Mysql进行复杂查询

1.查询“生物”课程比“物理”课程成绩高的所有学生的学号；

 思路： （1）获取所有选了 生物 课程的学生的成绩（学号，成绩） --临时表

　　    （2）获取所有选了 物理 课程的学生的成绩（学号，成绩） --临时表

　　　（3）根据学号连接两张临时表（学号，生物成绩，物理成绩），加条件进行查询

SELECT    A.student_id AS 学号,    sw AS 生物,    wl AS 物理FROM    (        SELECT            student_id,            num AS sw        FROM            score        LEFT JOIN course ON score.course_id = course.cid        WHERE            course.cname = '生物'    ) AS ALEFT JOIN (    SELECT        student_id,        num AS wl    FROM        score    LEFT JOIN course ON score.course_id = course.cid    WHERE        course.cname = '物理') AS B ON A.student_id = B.student_idWHERE    sw >IF (isnull(wl), 0, wl);

2.查询平均成绩大于60分的同学的学号和平均成绩；

思路：（1）根据学号分组

　　  （2）使用avg（）聚合函数计算平均成绩

　　  （3）通过having对平均成绩进行筛选

SELECT student_id,avg(num) FROM scoreLEFT JOIN courseON score.student_id=course.cidGROUP BY student_idHAVING avg(num)>60

3.查询所有同学的学号、姓名、选课数、总成绩；

思路：根据学号分组，使用count（）对选课数计数，sum（）计算总成绩

SELECT    student_id,    sname,    count(student_id),    sum(num)FROM  score LEFT JOIN student ON score.student_id = student.sidGROUP BY student_id

4.查询姓“李”的老师的个数；

思路：使用like及通配符匹配，count（）进行计数

SELECT count(tid) FROM teacherWHERE tname LIKE '李%'

5.查询没学过“李平”老师课的同学的学号、姓名；

思路：（1）连接成绩表 课程表 教师表得到选了李平老师课程的学生

         （2）再通过学生表筛选结果

SELECT sid,sname FROM student WHERE sid not IN (SELECT student_id FROM     (SELECT cid,teacher_id,student_id,course_id         FROM score         LEFT JOIN course ON score.course_id=course.cid     ) AS A     LEFT JOIN teacher     ON A.teacher_id=teacher.tid where teacher.tname='李平老师'  GROUP BY student_id)

6.查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

思路：（1）筛选出学过001课程的学生或学过002课程的学生

　　  （2）根据学生分组，如果学生数量等于2，则该学生选择了以上两门课程

SELECT student.sid,student.sname FROM (SELECT student_id,count(student_id) FROM score LEFT JOIN course ON score.course_id=course.cid  WHERE course.cid='001' or course.cid='002' GROUP BY student_id HAVING count(student_id)>1 ) AS A LEFT JOIN student ON A.student_id=student.sid;

7.查询学过“李平”老师所教的所有课的同学的学号、姓名；

思路：（1）查询李平老师所教的课程

         （2）在成绩表中筛选出学生选择的课程 in 李平老师的课程

SELECT student_id,sname FROM (SELECT student_id FROM scoreWHERE course_id IN(SELECT cid FROM teacher LEFT JOIN course ON teacher.tid=course.teacher_id WHERE tname='李平老师')GROUP BY student_id) AS BLEFT JOIN studentON B.student_id=student.sid

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；

思路：（1）分别获取选择了 课程001和002的学生和成绩；

　　 （2）连接两张表，筛选出001的成绩大于002成绩的学生

SELECT student_id,sname FROM(SELECT A.student_id FROM(SELECT student_id,num FROM score WHERE course_id=001) AS ALEFT JOIN(SELECT student_id,num FROM score WHERE course_id=002) AS BON A.student_id=B.student_idWHERE A.num>B.num) AS CLEFT JOIN studentON C.student_id=student.sid

9.查询有课程成绩小于60分的同学的学号、姓名；

思路：（1）筛选出成绩小于60的学生，并通过学生分组   --临时表

　　  （2）在学生表中筛选出 in 临时表中的学生

SELECT sid,sname FROM student WHERE sid IN(SELECT student_id FROM score WHERE num<60 GROUP BY student_id)

10.查询没有学全所有课的同学的学号、姓名；

思路：（1）统计出总课程数

         （2）成绩表中，通过学生分组，统计出每个学生的课程数，如果课程数等于总课程数，则表示选择了所有课程

SELECT sid,sname FROM student WHERE sid not IN(SELECT student_idFROM scoreGROUP BY student_id HAVING count(course_id) = (SELECT count(cid) FROM course))

11.查询至少有一门课与学号为“001”的同学所学课程相同的同学的学号和姓名；

思路：（1）查找学号001同学所学的所有课程　　--临时表

　　  （2）其他学生所学的课程如果在临时表中，则符合条件

SELECT student_id,sname FROM student LEFT JOIN score ON score.student_id=student.sidWHERE course_id in (SELECT course_id FROM score WHERE student_id=001) AND student_id != 001GROUP BY student_id

12.查询至少学过学号为“001”同学所选课程中任意一门课的其他同学学号和姓名；

13.查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；

 

