此篇博客内容引自“MySQL经典练习题及答案”
废话不不多说！！！
建表、插入数据。
--建表 --学生表 CREATE TABLE Student( s_id VARCHAR(20), s_name VARCHAR(20) NOT NULL DEFAULT '', s_birth VARCHAR(20) NOT NULL DEFAULT '', s_sex VARCHAR(10) NOT NULL DEFAULT '', PRIMARY KEY(s_id) ); --课程表 CREATE TABLE Course( c_id VARCHAR(20), c_name VARCHAR(20) NOT NULL DEFAULT '', t_id VARCHAR(20) NOT NULL, PRIMARY KEY(c_id) ); --教师表 CREATE TABLE Teacher( t_id VARCHAR(20), t_name VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(t_id) ); --成绩表 CREATE TABLE Score( s_id VARCHAR(20), c_id VARCHAR(20), s_score INT, PRIMARY KEY(s_id,c_id) ); --插入学生表测试数据 insert into Student values('01' , '赵雷' , '1990-01-01' , '男'); insert into Student values('02' , '钱电' , '1990-12-21' , '男'); insert into Student values('03' , '孙风' , '1990-05-20' , '男'); insert into Student values('04' , '李云' , '1990-08-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '女'); insert into Student values('06' , '吴兰' , '1992-03-01' , '女'); insert into Student values('07' , '郑竹' , '1989-07-01' , '女'); insert into Student values('08' , '王菊' , '1990-01-20' , '女'); --课程表测试数据 insert into Course values('01' , '语文' , '02'); insert into Course values('02' , '数学' , '01'); insert into Course values('03' , '英语' , '03'); --教师表测试数据 insert into Teacher values('01' , '张三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五'); --成绩表测试数据 insert into Score values('01' , '01' , 80); insert into Score values('01' , '02' , 90); insert into Score values('01' , '03' , 99); insert into Score values('02' , '01' , 70); insert into Score values('02' , '02' , 60); insert into Score values('02' , '03' , 80); insert into Score values('03' , '01' , 80); insert into Score values('03' , '02' , 80); insert into Score values('03' , '03' , 80); insert into Score values('04' , '01' , 50); insert into Score values('04' , '02' , 30); insert into Score values('04' , '03' , 20); insert into Score values('05' , '01' , 76); insert into Score values('05' , '02' , 87); insert into Score values('06' , '01' , 31); insert into Score values('06' , '03' , 34); insert into Score values('07' , '02' , 89); insert into Score values('07' , '03' , 98);
上题！！！
下面的内容中，mine下面的是我自己写的，mine再下面的一条引用的那篇博客自带的答案。（因为我用的是sqlserver，所以有的sql可能在mysql中不能执行，单道理都一样哈！）
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 --mine select s.*,t.s_score as score1,t.score2 from Student as s right join (select a.*,b.s_score as score2 from Score as a inner join Score as b on a.c_id in('01','02') and b.c_id in('01','02') and a.s_id = b.s_id and a.c_id != b.c_id and a.s_score > b.s_score and a.c_id = '01') t on s.s_id = t.s_id select a.* ,b.s_score as score_01,c.s_score as score_02 from student a join score b on a.s_id=b.s_id and b.c_id='01' left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数 --mine select s.*,t.s_score as score1,t.score2 from Student as s right join (select a.*,b.s_score as score2 from Score as a inner join Score as b on a.c_id in('01','02') and b.c_id in('01','02') and a.s_id = b.s_id and a.c_id != b.c_id and a.s_score < b.s_score and a.c_id = '01') t on s.s_id = t.s_id select a.* ,b.s_score as score_01,c.s_score as score_02 from student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score -- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩 --mine select s.*,t.avg_sc from Student as s inner join (select s_id as s_id,AVG(s_score) as avg_sc from Score group by s_id having AVG(s_score) > 60) t on s.s_id = t.s_id select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b join score a on b.s_id = a.s_id GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60; -- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 --mine select s.*,t.avg_sc from Student as s inner join (select s_id as s_id,AVG(s_score) as avg_sc from Score group by s_id having AVG(s_score) < 60) t on s.s_id = t.s_id union select *,0 as avg_sc from Student as s where not exists (select 1 from Score as a where a.s_id = s.s_id) --mine select a.s_id,a.s_name,AVG(b.s_score) as avg_sc from student as a left join score as b on a.s_id = b.s_id group by a.s_id,a.s_name having AVG(b.s_score) <60 or AVG(b.s_score) is null select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b left join score a on b.s_id = a.s_id GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60 union select a.s_id,a.s_name,0 as avg_score from student a where a.s_id not in ( select distinct s_id from