sql查询语句练习(解析版) BY DD
表情况
Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表
create table Student(S# varchar2(20),Sname varchar2(10),Sage int,Ssex varchar2(2)) ;
create table Course(C# varchar2(20),Cname varchar2(10),score varchar2(4)) ;
create table SC(S# varchar2(20),C# varchar2(20),score varchar2(4)) ;
create table Teacher(T# varchar2(20),Tname varchar2(10)) ;
insert into Student(S#,Sname,Sage,Ssex) values('1001','李五','15','男');
insert into Student(S#,Sname,Sage,Ssex) values('1002','张三','16','女');
insert into Student(S#,Sname,Sage,Ssex) values('1003','李四','15','女');
insert into Student(S#,Sname,Sage,Ssex) values('1004','陈二','14','男');
insert into Student(S#,Sname,Sage,Ssex) values('1005','小四','15','男');
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#;
解析:(select s#,score from SC where C#='001') a//从SC中查询C#=001的学生学号和分数,并定义为a表。
(select s#,score from SC where C#='002') b //从SC中查询C#=002的学生学号和分数,并定义为b表。
where a.score>b.score and a.s#=b.s# 条件,a表与b表中学号相同且分数比b中的高。
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) rom sc
group by S# having avg(score) >60;
解析: group by S# having avg(score) >60 这里是指平均分数大于60分的学号。使用HAVING 子句原因是,WHERE 关键字无法与合计函数一起使用。
合计函数 (比如 SUM) 常常需要添加 GROUP BY 语句,根据一个或多个列对结果集进行分组。
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
解析: Student left Outer join SC on Student.S#=SC.S# 左外连接,从student表返回所有的行放于左表中,从SC中返回的与左表匹配的行放于右表。Student表中有但SC表中没有的在右表中对应行留空。
group by Student.S#,Sname 以 Student.S#,Sname分组。
4、查询姓“李”的老师的个数;
select count(distinct(Tname))
from Teacher
where Tname like '李%';
解析:distinct(Tname) 在表中,可能会包含重复值。DISTINCT 用于返回唯一不同的值,这里的作用就是防止返回的Tname出现重复值。
where Tname like '李%' 模糊查询,"%"代表一或多个字符 。
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');
解析:not in 题目是查询没学过“叶平”老师课的同学的学号、姓名,这里使用not in 就表示所要查询的S#不在后面的结果中。
(select distinct(SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平');
这段语句是查询学过“叶平”老师课的同学的学号,这里其实查询的SC表,但可看到查的表是SC,Course,Teacher表,根据后面的条件语句,用来关联查询的。三个条件要同时满足,从而查询出“叶平”老师课的同学的学号
SC.C#=Course.C#,Teacher.T#=Course.T# 由于SC与Teacher并无直接联系,这里将SC与Teacher表联系起来, Teacher.Tname='叶平':教师名为叶平,从而对应到SC表。
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');
解析:select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001' 这一句查询到的是学过“001”课程的同学的学号、姓名;
and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002');
使用 exists关键字,用来判断是否存在的,当exists(查询)中的查询结果存在时则返回真,否则返回假。这里指查询学过002课程的同学,使用 and则将得到的结果与上一句匹配得上就都为真,返回结果。若不匹配,则不返回结果。
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'));
解析:这一道题与第5题接近,只是答案后面多了:group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平')。
Group by...having..的句式我们已经知道是因为WHERE 关键字无法与合计函数一起使用。合计函数 (比如 SUM) 常常需要添加 GROUP BY 语句,根据一个或多个列对结果集进行分组。
接下来就是理解count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平')的含义了,count(SC.C#)是基于前面条件叶平老师在SC表中课程号的数量,select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平',是从Course中查询叶平老师的所有课程号数量。
这里条件是说course中叶平老师的课程数与成绩表中的课程数相等。(SB我也不知为什么这么写)
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 60);
不解析!
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);
解析: having count(C#) <(select count(C#) from Course); 基于上面条件Student.S#=SC.S# 课程数量小于course表中所有课程数。
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 S#,Sname from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='1001');
不解析!
12、查询至少学过学号为“1001”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#='1001');
不解析!
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');
不解析!
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';
解析:delete from ..
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 into SC (S#,C#,score) values(select S#,'002',(Select avg(score) from SC where C#='002') from Student where S# not in (Select S# from SC where C#='003'));
解析:insert into table1(field1,field2) values(value1,value2)
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)
不解析!
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#
);
解析:不想解析!
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
解析:isnull(AVG(score),0) 使用指定的替换值替换 NULL,即是说当返回的值为空时使用0代替输出,当不为空时,则输出AVG(score)的值。
100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*)
CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END:当isnull(score,0)大于等于60时,输出1,否则为0,
计算CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END输出的总和,再除于Count(*)总个数,就是百分数了。
至于乘于100,神马?