14.删除学习“叶平”老师课的成绩表记录；

DELETE FROM score WHERE course_id IN(SELECT cid FROM course LEFT JOIN teacher ON course.teacher_id=teacher.tid WHERE tname='叶平老师')

15.向成绩表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；

16.按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；

select sc.student_id,        (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,        (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,        (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,        count(sc.course_id),        avg(sc.num)    from score as sc    group by student_id desc

17.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 course_id,max(num) as 最高分,min(num) as 最低分FROM scoreGROUP BY course_id

18.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；

思路：三元运算（三目运算），case .. when .. then .. else .. end

SELECT course_id,avg(num) AS 平均分,sum(CASE WHEN score.num>60 THEN 1 ELSE 0 END)/count(1)*100 AS 及格率 FROM score GROUP BY course_idORDER BY 平均分 ASC,及格率 DESC 

19.课程平均分从高到低显示（显示任课老师）；

SELECT course_id,avg(num),teacher.tname FROM score LEFT JOIN course ON score.course_id=course.cid LEFT JOIN teacher ON course.teacher_id=teacher.tidGROUP BY course_idORDER BY avg(num) DESC

20.查询各科成绩前三名的记录:(不考虑成绩并列情况) ;

 

21.查询每门课程被选修的学生数；

SELECT course_id,count(student_id) FROM scoreGROUP BY course_id

22.查询出只选修了一门课程的全部学生的学号和姓名；

SELECT student_id,student.sname FROM score LEFT JOIN student ON score.student_id=student.sidGROUP BY student_idHAVING count(student_id)=1

23.查询男生、女生的人数；

SELECT   (SELECT count(1) FROM student WHERE gender='男') AS 男,  (SELECT count(1) FROM student WHERE gender='女') As 女

24.查询姓“张”的学生名单；

SELECT * FROM student WHERE student.sname LIKE '张%'

25.查询同名同姓学生名单，并统计同名人数；

SELECT sname,count(sname) FROM student GROUP BY sname HAVING count(sname)>1

26.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 course_id,avg(num) FROM score GROUP BY course_id ORDER BY course_id ASC,course_id DESC

27.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 student_id,sname,avg(num) FROM score LEFT JOIN student ON score.student_id=student.sidGROUP BY student_id HAVING avg(num)>85

28.查询课程名称为“数学”，且分数低于60的学生姓名和分数；

SELECT sname,num FROM score LEFT JOIN course ON score.course_id=course.cid LEFT JOIN student ON score.student_id=student.sid  WHERE cname='数学' AND num>60

29.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 student_id,sname FROM score LEFT JOIN student ON score.student_id=student.sid WHERE course_id=003 AND num>80

30.求选了课程的学生人数

select count(distinct student_id) from score

31.查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩；

思路：根据学生排序，成绩按从大到小排序，limit取最高的成绩

SELECT sname,max(num) FROM score LEFT JOIN course ON score.course_id=course.cid LEFT JOIN teacher ON course.teacher_id=teacher.tidLEFT JOIN student ON score.student_id=student.sidWHERE tname='张磊老师'GROUP BY student_idORDER BY max(num) DESCLIMIT 1

32.查询各个课程及相应的选修人数；

SELECT cid,cname,count(student_id) FROM score LEFT JOIN course ON score.course_id=course.cidGROUP BY course_id

33.查询不同课程但成绩相同的学生的学号、课程号、学生成绩；

select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from   score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;

34.查询每门课程成绩最好的前两名；

SELECT * FROM score LEFT JOIN(SELECT course_id, (SELECT num FROM score WHERE score.course_id=A.course_id ORDER BY num DESC LIMIT 0,1) AS '第一名',    (SELECT num FROM score WHERE score.course_id=A.course_id ORDER BY num DESC LIMIT 1,1) AS '第二名'FROM score AS AGROUP BY course_id) AS BON score.course_id=B.course_id

35.检索至少选修两门课程的学生学号；

思路：根据学号分组，统计

SELECT student_id,count(course_id) FROM score GROUP BY student_id HAVING count(course_id)>=2

36.查询全部学生都选修的课程的课程号和课程名；

思路：从学生表中统计出学生总数，在成绩表中根据课程分组，如果选择没门课程的人数等于学生总数，则符合

SELECT course_id,cname FROM score LEFT JOIN course ON score.course_id=course.cid GROUP BY course_id HAVING count(student_id)=(SELECT count(sid) FROM student)

37.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 score LEFT JOIN student ON score.student_id=student.sid WHERE course_id NOT IN(SELECT cid FROM teacher LEFT JOIN course ON course.teacher_id=teacher.tid WHERE tname='叶平老师')

38.查询两门以上不及格课程的同学的学号及其平均成绩；

思路：通过学生分组，筛选出不及格课程数

SELECT student_id,avg(num) FROM score WHERE num<60 GROUP BY student_id HAVING count(1)>2

39.检索“004”课程分数小于60，按分数降序排列的同学学号；

SELECT student_id FROM score WHERE course_id=004 AND num<60 ORDER BY num DESC

40.删除“002”同学的“001”课程的成绩；

DELETE FROM score WHERE student_id=002 AND course_id=001