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[转]T-SQL_面试题
阅读量:5833 次
发布时间:2019-06-18

本文共 15922 字,大约阅读时间需要 53 分钟。

[转]T-SQL_面试题

2015-05-19 

1 创建表插入数据 

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

1.1 Student 

1 Create table Student 2 ( 3 S# number(4), 4 Sname varchar2(9) not null, 5 Ssex varchar2(3), 6 Sage date, 7 constraint Student_S#_Pk primary key(S#) 8 ); 9 10 insert into Student values(1001,'李志国','男',to_date('26-9-1985','dd-mm-yyyy'));11 insert into Student values(1002,'李甜甜','女',to_date('6-2-1987','dd-mm-yyyy'));12 insert into Student values(1003,'张小燕','女',to_date('18-11-1984','dd-mm-yyyy'));13 insert into Student values(1004,'王菲','男',to_date('3-6-1985','dd-mm-yyyy'));14 insert into Student values(1005,'杜宇','女',to_date('25-4-1986','dd-mm-yyyy'));15 insert into Student values(1006,'彭大生','男',to_date('28-12-1988','dd-mm-yyyy'));16 insert into Student values(1007,'王亮','男',to_date('1-8-1983','dd-mm-yyyy'));17 insert into Student values(1008,'赵婷婷','女',to_date('2-2-1984','dd-mm-yyyy'));
View Code

1.2 Course

Create table Course(C# number(3),Cname varchar2(12) not null,T# number(4),constraint Course_C#_Pk primary key(C#),constraint Course_T#_Fk foreign key (T#) references Teacher(T#));insert into Course values(001,'企业管理',2003);insert into Course values(002,'马克思',2001);insert into Course values(003,'OO'||'&'||'UML',2002);insert into Course values(004,'数据库',2004);insert into Course values(005,'英语',2005);insert into Course values(006,'大学语文',2005);insert into Course values(007,'现代企业',2003);
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1.3 SC

Create table SC(S# number(4),C# number(4),score number(3),constraint SC__S#_C#_Pk primary key(S#,C#),constraint Sc_S#_Fk foreign key (S#) references Student(S#),constraint Sc_C#_Fk foreign key (C#) references Course(C#));insert into SC values(1001,1,87);insert into SC values(1001,3,66);insert into SC values(1001,2,77);insert into SC values(1001,5,45);insert into SC values(1001,7,67);insert into SC values(1002,4,26);insert into SC values(1002,1,86);insert into SC values(1003,3,67);insert into SC values(1003,2,97);insert into SC values(1003,5,67);insert into SC values(1004,5,78);insert into SC values(1004,3,67);insert into SC values(1004,4,49);insert into SC values(1005,4,76);insert into SC values(1005,2,84);insert into SC values(1005,3,35);insert into SC values(1006,3,54);insert into SC values(1006,2,76);insert into SC values(1006,1,56);insert into SC values(1007,1,56);insert into SC values(1007,7,88);insert into SC values(1007,2,89);insert into SC values(1007,3,57);insert into SC values(1007,4,99);insert into SC values(1007,5,85);insert into SC values(1008,1,76);insert into SC values(1008,2,44);insert into SC values(1008,4,87);insert into SC values(1008,5,33);insert into SC values(1002,6,65);insert into SC values(1002,7,90);insert into SC values(1006,6,67);
View Code

1.4 Teacher

Create table Teacher(T# number(4),Tname varchar2(9) not null,constraint Teacher_T#_Pk primary key(T#));insert into Teacher values(2001,'李华');insert into Teacher values(2002,'王力宏');insert into Teacher values(2003,'叶平');insert into Teacher values(2004,'王倩倩');insert into Teacher values(2005,'李莉莉');
View Code