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 WHENC# = '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
解析:不就是计算平均分和百分比么,case when ...then...else ..then
平均分:SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE WHENC# = '001' THEN 1 ELSE 0 END) 课程为001的总分数/课程为001的总个数=平均分
百分数:100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END 课程为001的大于等于60分的个数/课程为001的总个数
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
不解析:max(Z.T#) ,MAX(Z.Tname)
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;
解析:SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) 分数在85到100的个数,以下类似方法。
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;
解析:
select s#,avg(score) as 平均成绩 from SC group by s#-查询学号和平均成绩
1+(select count(distinct 平均成绩) from T1。。。。。)计算出名次,随着循环加1。
(SELECT S#,AVG(score) 平均成绩 FROM SC GROUP BY S# ) AS T2
T2 在不断缩小范围 由平均成绩 > T2.平均成绩看出。
25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT t1.S# as 学生ID, t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.C#= C# ORDER BY score DESC
)
ORDER BY t1.C#;
解析:1.SELECT TOP 3 score as score3 FROM SC ORDER BY score DESC
2 .SELECT t1.S# as 学生ID, t1.C# as 课程ID,Score as 分数 FROM SC t1 WHERE score in score3 ORDER BY t1.C#;
26、查询每门课程被选修的学生数
select c#,count(S#) from sc group by C#;
不解析!
27、查询出只选修了一门课程的全部学生的学号和姓名
select SC.S#,Student.Sname,count(C#) AS 选课数
from SC ,Student
where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
不解析!
28、查询男生、女生人数
Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男';
Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女';
不解析!
29、查询姓“张”的学生名单
SELECT Sname FROM Student WHERE Sname like '张%';
不解析!
30、查询同名同性学生名单,并统计同名人数
select Sname,count(*) from Student group by Sname having count(*)>1;;
不解析!
31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
from student
where CONVERT(char(11),DATEPART(year,Sage))='1981';
解析:CONVERT(char (11),DATEPART(year,Sage)),
DATEPART() 函数用于返回日期/时间的单独部分,比如年、月、日、小时、分钟等等
32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ;
不解析!
33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
select Sname,SC.S# ,avg(score)
from Student,SC
where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>85;
不解析!
34、查询课程名称为“数据库”,且分数低于60的学生姓名和分数
Select Sname,isnull(score,0)
from Student,SC,Course
where SC.S#=Student.S# and SC.C#=Course.C#and Course.Cname='数据库'and score <60;
解析:isnull(score,0) 使用指定的替换值替换 NULL,即是说当返回的值为空时使用0代替输出,当不为空时,则返回score的值。(懂了吧!!!)
35、查询所有学生的选课情况;
SELECT SC.S#,SC.C#,Sname,Cname
FROM SC,Student,Course
where SC.S#=Student.S# and SC.C#=Course.C# ;
不解析!
36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
SELECT distinct student.S#,student.Sname,SC.C#,SC.score
FROM student,Sc
WHERE SC.score>=70 AND SC.S#=student.S#;
解析:当某一同学有一门课程成绩大于70,则distinct可无,但当某一同学有多门课程成绩大于70,返回的结果将是有多少个上70分就多少条记录,并不理是否同一个人,这是就要是用distinct函数。
37、查询不及格的课程,并按课程号从大到小排列
select c# from sc where scor e <60 order by C# ;
不解析!
38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
不解析!
39、求选了课程的学生人数
select count(*) from sc;
不解析!
40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
select Student.Sname,score from Student,SC,Course ,Teacher
where Student.S#=SC.S# and SC.C#=Course.C#
and Course.T#=Teacher.T# and Teacher.Tname='叶平'
and SC.score=(select max(score)from SC where C#=Course.C# );
解析:and SC.score=(select max(score)from SC where C#=Course.C# ) 课程分数最高的。
41、查询各个课程及相应的选修人数
select count(*) from sc group by C#;
不解析!
42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
select distinct A.S#,B.score ,from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
这里的课程号从哪里得?
43、查询每门功成绩最好的前两名
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
解析:分解-1.select top 2 score as score2 from sc where t1.c#=c# order by score desc;
2.select t1.s# as 学生ID,t1.c#as 课程ID,Score as 分数 from SC t1
Where score in score order by t1.c#;
44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
select C# as 课程号,count(*) as 人数
from sc
group by C#
order by count(*) desc,c#
不解析!
45、检索至少选修两门课程的学生学号
select S# from sc
group by s# having count(*) > = 2
不解析!
46、查询全部学生都选修的课程的课程号和课程名
select C#,Cname
from Course
where C# in (select c# from sc group by c#)
解析:有成绩便是有选修!
47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
select Sname from Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平');
不解析!
48、查询两门以上不及格课程的同学的学号及其平均成绩
select S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where
score <60 group by S# having count(*)>2)group by S#;
解析:avg(isnull(score,0)) 使用指定的替换值替换 NULL。
49、检索“004”课程分数小于60,按分数降序排列的同学学号
select S# from SC where C#='004'and score <60 order by score desc;
解析:Desc 降序。Asc升序。
50、删除“002”同学的“001”课程的成绩
delete from Sc where S#='001'and C#='001';
解析:delete from ...where...
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