score); -- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩 --mine select a.s_id,a.s_name,COUNT(c_id) as count_sc,SUM(s_score) as totle_sc from student as a left join score as b on a.s_id = b.s_id group by a.s_id,a.s_name select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from student a left join score b on a.s_id=b.s_id GROUP BY a.s_id,a.s_name; -- 6、查询"李"姓老师的数量 --mine select COUNT(t_id) from teacher where t_name like '李%' select count(t_id) from teacher where t_name like '李%'; -- 7、查询学过"张三"老师授课的同学的信息 --mine select * from student where s_id in( select s_id from Score where c_id = (select c_id from course where t_id = (select t_id from teacher where t_name = '张三'))) select a.* from student a join score b on a.s_id=b.s_id where b.c_id in( select c_id from course where t_id =( select t_id from teacher where t_name = '张三')); -- 8、查询没学过"张三"老师授课的同学的信息 --mine select * from student where s_id not in( select s_id from Score where c_id = (select c_id from course where t_id = (select t_id from teacher where t_name = '张三'))) select * from student c where c.s_id not in( select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in( select c_id from course where t_id =( select t_id from teacher where t_name = '张三'))); -- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息 --mine --法1 select s.* from student as s inner join (select a.s_id as s_id,a.c_id as c_id_1,b.c_id as c_id_2 from score as a inner join score as b on a.s_id = b.s_id and a.c_id = '01' and b.c_id = '02') t on s.s_id = t.s_id --法2 select s.* from student s inner join (select s_id from score where c_id = '01' and s_id in (select s_id from score where c_id = '02')) t on s.s_id = t.s_id --个人觉得下面这个sql写的不好（这是网上的答案），因为行程了“笛卡尔积”。 select * from student a,score b,score c where a.s_id = b.s_id and a.s_id = c.s_id and b.c_id='01' and c.c_id='02'; -- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息 --mine select s.* from student s inner join (select s_id from score where c_id = '01' and s_id not in (select s_id from score where c_id = '02')) t on s.s_id = t.s_id select a.* from student a where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02') -- 11、查询没有学全所有课程的同学的信息 --mine,说实话我觉得我的要更准确，下面的那个答案吧课程号都给写出来了，很不合适。 select * from student where s_id not in (select s_id from (select a.s_id as s_id from score as a inner join score as b on 1=1 inner join score as c on a.s_id = b.s_id and a.s_id= c.s_id and a.c_id != b.c_id and a.c_id != c.c_id and b.c_id != c.c_id and a.c_id in (select c_id from course) and b.c_id in (select c_id from course) and c.c_id in (select c_id from course)) t group by s_id) select s.* from student s where s.s_id in( select s_id from score where s_id not in( select a.s_id from score a join score b on a.s_id = b.s_id and b.c_id='02' join score c on a.s_id = c.s_id and c.c_id='03' where a.c_id='01')) -- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息 --mine,我觉得我的更好，因为下面的01这位同学也包含进去了，按照题目意思应该是不包含进去的。 select s.* from student as s inner join (select s_id as s_id from score where c_id in (select c_id from score where s_id = '01') and s_id != '01' group by s_id) t on s.s_id = t.s_id select * from student where s_id in( select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01') ); -- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息 --这个我没有做出来，但是看他的这个答案，我不敢苟同，因为它是用记录数是否相等来做判断的，我觉得这是不对的。 select a.* from student a where a.s_id in( select s_id,count(c_id) from score where s_id!='01' and c_id in(select c_id from score where s_id='01') group by s_id having count(s_id)=(select count(s_id) from score where s_id='01')); -- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 --mine select s_name from student where s_id not in (select s_id as s_id from score where c_id in ( select a.c_id from course as a inner join teacher as b on a.t_id = b.t_id and b.t_name = '张三' )) select a.s_name from student a where a.s_id not in ( select s_id from score where c_id = (select c_id from course where t_id =( select t_id from teacher where t_name = '张三')) group by s_id); -- 15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩 --mine select t.s_id,s.s_name,t.avg_sc from student as s inner join (select s_id as s_id,AVG(s_score) as avg_sc from score where s_score < 60 group by s_id having COUNT(1) >= 2) t on s.s_id = t.s_id --网上答案，但是报错了 select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from student a left join score b on a.s_id = b.s_id and a.s_id in( select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2) GROUP BY a.s_id,a.s_name --这里是我改过的网上答案sql，能正常运行，且答案和我写的sql一样。 