分享整理的12条sql语句连同数据_MsSql_脚本之家,sql语句练习50题

  1. 选出每门功课都及格的学号 select distinct `s#` from sc where `s#` not in (select `s#` from sc where score <60) 2. 查询“1”课程比“2”课程成绩高的所有学生的学号; SELECT c01.`s#` from (select `s#`, `score` from sc where `c#`=1) c01, (select `s#`, `score` from sc where `c#`=2) c02 where c01.`s#` = c02.`s#` and c01.score > c02.score 3. 查询平均成绩大于60分的同学的学号和平均成绩; select `s#`, avg from sc group by `s#` having avg > 60 4. 查询所有同学的学号、姓名、选课数、总成绩; select student.`s#`, student.`Sname`, count from student left outer join sc on student.`s#` = sc.`s#` group by student.`s#`, sc.`s#` 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 student.`s#` not in (select distinct from teacher, course, sc where Tname='叶平' and teacher.`t#` = course.`t#` and sc.`c#`= course.`c#` ) 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#` = 1 and exists (select * from sc sc_2 where sc_2.`c#`=2 and sc.`s#`=sc_2.`s#`) 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, teacher, course where tname='叶平' and teacher.`t#`=course.`t#` and course.`c#`= sc.`c#` group by `s#` having count= from teacher, course where tname='叶 平' and teacher.`t#`=course.`t#`) ) 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 student.`s#`=sc_2.`s#` and sc_2.`c#`=2) score2 from student , sc where sc.`s#`=student.`s#` and sc.`c#`=1) s_2 where score2 < score 9.查询没有学全所有课的同学的学号、姓名 select student.`S#`, Sname from student, sc where student.`s#` = sc.`s#` group by `s#`, sname having count < from course) 10. 查询至少有一门课与学号为“002”的同学所学相同的同学的学号和姓名; select distinct, sname from student, sc where student.`s#`=sc.`s#` and `c#` in (select `c#` from sc where `s#`=002) 11. 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 inner join (select sc2.`c#`, avg score from sc sc2, teacher, course where sc2.`c#`=course.`c#` and tname='叶平' and teacher.`t#` = course.`t#` and course.`c#`=sc2.`c#` group by course.`c#`) sc3 on sc.`c#`=sc3.`c#` set sc.score=sc3.score 12. 查询2号的同学学习的课程他都学了的同学的学号;(注意理解:where语句的 第一个条件过滤掉不满足c#的记录,再group by,就比较清晰) select `S#` from SC where `C#` in (select `C#` from SC where `S#`=2) group by `S#` having count from SC where `S#`=2); 作者 人在江湖

7.查询学过“叶平”老师所教的所有课的同学的学号、姓名;

俺觉得自 己试着写写sql,调试调试还是有帮助的,读人家sql例子好像读懂了,自己写就未 必思路正确,调试得通,写得简洁。 跟着网上流行的学生选课表的例子复习了一下: //www.jb51.net/article/30655.htm 这篇文字在网上被转载烂了,里面有些sql适合用在应用系统里,有些“报表”的感 觉更重些,主要是想复习前者。前20条大体还挺好,后30条明显偏报表风格了,而 且后面选例良莠不齐,选了12个例子做练习,(其实很多语法,case, any/all, union之类的都没包括),用mysql数据库,并共享自己造出来的数据。关于这12条 sql, 修正了原文中有纰漏的地方。 sql是基本技能,若能写得好也挺精彩的,还在继续练习。绝不提倡努力写复杂sql 解决业务问题。应用系统里如果存在很复杂的sql,往往揭示了业务逻辑向下泄露 到sql层的问题,不利于维护和扩展,虽然这样确实常能提高运行效率。具体情况 自行取舍。 下面的例子都是比较通用的sql, 其实针对特定的数据库,需要学的也挺多,比如 oracle db的decode函数, rowid, rownum, connect by 虽然不通用,但是很实用。 数据可以在这里下载,只是用作练习,没做任何外键关联: 整理的sql在下面: Student 学生表 Course 课程表 SC 成绩表 Teacher 教师表

3.查询所有同学的学号、姓名、选课数、总成绩

WHERE Student.sid=SC.sid GROUP BY  Student.sid,Student.Sname having count(cid) <(SELECT count(cid) FROM Course); 

