最近在複習資料庫SQL，所以開始在網上找了一些SQL的訓練。完整的50個題目及答案：

學生資訊表：

課程資訊：

選課記錄：

教師資訊：

1. 查詢01課程比"02"成績高的同學學號

select sc1.sid,sc1.score,sc2.score from sc sc1,sc sc2 where sc1.sid = sc2.sid and 
sc1.cid = "01" and sc2.cid = "02" and sc1.score>sc2.score; 

2. 查詢平均成績大於60分的同學的學號和平均成績

select sc.sid ,avg(sc.score) as avg from sc
group by sc.sid
having avg>60;
 

3. 查詢所有同學的學號，姓名，選課數，總成績

select s.sid,s.sname,(select count(*) from sc where sc.sid = s.sid) as "選課數",(select sum(sc.score) from sc where sc.sid = s.sid) as "總成績" 
from student s,sc
group by s.sid

4. 查詢姓李的老師個數

select count(*)from teacher t where t.tname like "李%";

5. 查詢沒有學過張三老師課的同學

寫法1：（使用in）

select s.sname from student s where s.sid not in
(select distinct sc.sid from sc,course c,teacher t where t.tname = "張三"
and t.tid = c.tid and c.cid = sc.cid);
 

寫法2：（使用exists）

select s.sname from student s where not exists
(select * from sc,course c,teacher t where t.tname = "張三" 
and t.tid = c.tid and c.cid = sc.cid and sc.sid = s.sid);

6. 查詢學過01也學過02的同學編號

寫法1：使用SQL除法

select s.sid,s.sname from student s where not exists(
	select distinct cid from (select c.cid from course c where c.cid = "01" or c.cid = "02") tmp where not exists(
		select * from sc where sc.sid = s.sid and sc.cid = tmp.cid));
 

寫法2：使用計數

select sc.sid,s.sname from student s ,sc where sc.cid in (select c.cid from course c where c.cid = "01" or c.cid = "02")
and s.sid = sc.sid
group by sc.sid
having count(*) = (select count(*) from course c where c.cid = "01" or c.cid = "02");

這裡解釋一下這段SQL：

首先我們在選課記錄中，選擇所有選中了01或者02的課程記錄

然後，我們進行分組，按照學號來分組，這樣每個組裡面的記錄都是某某同學選了01或者某某同學選了02

我們選擇這樣的分組，該分組的個數是01,02號課程的總和（當然就是2了）

7. 查詢學過“張三老師”所有課程的同學

同上，也分為兩種寫法。把6中的01,02課程的集合改為張老師教授的課程集合

8. 查詢課程“02”比“01”分數低同學學號

寫法同1題

9. 查詢所有課程成績小於60分的同學學號，姓名

select distinct sc.sid,s.sname from student s,sc where sc.sid not in 
(select distinct sc.sid from sc where sc.score>=60)
and sc.sid = s.sid ;

首先找出成績大於60分的選課記錄。該同學的學號不在集合中

10. 查詢沒有選全所有課的同學

select s.sid,(select count(*) from sc where sc.sid = s.sid) as cnt from student s,sc t
group by s.sid
having cnt != (select count(*) from course);

計數法

11. 查詢至少一門課和01學號同學相同的其他同學學號和姓名

select distinct sc.sid,s.sname from sc,student s where sc.cid in (select cid from sc where sc.sid = "01") and 
sc.sid != "01" and s.sid = sc.sid;

12.題乾和11相同

13. 把sc表中“張三”老師教的課的成績改為此課程平均數

update sc set score = (select avg(sc1.score) from sc sc1 where sc1.cid = sc.cid)
where sc.cid in (select sc2.cid from sc sc2,course c,teacher t where sc2.cid = c.cid and c.tid = t.tid and t.tname = "張三" )

mysql會報報錯，還沒找到解決辦法

14. 查詢和“01”號同學學習的課程完全相同的其他同學學號和姓名

select sc.sid from sc where sc.sid!="01" and sc.cid in (select cid from sc where sc.sid = "01")
group by sc.sid
having count(*) = (select count(*) from sc where sc.sid = "01");

15. 刪除學習“王五”老師課程的SC記錄

delete sc from sc,course c,teacher t 
where sc.cid = c.cid and c.tid = t.tid and t.tname = "王五";

16. 向SC表插入一些記錄，這些記錄符合：沒有上過“02”課程的同學學號，“02”，2號課平均成績

insert into sc
select distinct sc.sid,"02",(select avg(sc1.score) from sc sc1 where sc1.cid = "02")from sc where sc.sid not in (select sid from sc where cid = "02");