select a.s_id,a.s_name,AVG(s_score) as avg_sc from student a left join score b on a.s_id = b.s_id where a.s_id in( select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2) GROUP BY a.s_id,a.s_name -- 16、检索"01"课程分数小于60，按分数降序排列的学生信息 --mine select b.* from score as a inner join student as b on a.s_id = b.s_id and a.c_id = '01' and a.s_score < 60 order by a.s_score desc --网上这个答案有事用笛卡尔积开做的，我个人不是很推崇。 select a.*,b.c_id,b.s_score from student a,score b where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC; -- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩 --这个可能是我没理解好题目需求，始终不知道题目想要的是什么，只能做成一下这样了。看了下面的答案知道是怎么回事儿了。哎！！！ select a.s_id,a.s_name,AVG(s_score) as avg_sc from Student as a inner join Score as b on a.s_id = b.s_id group by a.s_id,s_name order by AVG(s_score) desc --网上这个sql可以好好揣摩一下，我以前没这么写过，很是经典 select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文, (select s_score from score where s_id=a.s_id and c_id='02') as 数学, (select s_score from score where s_id=a.s_id and c_id='03') as 英语, round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC; -- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率 --及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90 --mine select c_id,MAX(s_score) as 最高分,MIN(s_score) as 最低分,AVG(s_score) as 平均分 , round(100*((select COUNT(1) from Score where s_score >60 and c_id = s.c_id)*1.0/COUNT(1)),2) as 及格率, round(100*((select COUNT(1) from Score where s_score >70 and s_score <80 and c_id = s.c_id)*1.0/COUNT(1)),2) as 中等率, round(100*((select COUNT(1) from Score where s_score >80 and s_score <90 and c_id = s.c_id)*1.0/COUNT(1)),2) as 优良率, round(100*((select COUNT(1) from Score where s_score >=90 and c_id = s.c_id)*1.0/COUNT(1)),2) as 优秀率 from Score as s group by c_id select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2), ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率, ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率, ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率, ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率 from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name -- 19、按各科成绩进行排序，并显示排名 --mine这道题有点难度，没想明白，网上的那个答案也有问题。 select * from Score order by c_id,s_score desc -- 20、查询学生的总成绩并进行排名 --mine select *,RANK() over(order by t.总成绩 desc) as rank from (select a.s_id as s_id,a.s_name as 姓名,SUM(b.s_score) as 总成绩 from Student as a inner join Score as b on a.s_id = b.s_id group by a.s_id,a.s_name) t select a.s_id, @i:=@i+1 as i, @k:=(case when @score=a.sum_score then @k else @i end) as rank, @score:=a.sum_score as score from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a, (select @k:=0,@i:=0,@score:=0)s -- 21、查询不同老师所教不同课程平均分从高到低显示 --mine select a.t_id,a.t_name,b.c_name,AVG(s_score) as 平均分 from teacher a,Course b,Score c where a.t_id = b.t_id and b.c_id = c.c_id group by a.t_id,a.t_name,b.c_name order by 平均分 desc select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a left join score b on a.c_id=b.c_id left join teacher c on a.t_id=c.t_id GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC; -- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩 --mine这个题看自己怎么理解，我这里的理解为，如果有两个第一，这两个第一过后便是第三名，没有第二名。 select a.*,t.rang,c.c_name from student a, (select *,(select COUNT(1)+1 from Score where s_score > s.s_score and c_id = s.c_id) as rang from Score as s) t ,Course c where a.s_id = t.s_id and t.c_id = c.c_id and t.rang >= 2 and t.rang<=3 order by c.c_id --网上的这些是mysql做的，我是用sqlserver做的，不过语法都差不多，只是个别时候回有报错。 select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3 UNION select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3 UNION select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3; -- 23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 --mine select c_id, (select COUNT(1) from Score where s_score <= 100 and s_score > 85 and c_id = s.c_id) as [100-85], (select COUNT(1) from Score where s_score <= 85 and s_score > 70 and c_id = s.c_id) as [85-70], (select COUNT(1) from Score where s_score <= 70 and s_score > 60 and c_id = s.c_id) as [70-60], (select COUNT(1) from Score where s_score <= 60 and s_score >= 0 and c_id = s.c_id) as [60-0] from Score as s group by c_id select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`, ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)b on a.c_id=b.c_id