mysql

DROP TABLE IF EXISTS `student`;CREATE TABLE `student` (  `SId` int(11) unsigned NOT NULL AUTO_INCREMENT,  `Sname` varchar(255) NOT NULL,  `Ssex` varchar(255) DEFAULT NULL,  `Sage` date DEFAULT NULL COMMENT 'CURRENT_TIMESTAMP',  PRIMARY KEY (`SId`)) ENGINE=InnoDB AUTO_INCREMENT=1009 DEFAULT CHARSET=utf8;-- ------------------------------ Records of student-- ----------------------------INSERT INTO `student` VALUES ('1001', '李志国', '男', '1985-09-26');INSERT INTO `student` VALUES ('1002', '李甜甜', '女', '1987-02-06');INSERT INTO `student` VALUES ('1003', '张小燕', '女', '1984-11-18');INSERT INTO `student` VALUES ('1004', '王菲', '男', '1986-06-03');INSERT INTO `student` VALUES ('1005', '杜宇', '女', '1986-04-25');INSERT INTO `student` VALUES ('1006', '彭大生', '男', '1988-08-28');INSERT INTO `student` VALUES ('1007', '王亮', '男', '1983-08-01');INSERT INTO `student` VALUES ('1008', '赵婷婷', '女', '1984-02-02');DROP TABLE IF EXISTS `course`;CREATE TABLE `course` (  `CId` int(11) unsigned NOT NULL AUTO_INCREMENT,  `TId` int(11) DEFAULT NULL,  `Cname` varchar(255) DEFAULT NULL,  PRIMARY KEY (`CId`)) ENGINE=InnoDB AUTO_INCREMENT=1008 DEFAULT CHARSET=utf8;-- ------------------------------ Records of course-- ----------------------------INSERT INTO `course` VALUES ('1001', '2003', '企业管理');INSERT INTO `course` VALUES ('1002', '2001', '马克思');INSERT INTO `course` VALUES ('1003', '2002', 'OO&UML');INSERT INTO `course` VALUES ('1004', '2004', '数据库');INSERT INTO `course` VALUES ('1005', '2005', '英语');INSERT INTO `course` VALUES ('1006', '2005', '大学语文');INSERT INTO `course` VALUES ('1007', '2003', '现代企业');DROP TABLE IF EXISTS `sc`;CREATE TABLE `sc` (  `SCId` int(10) unsigned NOT NULL AUTO_INCREMENT,  `SId` int(11) DEFAULT NULL,  `CId` int(11) DEFAULT NULL,  `score` int(255) DEFAULT NULL,  PRIMARY KEY (`SCId`)) ENGINE=InnoDB AUTO_INCREMENT=33 DEFAULT CHARSET=utf8;-- ------------------------------ Records of sc-- ----------------------------INSERT INTO `sc` VALUES ('1', '1001', '1', '87');INSERT INTO `sc` VALUES ('2', '1001', '3', '66');INSERT INTO `sc` VALUES ('3', '1001', '2', '77');INSERT INTO `sc` VALUES ('4', '1001', '5', '45');INSERT INTO `sc` VALUES ('5', '1001', '7', '67');INSERT INTO `sc` VALUES ('6', '1002', '4', '26');INSERT INTO `sc` VALUES ('7', '1002', '1', '86');INSERT INTO `sc` VALUES ('8', '1003', '3', '67');INSERT INTO `sc` VALUES ('9', '1003', '2', '97');INSERT INTO `sc` VALUES ('10', '1003', '5', '67');INSERT INTO `sc` VALUES ('11', '1004', '5', '78');INSERT INTO `sc` VALUES ('12', '1004', '3', '67');INSERT INTO `sc` VALUES ('13', '1004', '4', '49');INSERT INTO `sc` VALUES ('14', '1005', '4', '76');INSERT INTO `sc` VALUES ('15', '1005', '2', '84');INSERT INTO `sc` VALUES ('16', '1005', '3', '35');INSERT INTO `sc` VALUES ('17', '1006', '3', '54');INSERT INTO `sc` VALUES ('18', '1006', '2', '76');INSERT INTO `sc` VALUES ('19', '1006', '1', '56');INSERT INTO `sc` VALUES ('20', '1007', '1', '56');INSERT INTO `sc` VALUES ('21', '1007', '7', '88');INSERT INTO `sc` VALUES ('22', '1007', '2', '89');INSERT INTO `sc` VALUES ('23', '1007', '3', '57');INSERT INTO `sc` VALUES ('24', '1007', '4', '99');INSERT INTO `sc` VALUES ('25', '1007', '5', '85');INSERT INTO `sc` VALUES ('26', '1008', '1', '76');INSERT INTO `sc` VALUES ('27', '1008', '2', '44');INSERT INTO `sc` VALUES ('28', '1008', '4', '87');INSERT INTO `sc` VALUES ('29', '1008', '5', '33');INSERT INTO `sc` VALUES ('30', '1002', '6', '65');INSERT INTO `sc` VALUES ('31', '1002', '7', '90');INSERT INTO `sc` VALUES ('32', '1006', '6', '67');DROP TABLE IF EXISTS `teacher`;CREATE TABLE `teacher` (  `TId` int(10) unsigned NOT NULL AUTO_INCREMENT,  `TName` varchar(255) DEFAULT NULL,  PRIMARY KEY (`TId`)) ENGINE=InnoDB AUTO_INCREMENT=2006 DEFAULT CHARSET=utf8;-- ------------------------------ Records of teacher-- ----------------------------INSERT INTO `teacher` VALUES ('2001', '李华');INSERT INTO `teacher` VALUES ('2002', '王力宏');INSERT INTO `teacher` VALUES ('2003', '叶平');INSERT INTO `teacher` VALUES ('2004', '王倩倩');INSERT INTO `teacher` VALUES ('2005', '李莉莉');
View Code