17. 按照平均成績顯示所有學生“01”，“02”，“03”三門課程的成績，按如下形式顯示：

學生ID，“01”，“02”，“03”，有效課程數，有效平均分

select t.sid,(select sc.score from sc where sc.cid = "01" and t.sid = sc.sid)
from sc t
group by t.sid;

18. 查詢各科成績最高和最低的分數

select sc.cid,max(sc.score),min(sc.score) from sc
group by sc.cid;

19.按各科平均成績從低到高和及格率的百分數從高到低排列.（不包含沒有選課記錄的課程）

select t.cid,avg(t.score) as avg, ((select count(*) from sc where sc.cid = t.cid and sc.score>=60 )/(select count(*) from sc where sc.cid = t.cid ))as rate
from sc t
group by t.cid
order by avg desc,rate asc;

20. 查詢如下課程的平均成績及合格率:01,02,03(一行表示)

select 
	(select avg(sc.score) from sc where sc.cid = "01") as avg,(select count(*) from sc where sc.cid = "01" and sc.score<60)/(select count(*) from sc where sc.sid = "01")*100 as rate,
	(select avg(sc.score) from sc where sc.cid = "02") as avg,(select count(*) from sc where sc.cid = "02" and sc.score<60)/(select count(*) from sc where sc.sid = "02")*100 as rate,
	(select avg(sc.score) from sc where sc.cid = "03") as avg,(select count(*) from sc where sc.cid = "03" and sc.score<60)/(select count(*) from sc where sc.sid = "03")*100 as rate;

21.查詢不同老師不同課程平均分從高到低顯示

select sc.cid,t.tname as tname,avg(sc.score) as avg from teacher t,course c,sc
where t.tid = c.tid and c.cid = sc.cid
group by sc.cid
order by avg desc;

22. 查詢如下課程第三名到第六名的同學：01,02,03

select cid,score,rank from
(select tmp.sid,tmp.cid,tmp.score,if(tmp.cid = @tmid,@rank:[email protected]+1,@rank:=1) as rank,@tmid:= tmp.cid
from (select sc.sid,sc.cid,sc.score from sc group by sc.cid,sc.score order by sc.cid asc,sc.score desc,sc.sid) tmp,(select @rank:=0,@tmid=null) t
)r
where r.cid = "01" or r.cid = "02" or r.cid = "03"
having rank>=3 and rank<=6;

23. 統計列印各科成績，各分數段人數，課程id，課程名稱，[100-85],[85-70],[70-60],[<60]

select r.cid,cname,`[100-85]`,`[85-70]`,`[70-60]`,`[<=60]` from
(select t.cid,
(select count(*) from sc where sc.cid = t.cid and sc.score<=100 and sc.score>85) as "[100-85]",
(select count(*) from sc where sc.cid = t.cid and sc.score<=85 and sc.score>70) as "[85-70]",
(select count(*) from sc where sc.cid = t.cid and sc.score<=70 and sc.score>60) as "[70-60]",
(select count(*) from sc where sc.cid = t.cid and sc.score<=60) as "[<=60]"
from sc t
group by t.cid) as r,course c
where c.cid = r.cid;

這個題目要注意的地方是，當列名中含有[]、（）的時候，在查詢時要加上` `，即反引號（tab上邊）

24. 查詢學生平均成績及其名次

select sid,sname,avg,@rownum:[email protected]+1 as rank from
(select tmp.sid,sname,avg from
(select sid ,avg(sc.score) as avg from sc group by sc.sid order by avg desc)tmp,student s
where tmp.sid = s.sid
order by avg desc)
r,(select @rownum:=0)t

這個題目，解釋瞭如何在Mysql中實現rank函式的實現

25. 查詢各科成績前三名記錄

解法同22題

26. 查詢每門課程被選的人數

select sc.cid ,count(*) from sc 
group by sc.cid;

27. 查詢只選修了一門課程的學生姓名學號

select s.sname,s.sid from student s,sc
where s.sid = sc.sid 
group by sc.sid
having count(sc.cid) = 1;

28. 查詢男女生人數

select (select count(*) from student s where s.ssex = "男") as "男",
(select count(*) from student s where s.ssex = "女") as "女";

29. 查詢姓王的同學資訊

同4題

30. 查詢同名同姓的學生名單，並統計同名人數

select s.sname,count(*) as cnt from student s
group by s.sname
having cnt >1;

31. 查詢1991年出生的學生

select s.sname from student s where s.sbirth = "1991";

32. 查詢每門成績的平均成績，結果按照平均成績升排列，平均成績相同時，按照課程號
降序排列

select sc.cid,avg(sc.score) as avg
from sc
group by sc.cid
having avg(sc.score) is not null
order by avg asc,sc.cid desc;

