Student 学生表 ，Course 课程表 ，SC成绩表 ，Teacher 教师表，sql操作运用 数据库面试题 有四个表Student表，Course表，Sc...
\u6709\u56db\u5f20\u8868\uff1aStudent \u5b66\u751f\u8868 \uff0cCourse \u8bfe\u7a0b\u8868 \uff0cSC\u6210\u7ee9\u8868 \uff0cTeacher \u6559\u5e08\u8868\uff0c\u6c42SQL\u8bed\u53e5select sc.* from sc,course where sc.\u8bfe\u7a0b\u53f7=course.\u8bfe\u7a0b\u53f7 and course.\u8bfe\u7a0b\u540d='3-105' and sc.\u6210\u7ee9 between 60 and 80;
select sc.* from sc where sc.\u6210\u7ee9 in(85,56,88);
3. \u8fd9\u4e2a\u6709\u7591\u95ee
\u6211\u7684\u7406\u89e3\u662f\uff0c\u5b66\u751f\u53c2\u52a0\u4e86\u591a\u95e8\u8003\u8bd5\uff0c\u6240\u6709\u6210\u7ee9\u90fd\u572870-90\u4e4b\u95f4\u7684\u3002\u53d6\u8fd9\u90e8\u5206\u4eba\u7684\u5b66\u53f7\u3002
select \u5b66\u53f7 from (select \u5b66\u53f7\uff0cmax(\u6210\u7ee9) \u6700\u9ad8\u5206\uff0cmin(\u6210\u7ee9) \u6700\u4f4e\u5206 from sc group by \u5b66\u53f7) a where a.\u6700\u9ad8\u5206 70
1.
select \u82f1\u8bed.\u5b66\u53f7
from Score \u82f1\u8bed
left join
(
select \u5b66\u53f7,\u8bfe\u7a0b\u4ee3\u7801,\u6210\u7ee9
from Score where \u8bfe\u7a0b\u4ee3\u7801='\u6570\u5b66\u8bfe\u7a0b\u4ee3\u7801'
) as \u6570\u5b66
on \u82f1\u8bed.\u5b66\u53f7=\u6570\u5b66.\u5b66\u53f7
where \u82f1\u8bed.\u8bfe\u7a0b\u4ee3\u7801='\u82f1\u8bed\u8bfe\u7a0b\u4ee3\u7801' and \u82f1\u8bed.\u6210\u7ee9>\u6570\u5b66.\u6210\u7ee9
2
select Student.\u5b66\u53f7,Student.\u59d3\u540d,AVG(\u6210\u7ee9) as \u5e73\u5747\u6210\u7ee9
from Score
left join Student on Student.\u5b66\u53f7 =Score.\u5b66\u53f7
group by Student.\u5b66\u53f7,Student.\u59d3\u540d
having AVG(Score.\u6210\u7ee9)>30
3
select Student.\u5b66\u53f7,Student.\u59d3\u540d, ISNULL(\u9009\u8bfe_\u6210\u7ee9.\u9009\u8bfe\u6570,0) as \u9009\u8bfe\u6570,ISNULL(\u9009\u8bfe_\u6210\u7ee9.\u603b\u6210\u7ee9,0) as \u603b\u6210\u7ee9
from Student
left join
(
select \u5b66\u53f7, COUNT(Score.\u8bfe\u7a0b\u4ee3\u7801) as \u9009\u8bfe\u6570,SUM(Score.\u6210\u7ee9) as \u603b\u6210\u7ee9
from Score group by \u5b66\u53f7
) as \u9009\u8bfe_\u6210\u7ee9
on Student.\u5b66\u53f7= \u9009\u8bfe_\u6210\u7ee9.\u5b66\u53f7
4
select a.\u5b66\u53f7,a.\u59d3\u540d from Student as a
where a.\u5b66\u53f7 not in(
select distinct(Student.\u5b66\u53f7) as \u5b66\u53f7
from Student
left join Score on Score.\u8bfe\u7a0b\u4ee3\u7801 =
(
select Course.\u8bfe\u7a0b\u4ee3\u7801 from Course
where Course.\u6559\u5e08\u7f16\u53f7 =
(
select Teacher.\u6559\u5e08\u7f16\u53f7 from Teacher
where Teacher.\u6559\u5e08\u59d3\u540d='\u738b\u519b'
)
)
)
\u4ec5\u4f9b\u5b66\u4e60\u53c2\u8003
CREATE TABLE student
(
s# INT,
sname nvarchar(32),
sage INT,
ssex nvarchar(8)
)
CREATE TABLE course
(
c# INT,
cname nvarchar(32),
t# INT
)
CREATE TABLE sc
(
s# INT,
c# INT,
score INT
)
CREATE TABLE teacher
(
t# INT,
tname nvarchar(16)
)
插入测试数据语句
insert into Student select 1,N'刘一',18,N'男' union all
select 2,N'钱二',19,N'女' union all
select 3,N'张三',17,N'男' union all
select 4,N'李四',18,N'女' union all
select 5,N'王五',17,N'男' union all
select 6,N'赵六',19,N'女'
insert into Teacher select 1,N'叶平' union all
select 2,N'贺高' union all
select 3,N'杨艳' union all
select 4,N'周磊'
insert into Course select 1,N'语文',1 union all
select 2,N'数学',2 union all
select 3,N'英语',3 union all
select 4,N'物理',4
insert into SC
select 1,1,56 union all
select 1,2,78 union all
select 1,3,67 union all
select 1,4,58 union all
select 2,1,79 union all
select 2,2,81 union all
select 2,3,92 union all
select 2,4,68 union all
select 3,1,91 union all
select 3,2,47 union all
select 3,3,88 union all
select 3,4,56 union all
select 4,2,88 union all
select 4,3,90 union all
select 4,4,93 union all
select 5,1,46 union all
select 5,3,78 union all
select 5,4,53 union all
select 6,1,35 union all
select 6,2,68 union all
select 6,4,71
问题