 

2 T-SQL试题

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#;
-- mysqlSELECT a.SId from (select SId,score from sc where CId='1') a,(select SId,score from sc where CId='2') b where a.SId=b.SId and a.score>b.score;

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;
-- mysqlselect SID, AVG(score)from SCGROUP BY SIDHAVING 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
select student.SId,student.Sname,count(SC.Cid),sum(SC.score) from student,SCwhere student.SId=SC.SIdgroup by student.SId

 

4、查询姓“李”的老师的个数;

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
select COUNT(DISTINCT TName) from Teacherwhere 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');
select SId,Sname from Student where SId not in(SELECT DISTINCT SC.SId from SC,Course,Teacher where SC.CId=Course.CId and Course.TId=Teacher.TId 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  SC_2 where SC_2.S#=SC.S# and SC_2.C#='002');

解析: 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。

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:跟第5题的要求正好相反,且难度提高了。所有课程通过groupby 后的having条件实现

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 

解析: 跟第1题基本一样的问题,只是实现方式不一样。第1题的实现方式更适合我。

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 DISTINCT(Student.S#) from Student,SC where S.S#=SC.S# and score>=60);

解析:反过来说:没有一门课程大于等于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 S#,Sname from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='1001');

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');
-- mysqlupdate SC ,(SELECT SC.CId,AVG(SC.score) avgScore from SC,Course,Teacher where SC.CId=Course.CId and Course.TId=Teacher.TId and Teacher.Tname='叶平' GROUP BY SC.CId) aset SC.score=a.avgScorewhere SC.CId=a.CId

 

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;

Delete 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、“002”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'));

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#
                    );

-- mysqlSELECT DISTINCTCId '课程ID',(select Max(score) from SC SC1 where SC1.CId=SC.CId group by CId) '最高分',(select Min(score) from SC SC2 where SC2.CId=SC.CId group by CId) '最低分 'from SC order by CId;

 

 

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
SELECT t.CId AS '课程号',max(course.Cname)AS '课程名',ifnull(AVG(score),0) AS '平均成绩' ,100 * SUM(CASE WHEN ifnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS '及格百分数' FROM SC T,Course where t.CId=course.CId GROUP BY t.CId ORDER BY 100 * SUM(CASE WHEN ifnull(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

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,总分 

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 SCGROUP 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;

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#;

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';

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# ;

 

转载于:https://www.cnblogs.com/Ming8006/p/4515211.html

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