标准答案(但是运行不好使)SELECT L.cid As 课程ID,L.score AS 最高分,R.score AS 最低分 

 

1.查询“某1”课程比“某2”课程成绩高的所有学生的学号;

12.查询至少学过学号为“”同学所有一门课的其他同学学号和姓名;

45.检索至少选修两门课程的学生学号

此题知识点,用数量来判断 

此题知识点,GROUP BY 语句用于结合合计函数,根据一个或多个列对结果集进行分组。group by后面不能接where,having代替了where

SELECT Student.sid,Student.Sname,count(SC.cid),sum(score)FROM Student left Outer JOIN SC on Student.sid=SC.cid GROUP BY Student.sid,Sname

SELECT cid,sid FROM sc WHERE score <60 ORDER BY cid 

此题知识点,SELECT sid,Sname FROM Student,SC WHERE Student.sid=SC.sid AND cid in (SELECT cid FROM SC WHERE sid=1)这样写是错误的,因为from后面是两个表,不能明确是哪个表里面的sid和sname所以错误提示是“未明确定义列”

SELECT Student.sid,Student.Sname FROM Student WHERE sid not in (SELECT distinct( SC.sid) FROM SC,Course,Teacher WHERE  SC.cid=Course.cid AND Teacher.id=Course.tid AND Teacher.Tname='叶平');

44.统计每门课程的学生选修人数(超过人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数降序排列,若人数相同,按课程号升序排列

练习内容:

GROUP BY IL.cid)  

SELECT student.sid,student.Sname FROM Student,SC WHERE Student.sid=SC.sid AND cid in (SELECT cid FROM SC WHERE sid=1)

标准答案:SELECT sid,Sname FROM Student WHERE sid in (SELECT sid FROM SC ,Course ,Teacher WHERE SC.cid=Course.cid AND Teacher.tid=Course.tid AND Teacher.Tname='杨巍巍' GROUP BY sid having count(SC.cid)=(SELECT count(cid) FROM Course,Teacher  WHERE Teacher.tid=Course.tid AND Tname='杨巍巍'))

FROM SC L ,SC AS R  

37.查询不及格的课程,并按课程号从大到小排列 

49.检索“”课程分数小于,按分数降序排列的同学学号

select student.sname,sc.score from sc,student,teacher,course c where teacher.tname='李子'

 

SELECT Student.sid,Student.Sname  FROM Student,SC  

SELECT sid,Sname FROM Student WHERE sid not in (SELECT Student.sid FROM Student,SC WHERE Student.sid=SC.sid AND score>60); 

38.查询课程编号为且课程成绩在分以上的学生的学号和姓名;

select a.sid,a.sname from (select student.sid,student.sname from student,teacher,course,sc 

8.查询课程编号“”的成绩比课程编号“”课程低的所有同学的学号、姓名;

where teacher.TNAME='杨巍巍' and teacher.tid=course.tid and course.cid=sc.cid and student.sid=sc.sid) a

WHERE SC.sid=Student.sid GROUP BY SC.sid ,Student.Sname having count(cid)=1;

14.查询和“”号的同学学习的课程完全相同的其他同学学号和姓名;

9.查询所有课程成绩小于分的同学的学号、姓名;

select distinct sid from sc where sid not in(select sc.sid from sc,course,teacher where sc.cid=course.cid and course.tid=teacher.tid and 

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 sc.sid from sc,course where sc.cid=course.cid and course.cname='java' and sc.score<90

select student.sid,student.sname from sc,student where sc.cid=1 and sc.score>60 and sc.sid=student.sid

5.查询没学过“叶平”老师课的同学的学号、姓名;

Student(Sid,Sname,Sage,Ssex) 学生表

(SELECT score FROM SC SC_2 WHERE SC_2.sid=Student.sid AND SC_2.cid=1) score2 FROM Student,SC

6.查询学过“```”并且也学过编号“```”课程的同学的学号、姓名;

SELECT sid FROM SC WHERE cid in (SELECT cid FROM SC WHERE sid=6) GROUP BY sid having count(*)=(SELECT count(*) FROM SC WHERE sid=6); 

SC(Sid,Cid,score) 成绩表

 

