由于被拉去進行建模比賽,小生的sql練習計劃直接被延后了三天。趁著今晚沒課(逃了)將這一套
題集進行一下練習,鞏固下之前的學習。需要說明的是,練習所針對的數據都是來自上一盤文章中的
http://blog.csdn.net/kiritor/article/details/8790214。數據不多,也就是學生、課程的相關信息。
1、查詢“C001”課程比“C002”課程成績高的所有學生的信息。
☞ 在from子句后面使用子查詢
[sql] view plaincopyprint?
select s.*,a.cno,a.score from student s,
(
select * from score where cno='C001'
)a,
(
select * from score where cno='C002'
)b
where a.sno=b.sno and a.score>=b.score and s.SNO = a.SNO
;
select s.*,a.cno,a.score from student s,
(
select * from score where cno='C001'
)a,
(
select * from score where cno='C002'
)b
where a.sno=b.sno and a.score>=b.score and s.SNO = a.SNO
;
☞ 使用相關子查詢方式實現
[sql] view plaincopyprint?
select * from student s, score a
where a.cno ='C001'
and exists
(
select * from score b
where b.cno='C002'
and a.score >=b.score
and a.sno = b.sno
)
and s.sno =a.sno
;
select * from student s, score a
where a.cno ='C001'
and exists
(
select * from score b
where b.cno='C002'
and a.score >=b.score
and a.sno = b.sno
)
and s.sno =a.sno
;
2、查詢平均成績大于60分的所有學生的信息,包括其平均成績
☞ 第一種方式
[sql] view plaincopyprint?
select s.sno,t.sname, avg(s.score) from student t, score s
where t.sno=s.sno
group by (s.sno,t.sname)
having avg(s.score)>60
;
select s.sno,t.sname, avg(s.score) from student t, score s
where t.sno=s.sno
group by (s.sno,t.sname)
having avg(s.score)>60
;
☞ 第二種方式
[sql] view plaincopyprint?
select t.sno ,t.sname, m.avg_s from student t,
(
select s.sno, avg(s.score) avg_s from score s
group by s.sno having avg(s.score)>60
) m
where m.sno = t.sno
;
select t.sno ,t.sname, m.avg_s from student t,
(
select s.sno, avg(s.score) avg_s from score s
group by s.sno having avg(s.score)>60
) m
where m.sno = t.sno
;
3、查詢所有同學的姓名、學號、總成績、選課數
[sql] view plaincopyprint?
/*思路:可以知道的是選課數、總成績可以通過
子查詢中的內置函數查詢出來的*/
--首先查詢出總成績與選課數
select sum(score) ,count(cno) from score group by sno;
--之后查詢學生的學號姓名等信息就順理成章了
select t.* ,tmp.sum_sc,tmp.sum_cn from student t,
(
select sno, sum(score) sum_sc ,count(cno) sum_cn from score group by sno
) tmp
where t.sno = tmp.sno
;
/*思路:可以知道的是選課數、總成績可以通過
子查詢中的內置函數查詢出來的*/
--首先查詢出總成績與選課數
select sum(score) ,count(cno) from score group by sno;
--之后查詢學生的學號姓名等信息就順理成章了
select t.* ,tmp.sum_sc,tmp.sum_cn from student t,
(
select sno, sum(score) sum_sc ,count(cno) sum_cn from score group by sno
) tmp
where t.sno = tmp.sno
;
4、查詢姓為“劉”的老師的信息
[sql] view plaincopyprint?
select * from teacher where tname like '劉%';
select count(*) from teacher where tname like '劉%';
select * from teacher where tname like '劉%';
select count(*) from teacher where tname like '劉%';
5、查詢學過“王燕”老師課的同學的學號、姓名
思考:首先查詢出王燕老師教授的課程的編號
☞ 第一種方式
[sql] view plaincopyprint?
select t.* ,s.cno,s.score from student t, score s
where s.cno in
(
select distinct cno from course c,teacher t
where c.tno =
(
select tno from teacher where tname='王燕'
原文轉自:http://blog.csdn.net/kiritor/article/details/8805310