33. 查詢平均分數大於85的同學學號，姓名和平均成績

select s.sid,s.sname,avg(sc.score) as “Avg” from student s,sc
where s.sid = sc.sid
group by sc.sid
having  avg(sc.score) > 85;

34. 查詢課程名稱為“語文”，且分數低於60分的學生姓名和分數

select s.sname,sc.score from student s,sc,course c
where s.sid = sc.sid and sc.cid = c.cid and c.cname = "語文"
and sc.score <60;

35. 查詢所有學生的選課情況

select s.sid,s.sname,c.cname from sc,student s ,course c
where s.sid = sc.sid and c.cid = sc.cid ; 

36. 查詢任何一門成績在70分以上的姓名，課程編號和分數

select s.sname,sc.cid,sc.score from sc,student s
where s.sid = sc.sid and sc.score>=70;

37. 查詢不及格的課程成績並且按照課程號從小到大排列

select sc.cid,sc.score from sc where sc.score <60
order by sc.cid desc;

38.查詢課程編號為003且課程成績在80分以上的學生學號和姓名

select s.sid,s.sname from student s,sc where s.sid = sc.sid and
sc.cid = "03" and sc.score >=80;

39. 求選了課程的人

select count(*) from sc;

40.查詢選修“張三”老師的所有課程中，成績最高的學生姓名及其成績

select cid,tmp.sid,sname,score from
(select s.sname,sc.cid as cid,sc.sid as sid,sc.score as score from
sc,course c,teacher t,student s where 
t.tname = "張三" and 
t.tid = c.tid and 
sc.cid = c.cid and 
sc.sid = s.sid order by sc.score desc) tmp
limit 1;

41.查詢各個課程及相應選修人數

select sc.cid,c.cname,count(sc.sid) from sc,course c where c.cid = sc.cid
group by sc.cid;

42. 查詢不同課程成績相同的學生的學號，課程號，學生成績

select sc.sid,sc.cid,sc.score from sc,sc sc1 where sc.sid != sc1.sid and sc.score = sc1.score;

43.查詢每門課成績最好的前兩名：

select r.cid,c.cname,score,rank from
(select tmp.cid,tmp.score,if([email protected],@rank:[email protected]+1,@rank:=1) as rank,@tmid:=tmp.cid
from (select cid,score from sc group by cid,score order by cid asc,score desc) tmp,
(select @rank:=0,@tmid:=null)tmp1
) r,course c
where r.cid = c.cid and score is not null and rank <=2;

同樣也是rank的實現

44. 統計每門課學生選修人數(超過三個人才統計)。要求輸出課程號和選修人數，平均分
查詢結果按照人數降序排列，查詢結果按照人數降序排列。若人數相同按照課程號升序排列

select sc.cid,c.cname,count(sc.sid) as cnt ,avg(sc.score) from sc,course c
where c.cid = sc.cid
group by sc.cid
having count(sc.sid)>3
order by cnt desc,sc.sid asc;

45. 檢索至少選修三門課的學生學號

select sc.sid from sc group by sc.sid having count(sc.cid)>2;

46.查詢全部學生都選修的課程課程號和課程名

select c.cid,c.cname from course c where not exists
(select s.sid from student s where not exists
(select sc.sid from sc where sc.cid = c.cid and sc.sid = s.sid));

使用SQL除法，找出覆蓋了所有學生集合的課程

47.查詢沒有學過“王五”老師任一門課的學生姓名

同5題

48. 查詢兩門以上不及格課程的同學的學號及平均成績

select sc.sid,avg(sc.score) as avg from sc where exists 
(select * from sc sc1 where sc1.sid = sc.sid and sc1.score < 60 
having count(*) >= 2)
group by sc.sid;

49. 檢索04課程分數大於60的同學學號，按照分數降序排列

select sc.sid from sc  where sc.cid = "04" and sc.score > 60
order by sc.score desc;

50. 刪除“02”同學“01”課程成績

delete from sc where sc.sid = "02" and sc.cid = "01";

=======================================================================================================================

整理的其他題目

=======================================================================================================================

1. 統計選課但是沒有選02課程的學生姓名、學號

select distinct s.sid,s.sname from student s,sc where s.sid = sc.sid and not exists
(select sc.sid from sc where sc.cid = '02');

挑選這樣的學生，選課記錄中不存在02課程

2. 統計各個學科的平均分

select sc.cid,c.cname,avg(sc.score) from course c,sc where sc.cid = c.cid 
group by sc.cid

3. 挑選年齡比平均年齡大的同學資訊

select s.sid,s.sname,s.sage from student s where s.sage >
(select avg(s1.sage) from student s1)