问题： 1、查询“001”课程比“002”课程成绩高的所有学生的学号； select a.S# from (select s#,score from SC where C#='001') a,(select s#,score from SC where C#='002') b where a.score>b.score and a.s#=b.s#; 2、查询平均成绩大于60分的同学的学号和平均成绩； select S#,avg(score) from sc group by S# having avg(score) >60; 3、查询所有同学的学号、姓名、选课数、总成绩； select Student.S#,Student.Sname,count(SC.C#),sum(score) from Student left Outer join SC on Student.S#=SC.S# group by Student.S#,Sname 4、查询姓“李”的老师的个数； select count(distinct(Tname)) from Teacher where Tname like '李%'; 5、查询没学过“叶平”老师课的同学的学号、姓名； select Student.S#,Student.Sname from Student where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平'); 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名； select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002'); 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名； select S#,Sname from Student where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平')); 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名； Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2 from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score; 9、查询所有课程成绩小于60分的同学的学号、姓名； select S#,Sname from Student where S# not in (select S.S# from Student AS S,SC where S.S#=SC.S# and score>60); 10、查询没有学全所有课的同学的学号、姓名； select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course); 11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名； select distinct S#,Sname from Student,SC where Student.S#=SC.S# and SC.C# in (select C# from SC where S#='1001'); 12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名； select distinct SC.S#,Sname from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='001'); 13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩； update SC set score=(select avg(SC_2.score) from SC SC_2 where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名； select S# from SC where C# in (select C# from SC where S#='1002') group by S# having count(*)=(select count(*) from SC where S#='1002'); 15、删除学习“叶平”老师课的SC表记录； Delect SC from course ,Teacher where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平'; 16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、 号课的平均成绩； Insert SC select S#,'002',(Select avg(score) from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002'); 17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分 SELECT S# as 学生ID ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语 ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩 FROM SC AS t GROUP BY S# ORDER BY avg(t.score) 18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分 SELECT L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分 FROM SC L ,SC AS R WHERE L.C# = R.C# and L.score = (SELECT MAX(IL.score) FROM SC AS IL,Student AS IM WHERE L.C# = IL.C# and IM.S#=IL.S# GROUP BY IL.C#) AND R.Score = (SELECT MIN(IR.score) FROM SC AS IR WHERE R.C# = IR.C# GROUP BY IR.C# );
自己写的:select c# ,max(score)as 最高分 ,min(score) as 最低分 from dbo.sc group by c# 19、按各科平均成绩从低到高和及格率的百分数从高到低顺序 SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩 ,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数 FROM SC T,Course where t.C#=course.C# GROUP BY t.C# ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004） SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分 ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数 ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数 FROM SC
21、查询不同老师所教不同课程平均分从高到低显示
SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C ,Teacher AS Z
where T.C#=C.C# and C.T#=Z.T#
GROUP BY C.C#
ORDER BY AVG(Score) DESC
22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004）
[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
SELECT DISTINCT top 3
SC.S# As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC LEFT JOIN SC AS T1
ON SC.S# = T1.S# AND T1.C# = '001'
LEFT JOIN SC AS T2
ON SC.S# = T2.S# AND T2.C# = '002'
LEFT JOIN SC AS T3
ON SC.S# = T3.S# AND T3.C# = '003'
LEFT JOIN SC AS T4
ON SC.S# = T4.S# AND T4.C# = '004'
WHERE student.S#=SC.S# and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.S# = T1.S# AND T1.C# = 'k1'
LEFT JOIN sc AS T2
ON sc.S# = T2.S# AND T2.C# = 'k2'
LEFT JOIN sc AS T3
ON sc.S# = T3.S# AND T3.C# = 'k3'
LEFT JOIN sc AS T4
ON sc.S# = T4.S# AND T4.C# = 'k4'
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);
23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
SELECT SC.C# as 课程ID, Cname as 课程名称
,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]
,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]
,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]
,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]
FROM SC,Course
where SC.C#=Course.C#
GROUP BY SC.C#,Cname;
24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT S#,AVG(score) AS 平均成绩
FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩
FROM SC
GROUP BY S#
) AS T2
ORDER BY 平均成绩 desc;
原文地址：http://www.cnblogs.com/qixuejia/p/3637735.html
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