网上一些sql题目的解决（网上答案+自己答案）

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('' , '赵雷' , '1990-01-01' , '男');
insert into Student values('' , '钱电' , '1990-12-21' , '男');
insert into Student values('' , '孙风' , '1990-05-20' , '男');
insert into Student values('' , '李云' , '1990-08-06' , '男');
insert into Student values('' , '周梅' , '1991-12-01' , '女');
insert into Student values('' , '吴兰' , '1992-03-01' , '女');
insert into Student values('' , '郑竹' , '1989-07-01' , '女');
insert into Student values('' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('' , '语文' , '');
insert into Course values('' , '数学' , '');
insert into Course values('' , '英语' , '');

--教师表测试数据
insert into Teacher values('' , '张三');
insert into Teacher values('' , '李四');
insert into Teacher values('' , '王五');

--成绩表测试数据
insert into Score values('' , '' , 80);
insert into Score values('' , '' , 90);
insert into Score values('' , '' , 99);
insert into Score values('' , '' , 70);
insert into Score values('' , '' , 60);
insert into Score values('' , '' , 80);
insert into Score values('' , '' , 80);
insert into Score values('' , '' , 80);
insert into Score values('' , '' , 80);
insert into Score values('' , '' , 50);
insert into Score values('' , '' , 30);
insert into Score values('' , '' , 20);
insert into Score values('' , '' , 76);
insert into Score values('' , '' , 87);
insert into Score values('' , '' , 31);
insert into Score values('' , '' , 34);
insert into Score values('' , '' , 89);
insert into Score values('' , '' , 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('','') and b.c_id in('','') 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 = '') 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=''
    left join score c on a.s_id=c.s_id and c.c_id='' 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('','') and b.c_id in('','') 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 = '') 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='' or b.c_id=NULL
     join score c on a.s_id=c.s_id and c.c_id='' 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 = '' and b.c_id = '') 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 = ''
and s_id in
(select s_id from score where c_id = '')) 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='' and c.c_id='';

-- 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 = ''
and s_id not in
(select s_id from score where c_id = '')) 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='' ) and a.s_id not in(select s_id from score where c_id='')

-- 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=''
                join score c on a.s_id = c.s_id and c.c_id=''
            where a.c_id=''))

-- 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 = '') and s_id != ''
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='')
    );

-- 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!='' and c_id in(select c_id from score where s_id='')
    group by s_id
    having count(s_id)=(select count(s_id) from score where s_id=''));

-- 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 = '' 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='' 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='') as 语文,
                (select s_score from score where s_id=a.s_id and c_id='') as 数学,
                (select s_score from score where s_id=a.s_id and c_id='') 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=''
            )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=''
            )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=''
            )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 = '') as 语文,
(select s_score from Score where s_id = s.s_id and c_id = '') as 数学,
(select s_score from Score where s_id = s.s_id and c_id = '') 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 = '' 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 = '' 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=''
-- 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='')

-- 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。