首先在子查詢計算學生平均年齡，然後檢索

4.查詢至少選修了語文數學的學生學號

select distinct sc1.sid from student sc1
	where not exists
		(select distinct c.cid from (select * from course t where t.cid = "01" or t.cid = "02" ) as c
			where not exists 
					(select * from sc sc2
						where sc1.sid = sc2.sid
						and   c.cid = sc2.cid))

使用SQL除法表示，詳細檢視SQL除法這篇文章

5.選修了所有課程的同學學號（非計數寫法）

select distinct sc1.sid from student sc1  
	where not exists    
		(select distinct c.cid from course c   
			where not exists      
					(select * from sc sc2        
						where sc1.sid = sc2.sid
						and   c.cid = sc2.cid))

解釋同上。同時也可以用計數的方法，就是看同學選課的個數。也有很多習題使用這種簡單的寫法

6.王菊同學不學的課程名

select cid from course where cid not in (select sc.cid from sc ,student s where sc.sid = s.sid and s.sname = "王菊")

在子查詢中選出王菊的課程，然後減去該集合

7.至少選了兩門課的同學資訊

select s.sname, sc.sid,count(sc.cid) from student s, sc where s.sid = sc.sid group by sc.sid having count(sc.cid)>2;

按照學號分組，使用聚集函式計算每一組學生選課數目，檢索大於2的學生資訊

8.所有學生都選的課

select cid,cname from course c 
	where not exists
		(select sid from student s 
			where not exists 
				(select * from sc where sc.cid = c.cid and sc.sid = s.sid))

使用SQL除法，選出這樣一門課，不存在一個學生沒有選該課

9.選修了語文的同學的平均年齡

select avg(s.sage) from student s,sc where s.sid in (select distinct sc.sid from sc,course c where sc.cid = c.cid and c.cname = "語文")

首先選出選修語文的同學學號，計算平均年齡

10.各科選修同學的平均年齡

select c.cname,avg(s.sage) from course c,student s,sc where sc.sid =s.sid and sc.cid = c.cid
group by sc.cid;

按照課程號分組，計算每一組的學生平均年齡

11.求張三教授的每門課的學生平均成績

select c.tid,c.cname,avg(sc.score)
from course c,teacher t,sc
where c.tid = t.tid and t.tname = "張三" and sc.cid = c.cid
group by sc.cid

和10很相似，新增課程老師是張三這個條件

12.統計每門課的選課人數（超過3人），要求輸出課程號和選課人數，查詢結果按照人數降序排列，若人數相同按照課程號排列

select c.cname,sc.cid,count(sc.sid) as cnt from sc,course c where c.cid = sc.cid
group by sc.cid having count(sc.sid) > 2 order by cnt desc,sc.cid

首先按照課程號分組，選擇人數在三個以上，計算課程選課人數

13.學號比趙雷大，年齡比趙雷小的學生資訊

select * from student s
	where exists 
		(select * from student s1 where  s1.sname = "趙雷" and s1.sage>s.sage and s1.sid<s.sid )

14.所有姓王的同學資訊

select * from student s where s.sname like "王%"

使用like 和%符號，比使用正則表示式更為簡單。%表示0個或多個任意符號

15.sc表中成績為空的學生學號和課號

select sc.sid,sc.cid from sc where sc.score is null;

使用isnull判斷空值

16.年齡大於女同學平均年齡的男同學姓名年齡

select s.sname,s.sage from student s where s.ssex = "男" and s.sage > 
	(select avg(s1.sage) from student s1 where s1.ssex = "女");

17.年齡大於所有男同學年齡的女同學姓名年齡

select s.sname,s.sage from student s where s.ssex = "女" and 
s.sage > all (select s.sage from student s where s.ssex = "男");

18.在student 中檢索所選選課程每一門都大於等於80的學生的姓名學號性別，儲存到新的表youxiu中

首先建立表youxiu：

create table youxiu(
sid char(10),
sname char(10),
ssex char(10),
primary key(sid)
)

將select語句直接放在insert之後（沒有新增values）

insert into youxiu(sid,sname,ssex)
select s.sid,s.sname,s.ssex from student s
where not exists
(select sc.cid from sc
where sc.sid = s.sid and (sc.score <80 or sc.score is null))

注意，這裡為什麼要加上 isnull 呢？

主要 我們有一條記錄 王二

09

05

所以，王二同學在選課表中05課程沒有成績，在我們的select語句中，當檢索到王二的時候：

由於null 和 80相比結果為false，所以not exists 後邊的select子查詢沒有返回結果，所以not exists 為真，返回了王二這一條記錄

而我們不需要王二的記錄，所以新增上is null

P.S.文章不妥之處還望指正