left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`, ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)c on a.c_id=c.c_id left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`, ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)d on a.c_id=d.c_id left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`, ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)e on a.c_id=e.c_id left join course f on a.c_id = f.c_id -- 24、查询学生平均成绩及其名次 --mine select * from student a,(select s_id,AVG(s_score) as avg_sc, RANK() over(order by AVG(s_score) desc) as rank_avg_sc from Score group by s_id) t where a.s_id = t.s_id select a.s_id, @i:=@i+1 as '不保留空缺排名', @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名', @avg_score:=avg_s as '平均分' from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b; -- 25、查询各科成绩前三名的记录 --mine select c.c_name,t.s_score,t.rank_sc from course as c,(select *,(select COUNT(1)+1 from Score where s_score>s.s_score and c_id = s.c_id and s_id != s.s_id) as rank_sc from Score as s where (select COUNT(1)+1 from Score where s_score>s.s_score and c_id = s.c_id and s_id != s.s_id) <= 3) t where c.c_id = t.c_id order by t.c_id,t.s_score desc select a.s_id,a.c_id,a.s_score from score a left join score b on a.c_id = b.c_id and a.s_score<b.s_score group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3 ORDER BY a.c_id,a.s_score DESC -- 26、查询每门课程被选修的学生数 --mine select c.*,t.count_num from Course c,(select c_id,COUNT(1) as count_num from Score group by c_id) t where c.c_id = t.c_id select c_id,count(s_id) from score a GROUP BY c_id -- 27、查询出只有两门课程的全部学生的学号和姓名 --mine select s.s_id,s.s_name,t.count_course from student as s,(select s_id,COUNT(1) as count_course from Score group by s_id having COUNT(1) = 2) t where s.s_id = t.s_id select s_id,s_name from student where s_id in( select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2); -- 28、查询男生、女生人数 --mine select s_sex as 性别,COUNT(1) as 人数 from Student group by s_sex select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex -- 29、查询名字中含有"风"字的学生信息 --mine select * from Student where s_name like '%风%' select * from student where s_name like '%风%'; -- 30、查询同名同性学生名单，并统计同名人数 --mine这道题，我是没有做出来的。看了答案才知道,下面是网上答案。 --心得体会：如果你需要统计两个字段，那么久group by这两个字段 select a.s_name,a.s_sex,COUNT(1) from Student a,Student b where a.s_id != b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex group by a.s_name,a.s_sex -- 31、查询1990年出生的学生名单 --mine select * from Student where s_birth like '1990%' select s_name from student where s_birth like '1990%' -- 32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列 --mine select c_id,AVG(s_score) from Score group by c_id order by AVG(s_score) desc,c_id select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC -- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 --mine select a.s_id,a.s_name,t.avg_sc from student a inner join (select s_id as s_id,AVG(s_score) as avg_sc from Score group by s_id having AVG(s_score) >= 85) t on a.s_id = t.s_id select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85 -- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 --mine select s.s_name,t.s_score from student s inner join (select * from score where c_id in (select c_id from Course where c_name = '数学') and s_score < 60) t on s.s_id = t.s_id select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=( select c_id from course where c_name ='数学') and b.s_score<60 -- 35、查询所有学生的课程及分数情况； --mine按照我的理解，我做成了下面这样。但是网上好像不太一样 select a.s_name,c.c_name,b.s_score from Student a,Score b,Course c where a.s_id = b.s_id and b.c_id = c.c_id --看了答案后，知道了，所以我修改如下 select *, (select s_score from Score where s_id = s.s_id and c_id = '02') as 语文, (select s_score from Score where s_id = s.s_id and c_id = '01') as 数学, (select s_score from Score where s_id = s.s_id and c_id = '03') as 英语, (select SUM(s_score) from Score where s_id = s.s_id) as 总分 from Student s select a.s_id,a.s_name, SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文', SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学', SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语', SUM(b.s_score) as '总分' from student a left join score b on a.s_id = b.s_id left join course c on b.c_id = c.c_id GROUP BY a.s_id,a.s_name -- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数； --mine select a.s_name,c.c_name,b.s_score from Student a,Score b,Course c where a.s_id = b.s_id and b.c_id = c.c_id and b.s_score >= 70 select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id left join student a on a.s_id=c.s_id where c.s_score>=70 -- 37、查询不及格的课程 --mine select a.s_name,b.c_name,c.s_score from Student a,Course b,Score c where a.s_id = c.s_id and b.c_id = c.c_id and c.s_score < 60 select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id where a.s_score<60 --38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名； --mine select a.s_id,a.s_name from student a,Score b where a.s_id = b.s_id and b.c_id = '01' and b.s_score >= 80 select a.s_id,a.s_name from score b inner JOIN student a on a.s_id = b.s_id and b.c_id = '01' and b.s_score >= 80 -- 39、求每门课程的学生人数 --mine select c.c_name as 课程,t.人数 from Course c, (select c_id as c_id,COUNT(1) as 人数 from Student a,Score b where a.s_id = b.s_id group by c_id) t where c.c_id = t.c_id select count(*) from score GROUP BY c_id; -- 40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩 --mine网上的答案有失偏颇，他在sql中用了'02'但是题目中并没有提供这个信息 select y.s_name,t.s_score from student y, (select * from score where c_id = (select a.c_id from Course a where a.t_id = (select t_id from Teacher where t_name = '张三')) and s_score = (select MAX(s_score) as max_sc from score where c_id = (select a.c_id from Course a where a.t_id = (select t_id from Teacher where t_name = '张三')))) t where y.s_id = t.s_id -- 查询老师id select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三' -- 查询最高分（可能有相同分数） select MAX(s_score) from score where c_id='02' -- 查询信息 select a.*,b.s_score,b.c_id,c.c_name from student a LEFT JOIN score b on a.s_id = b.s_id LEFT JOIN course c on b.c_id=c.c_id where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三') and b.s_score in (select MAX(s_score) from score where c_id='02') -- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 --mine我的理解就只能做成这样了，我觉得这道题是我没理解好。 select * from score where s_score in (select s_score from Score group by s_score having COUNT(1) >=2) order by s_score desc select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score -- 42、查询每门功成绩最好的前两名 --mine select s.s_name,c.c_name,t.s_score,t.名次 from student s, course c, (select *,(select COUNT(1)+1 from Score where c_id = s.c_id and s_score > s.s_score) as 名次 from Score as s where (select COUNT(1)+1 from Score where c_id = s.c_id and s_score > s.s_score) <=2) t where s.s_id = t.s_id and c.c_id = t.c_id order by t.c_id,t.s_score desc --网上的这个方法也很好，值得品味。 select a.s_id,a.c_id,a.s_score from score a where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id -- 43、统计每门课程的学生选修人数（超过5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， --若人数相同，按课程号升序排列 --mine select c_id as 课程号,COUNT(1) as 选修人数 from Score group by c_id having COUNT(1) > 5 order by COUNT(1) desc,c_id select c_id,count(*) as total from score GROUP BY c_id HAVING count(*)>5 ORDER BY total,c_id ASC -- 44、检索至少选修两门课程的学生学号 --mine select s_id from Score group by s_id having COUNT(1) >= 2 select s_id,count(*) as sel from score GROUP BY s_id HAVING count(*)>=2 -- 45、查询选修了全部课程的学生信息 --mine select s.* from Student as s inner join (select s_id from Score group by s_id having COUNT(1) = (select COUNT(1) from Course)) t on s.s_id = t.s_id select * from student where s_id in( select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course)) --46、查询各学生的年龄 -- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一 --mine select *,convert(int,(DATENAME(YYYY,GETDATE()))-convert(int,DATENAME(YYYY,s_birth))- (case when CONVERT(datetime,GETDATE(),120)>CONVERT(datetime,s_birth,120) then 0 else 1 end) ) as age from student select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age from student; -- 47、查询本周过生日的学生 --min不要把事情想复杂了 select * from Student where datepart(week,s_birth) = datepart(week,getdate()) select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth) -- 48、查询下周过生日的学生 --mine select * from Student where datepart(week,s_birth) = datepart(week,getdate())+1 select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth) -- 49、查询本月过生日的学生 --mine select * from Student where datepart(month,s_birth) = datepart(month,getdate()) select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth) -- 50、查询下月过生日的学生 --mine 这个确实网上的比我写的要强一些，这个必须得承认 select * from Student where datepart(month,s_birth) = datepart(month,getdate())+1 --网上的我改写了一下，改成了sqlserver的格式 select * from student where DATEPART(MM,GETDATE())+1 =MONTH(s_birth) select MONTH('1990/08/08')
未完，待续！
此篇博客的主要作用就是记录自己的一些sql想法，不至于以后忘记。