SELECT a.sid FROM (SELECT sid,score FROM SC WHERE cid=1) a,(SELECT sid,score FROM SC WHERE cid=3) b WHERE a.score>b.score AND a.sid=b.sid;

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

Teacher(Tid,Tname) 教师表

 此题知识点,distinct是去重的作用

SELECT sid,avg(score)  FROM sc  GROUP BY sid having avg(score) >60;

select count(teacher.tid)from teacher where teacher.tname like'李%'

delete from sc where sid=1 and cid=1

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;

and teacher.tid=c.tid and c.cid=sc.cid and sc.sid=student.sid and sc.score=(select max(score)from sc where sc.cid=c.cid)

标准答案(但是好像不好使)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');  

48.查询两门以上不及格课程的同学的学号及其平均成绩

delete from sc s where s.cid in (select c.cid from teacher t,course c where t.tid = c.tid and tname='李子')

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

Insert into SC SELECT sid,2,(SELECT avg(score) FROM SC WHERE cid=2) FROM Student WHERE sid not in (SELECT sid FROM SC WHERE cid=2); 

(select student.SNAME,student.SID from student,course,sc where cname='english'and sc.sid=student.sid and sc.cid=course.cid) b where a.sid=b.sid;

L.score = (SELECT MAX(IL.score)  

(select student.SID,student.sname,sc.score from student,sc where student.sid=sc.sid and sc.cid=2) b where a.score<b.score and a.sid=b.sid

teacher.tname='杨巍巍')

Course(Cid,Cname,Tid) 课程表

26.查询每门课程被选修的学生数

select cid as 课程号,max(score)as 最高分,min(score) as 最低分 from sc group by cid

FROM SC AS IL,Student AS IM  

rownum的用法

update sc set score=(select avg(score) from sc,course,teacher where course.cid=sc.cid and course.tid=teacher.tid and teacher.tname='杨巍巍')

查询所有成绩第二名到第四名的成绩

此题知识点,嵌套查询可以分布考虑,先查出李子老师都交了什么课的id,然后再删除那些id的值

11.查询至少有一门课与学号为“”的同学所学相同的同学的学号和姓名;

50.删除“”同学的“”课程的成绩 

AND  R.Score = (SELECT MIN(IR.score) FROM SC AS IR WHERE R.cid = IR.cid  GROUP BY IR.cid ); 

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;

标准答案:SELECT sid,Sname FROM (SELECT Student.sid,Student.Sname,score ,

select sc.cid,count(sc.cid)from sc,course where sc.cid=course.cid group by sc.cid  order by sc.cid desc

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“”课程的同学学号、课程的平均成绩;

27.查询出只选修了一门课程的全部学生的学号和姓名

SELECT Cid,Avg(score) FROM SC GROUP BY cid ORDER BY Avg(score),cid DESC ;

select * from (select rownum p,t.score from(SELECT s.score score FROM sc s ORDER BY score desc)t )tt where tt.p>1 and tt.p<5

10.查询没有学全所有课的同学的学号、姓名;

此题知识点,exists是在集合里找数据,as就是起别名

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

4.查询姓“李”的老师的个数;

WHERE Student.sid=SC.sid AND cid=1) S_2 WHERE score2 <score;

select a.SID,a.SNAME from (select student.SNAME,student.SID from student,course,sc where cname='c++'and sc.sid=student.sid and sc.cid=course.cid) a,

41.查询各个课程及相应的选修人数

select a.sid,a.sname from(select student.SID,student.sname,sc.SCORE  from student,sc where student.sid=sc.sid and sc.cid=1) a,

15.删除学习“叶平”老师课的SC表记录; 

此题知识点,先查出大于60分的,然后not in 就是小于60分的了

select sc.cid ,count(sc.sid)from sc,student where sc.sid=student.sid group by sc.cid 

43.查询每门功成绩最好的前两名

WHERE L.cid = IL.cid AND IM.sid=IL.sid  

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;(没做出来)

此题知识点,嵌套查询和给查出来的表起别名

WHERE L.cid = R.cid AND  

SELECT SC.sid,Student.Sname,count(cid) AS 选课数 FROM SC ,Student  

SELECT sid FROM  sc group  by  sid having  count(*)  >  =  2  

47.查询没学过“叶平”老师讲授的任一门课程的学生姓名

2.查询平均成绩大于60分的同学的学号和平均成绩;

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