問題描述: 已知關系模式: S(SNO,SNAME)學生關系。SNO為學號,SNAME為姓名 C(CNO,CNAME,CTEACHER)課程關系。CNO為課程號,CNAME為課程名,CTEACHER為任課教師 SC(SN" name="description" />
題目2
MILY: " BACKGROUND-COLOR: #fafafa; Verdana: ; quot: ; Courier: ; mono: ">問題描述:
已知關系模式:
S (SNO,SNAME) 學生關系。SNO 為學號,SNAME 為姓名
C (CNO,CNAME,CTEACHER) 課程關系。CNO 為課程號,CNAME 為課程名,CTEACHER 為任課教師
SC(SNO,CNO,SCGRADE) 選課關系。SCGRADE 為成績
1. 找出沒有選修過“李明”老師講授課程的所有學生姓名
--實現代碼:
SELECT SNAME FROM S
WHERE NOT EXISTS(
SELECT * FROM SC,C
WHERE SC.CNO=C.CNO
AND CNAME='李明'
AND SC.SNO=S.SNO)
2. 列出有二門以上(含兩門)不及格課程的學生姓名及其平均成績
--實現代碼:
SELECT S.SNO,S.SNAME,AVG_SCGRADE=AVG(SC.SCGRADE)
FROM S,SC,(
SELECT SNO
FROM SC
WHERE SCGRADE<60
GROUP BY SNO
HAVING COUNT(DISTINCT CNO)>=2
)A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO
GROUP BY S.SNO,S.SNAME
3. 列出既學過“1”號課程,又學過“2”號課程的所有學生姓名
--實現代碼:
SELECT S.SNO,S.SNAME
FROM S,(
SELECT SC.SNO
FROM SC,C
WHERE SC.CNO=C.CNO
AND C.CNAME IN('1','2')
GROUP BY SNO
HAVING COUNT(DISTINCT CNO)=2
)SC WHERE S.SNO=SC.SNO
4. 列出“1”號課成績比“2”號同學該門課成績高的所有學生的學號
--實現代碼:
SELECT S.SNO,S.SNAME
FROM S,(
SELECT SC1.SNO
FROM SC SC1,C C1,SC SC2,C C2
WHERE SC1.CNO=C1.CNO AND C1.NAME='1'
AND SC2.CNO=C2.CNO AND C2.NAME='2'
AND SC1.SCGRADE>SC2.SCGRADE
)SC WHERE S.SNO=SC.SNO
5. 列出“1”號課成績比“2”號課成績高的所有學生的學號及其“1”號課和“2”號課的成績
--實現代碼:
SELECT S.SNO,S.SNAME,SC.[1號課成績],SC.[2號課成績]
FROM S,(
SELECT SC1.SNO,[1號課成績]=SC1.SCGRADE,[2號課成績]=SC2.SCGRADE
FROM SC SC1,C C1,SC SC2,C C2
WHERE SC1.CNO=C1.CNO AND C1.NAME='1'
AND SC2.CNO=C2.CNO AND C2.NAME='2'
AND SC1.SCGRADE>SC2.SCGRADE
)SC WHERE S.SNO=SC.SNO