复杂查询优质习题

#1)查询选修了 Oracle 的学生姓名；

SELECT t1.`name`
FROM stu t1,course t2,score t3
WHERE t1.`id`=t3.`stu_id` AND t2.`id`=t3.`course_id`
AND t2.`name`=‘Oracle‘;

#2)查询 姜振国 同学选修了的课程名字；
SELECT t2.`name`
FROM stu t1,course t2,score t3
WHERE t1.`id`=t3.`stu_id` AND t2.`id`=t3.`course_id`
AND t1.`name`=‘姜振国‘;


#3)查询只选修了1门课程的学生学号和姓名；
#子句
SELECT stu_id,COUNT(id)
FROM score
GROUP BY stu_id HAVING COUNT(id)=1;

SELECT id,NAME
FROM stu t1,(SELECT stu_id,COUNT(id)
FROM score
GROUP BY stu_id HAVING COUNT(id)=1) t2
WHERE t1.`id`=t2.stu_id;


#4)查询选修了至少3门课程的学生信息；
SELECT stu_id,COUNT(id)
FROM score
GROUP BY stu_id HAVING COUNT(id)>=3;

SELECT t1.`id`,t1.`name`,t1.`sex`,t1.`bj_id`
FROM stu t1,(SELECT stu_id,COUNT(id)
FROM score
GROUP BY stu_id HAVING COUNT(id)>=3) t2
WHERE t1.`id`=t2.stu_id;

#5)查询选修了所有课程的学生；
SELECT COUNT(id) FROM course;

SELECT stu_id
FROM score
GROUP BY stu_id HAVING COUNT(course_id)=(SELECT COUNT(id) FROM course);

SELECT *
FROM stu
WHERE stu_id IN();


#6)查询选修课程的学生人数；
SELECT COUNT(DISTINCT stu_id) FROM score;

#7)查询所学课程至少有一门跟 姜振国 所学课程相同的学生信息。

SELECT t2.`course_id`
FROM stu t1,score t2
WHERE t1.`id`=t2.`stu_id` AND t1.`name`=‘姜振国‘;

SELECT t1.*
FROM stu t1,score t2
WHERE t1.`id`=t2.`stu_id` AND t2.`course_id`=ANY(?);

SELECT DISTINCT t1.*
FROM stu t1,score t2
WHERE t1.`id`=t2.`stu_id` AND t2.`course_id`=ANY(SELECT t2.`course_id`
FROM stu t1,score t2
WHERE t1.`id`=t2.`stu_id` AND t1.`name`=‘姜振国‘) AND t1.`name`<>‘姜振国‘;

#8)查询两门及两门以上不及格同学的平均分(<90)
#子句 查询所有学生的id和不及格课程的数量（小余90）
SELECT stu_id,COUNT(id)
FROM score t1
WHERE t1.grade<90
GROUP BY stu_id;

#子句 查询所有学生的id和不及格课程的数量大于等于2的（小余90）
SELECT stu_id,COUNT(id),AVG(grade) temp
FROM score t1
WHERE t1.grade<90
GROUP BY stu_id HAVING COUNT(id)>=2;

SELECT t1.stu_id,t1.temp
FROM (SELECT stu_id,COUNT(id),AVG(grade) temp
FROM score t1
WHERE t1.grade<90
GROUP BY stu_id HAVING COUNT(id)>=2) t1;

#可以在关联学生姓名