c.cid and c.cname = '数学'
join student st
on sc.sid = st.sid
where sc.score < 60
select st.sid, st.sname,sc.* from student st
left join sc
on st.sid = sc.sid
select sc.sid, st.sname, c.cname, sc.score from sc
join student st
on sc.sid = st.sid
join course c
on sc.cid = c.cid
where sc.score > 70
select sc.sid, st.sname, c.cname, sc.score from sc
join student st
on sc.sid = st.sid
join course c
on sc.cid = c.cid
where sc.score < 60
select sc.cid, sc.sid, st.sname,sc.score from sc
join student st
on sc.sid = st.sid
where sc.cid = '01' and sc.score >=80
select sc.cid, count(sc.sid) from sc
group by sc.cid
select st.*,max(sc.score) from sc
join course c
on sc.cid = c.cid
join teacher t
on c.tid = t.tid and t.tname = '张三'
join student st
on sc.sid = st.sid
select t1.* from
(select st.*, sc.score from sc
join course c
on sc.cid = c.cid
join teacher t
on c.tid = t.tid and t.tname = '张三'
join student st
on sc.sid = st.sid) t1,
(select st.*,max(sc.score) as max_score from sc
join course c
on sc.cid = c.cid
join teacher t
on c.tid = t.tid and t.tname = '张三'
join student st
on sc.sid = st.sid) t2
where t1.score = t2.max_score
select sc.* from sc,
(select sid,cid,score from sc
group by sid
having count(distinct cid) > count(distinct score)) t1
where sc.sid = t1.sid
select * from
(select cid, sid, row_number() over (partition by cid order by score desc) as rank from sc)
t1
where rank<=2
select * from
(select cid, count(distinct sid) as num from sc
group by cid) t1
where num > 5
select sid from
(select sid, count(cid) as num from sc group by sid) t1
where num > 2
select sid from sc
group by sid
having count(distinct cid) = (select count(distinct cid) from course)
select sid, sname, timestampdiff(year, sage, current_date()) age from student
select sid, sname, timestampdiff(year, sage, current_date()) age from student
select *
from student
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(current_date())
select *
from student
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(current_date())+1
select *
from student
where MONTH(student.Sage)=MONTH(current_date())
select *
from student
where MONTH(student.Sage)=MONTH(current_date())+1