MYSQL之SQL語句練習及思路_1

關係模式

表格需求：

1.student表

sno	sname	age	gender
1	liqiang	23	male
2	liuli	22	female
3	zhangyou	22	male

2.course表

cno	cname	teacher
k1	c	wanghua
k5	database	chengjun
k8	complie	chengjun

3.score表

sno	cno	score
1	k1	83
2	k1	85
5	k1	92
2	k5	90
5	k5	84
5	k8	80

問題：

1.用SQL語句實現以下檢索：

（1）查詢‘程軍’老師所教授的全部課程

select cname from course where teacher like 'chengjun';

（2）查詢‘李強’同窗全部課程的成績

select course.cname,score from score,student,course wre here student.sno=score.sno and score.cno=course.cno and sname='liqiang';

（3)查詢課程名爲‘C’的平均成績

select avg(score) from score where cno=(select cno from course where cname='c');

（4）查詢選修了全部課程的同窗的信息

select sno from score group by sno  having count(sno) >= (select  count(cno) from course);

分析:該問題中首先是要對score表進行統計，統計出sno出現的次數，理論上sno的爲學生所報的課程數，若該值等於course表中課程統計數量，即爲結果。

（5）檢索王老師所授課程的課程號和課程名

select cno,cname from course where teacher like 'wang%';

（6）檢索年齡大於23歲的男學生的學號和姓名

select sno,sname from student where age > 23;

（7）檢索至少選修王老師所授課程中一門課程的女學生姓名

select  distinct sname from student,score where score.cno  in (select cno from course where teacher like 'wang%') and gender='female';

select sname from student,course,score where student.sno=score.sno and course.cno=score.cno and course.teacher like 'wang%' and student.gender='female';

(8)檢索李同窗不學的課程的課程號

select cno from course where cno not in (select cno from rom score,student  where student.sno=score.sno and  sname like 'li');

（9）檢索至少選修兩門課程的學生學號

select sno from score group by sno having count(sno) > 2;

（10）檢索所有學生都選修的課程的課程號和課程名

select cno,cname from course where cno in (select cno from score group by cno having count(*) = (select count(*) from student));

思路解析：學生數量=score表中單科目總數

（11）檢索選修課程包含王老師所授課程的學生學號

select sno from score where cno in (select cno from course where teacher like 'wang%');

select distinct sno from score,course where score.cno=course.cno and teacher like 'wang%';

（12）統計有學生選修的課程門數

select count(distinct cno) from score;

（13）統計選修k1課程的學生的平均成績

select avg(score) from score where cno='k1';

（14）求王老師所授課程的每門課程的學生的平均成績

select avg(score),score.cno from score,course where course.cno=score.cno and teacher like 'wang%' group by score.cno;

（15）統計每門課程的學生選修人數（超過2人的課程才統計）。要求輸出課程號和選修人數，查詢結果按人數降序排列，若人數相同，按課程號升序排列。

select cno,count(sno) from score group by cno having count(sno) > 2 order by count(sno) desc,cno asc;

（16）檢索學號比李同窗打，而年齡比她小的學生姓名

select sname from student where sno > (select sno from student where sname like 'liq%' ) and age < (select age from student where sname likke 'liq%');

（17）檢索姓名以li開頭的全部學生的姓名和年齡

select sname,age from student where sname like 'li%';

（18）在score中檢索成績爲空值的學生學號和課程號

select sno,cno from score where score is null;

（19）檢索年齡大於女同窗平均年齡的男學生姓名和年齡

select sname,age from student where age > (select avg(age) from student where gender='female') and gender='male';

（20）檢索年齡大於全部女同窗年齡的男學生姓名和年齡

select sname,age from student where age > any (select age from student where gender='female');

（21）檢索至少選修程軍老師所授所有課程的學生姓名

select sname from student,course,score  where student.sno=score.sno and course.cno=score.cno and teacher='chengjun' group by sname having count(*) >= (select count(*) from course where teacher='chengjun');

思路分析，首先找出選修程軍老師所有課程的學生，並統計該值在score表中出現的次數，該數字大於等於程軍老師教授課程總數量。