sql语句练习50题(Mysql版-详加注释)
表名和字段
1.学生表
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数
测试数据
--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(3),
PRIMARY KEY(`s_id`,`c_id`)
);--插入测试数据
--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);练习题和sql语句
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score 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
--也可以这样写，个人觉得第二种写法容易理解些。对于不懂join的同学来说。
select a.*,b.s_score as 01_score,c.s_score as 02_score 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'
and b.s_score>c.s_score关于"join"请查看：图解 SQL 里的各种 JOIN
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score 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 s2.`s_score`
AND
s1.`c_id` = '01'
AND
s2.`c_id` = '02' -- 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
-- 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
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 avg_score >=60;ROUND 函数用于把数值字段舍入为指定的小数位数。
SELECT ROUND(column_name,decimals) FROM table_namecolumn_name，必需，要处理的字段；decimals，必需，需要返回的小数位数；
在 SQL 中增加 HAVING 子句原因是，WHERE 关键字无法与合计函数一起使用。如AVG, Group By。
SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name
HAVING aggregate_function(column_name) operator value
-- 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
-- 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
-- (包括有成绩的和无成绩的)
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 avg_score <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);
-- 无成绩：没有记录进入score，故选择student表中有的而score表中没有的。在表中，可能会包含重复值。这并不成问题，不过，有时您也许希望仅仅列出不同（distinct）的值。
关键词 DISTINCT 用于返回唯一不同的值。SELECT DISTINCT 列名称 FROM 表名称
-- 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
-- 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
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
-- 6、查询"李"姓老师的数量
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
-- 7、查询学过"张三"老师授课的同学的信息
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'));
--另一版本，估计是错的
select a.* from
student a where a.s_id in
select b.s_id from score b, teacher c
where b.c_id=c.c_id and c.t_name='张三'-- 8、查询没学过"张三"老师授课的同学的信息
-- 8、查询没学过"张三"老师授课的同学的信息
select * from
student
where 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.c_id from course c join teacher b on c.t_id = b.t_id where b.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
-- 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
select a.* 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
-- 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
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
-- 11、查询没有学全所有课程的同学的信息
--@wendiepei的写法
select s.* from student s
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count(*) from course)COUNT() 函数返回匹配指定条件的行数。
SQL COUNT(column_name) 语法
COUNT(column_name) 函数返回指定列的值的数目（NULL 不计入）：
SELECT COUNT(column_name) FROM table_nameSQL COUNT(*) 语法
COUNT(*) 函数返回表中的记录数：
SELECT COUNT(*) FROM table_nameSQL COUNT(DISTINCT column_name) 语法
COUNT(DISTINCT column_name) 函数返回指定列的不同值的数目：
SELECT COUNT(DISTINCT column_name) FROM table_name注释：COUNT(DISTINCT) 适用于 ORACLE 和 Microsoft SQL Server，但是无法用于 Microsoft Access。
-- 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
-- 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
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
-- 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
--@ouyang_1993的写法
SELECT Student.*
FROM Student
WHERE s_id IN
(SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
--#下面的语句是找到'01'同学学习的课程数
SELECT COUNT(c_id) FROM Score WHERE s_id = '01')
)
AND s_id NOT IN (
--#下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
SELECT s_id FROM Score
WHERE c_id IN(
--#下面的语句是找到‘01’同学没学过的课程
SELECT DISTINCT c_id FROM Score
WHERE c_id NOT IN (
--#下面的语句是找出‘01’同学学习的课程
SELECT c_id FROM Score WHERE s_id = '01')
) GROUP BY s_id
) --#下面的条件是排除01同学
AND s_id NOT IN ('01')--@k1051785839的写法
SELECT t3.* FROM
(
SELECT s_id, group_concat(c_id ORDER BY c_id) group1
FROM score
WHERE s_id > '01'
GROUP BY s_id
) t1
INNER JOIN
(
SELECT
group_concat(c_id ORDER BY c_id) group2
FROM score
WHERE s_id = '01'
GROUP BY s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN
student t3 ON t1.s_id = t3.s_idMySQL教程之concat以及group_concat的用法
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
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')));
-- 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
-- 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
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
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
-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
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
-- 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
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;
--@喝完这杯还有一箱的写法
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,
avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 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
-- 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
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
-- 19、按各科成绩进行排序，并显示排名
-- mysql没有rank函数
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
--@k1051785839的写法
select * from
(select t1.c_id, t1.s_score, (select count(distinct t2.s_score)
from score t2 where t2.s_score>=t1.s_score and t2.c_id='01')
rank FROM score t1 where t1.c_id='01'
order by t1.s_score desc) t1)
union
(select * from (select t1.c_id, t1.s_score, (select count(distinct t2.s_score)
from score t2 where t2.s_score>=t1.s_score and t2.c_id='02')
rank FROM score t1 where t1.c_id='02' order by t1.s_score desc) t2)
union
(select * from (select t1.c_id, t1.s_score,(select count(distinct t2.s_score)
from score t2 where t2.s_score>=t1.s_score and t2.c_id='03')
rank FROM score t1 where t1.c_id='03'
order by t1.s_score desc) t3Mysql中（@i:=@i+1）的作用
mysql @value := 用法
-- 20、查询学生的总成绩并进行排名
-- 20、查询学生的总成绩并进行排名
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
-- 21、查询不同老师所教不同课程平均分从高到低显示
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
-- 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
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'
ORDER BY a.s_score DESC
)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'
ORDER BY a.s_score DESC
)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'
ORDER BY a.s_score DESC
)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
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
-- 24、查询学生平均成绩及其名次
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 ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;-- 25、查询各科成绩前三名的记录
-- 25、查询各科成绩前三名的记录
-- 1.选出b表比a表成绩大的所有组
-- 2.选出比当前id成绩大的 小于三个的
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-- 26、查询每门课程被选修的学生数
-- 26、查询每门课程被选修的学生数
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
-- 27、查询出只有两门课程的全部学生的学号和姓名
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
-- 28、查询男生、女生人数
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
-- 29、查询名字中含有"风"字的学生信息
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
-- 30、查询同名同性学生名单，并统计同名人数
select a.s_name,a.s_sex,count(*) from student a JOIN
student b on 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
-- 31、查询1990年出生的学生名单
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
-- 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
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
-- 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
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
-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select a.s_name,b.s_score from score b 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；
-- 35、查询所有学生的课程及分数情况；
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；
-- 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；
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
-- 37、查询不及格的课程
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；
--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；
select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80
-- 39、求每门课程的学生人数
-- 39、求每门课程的学生人数
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
-- 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
-- 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
-- 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
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
-- 42、查询每门功成绩最好的前两名
-- 牛逼的写法
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
-- 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
select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC-- 44、检索至少选修两门课程的学生学号
-- 44、检索至少选修两门课程的学生学号
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2-- 45、查询选修了全部课程的学生信息
-- 45、查询选修了全部课程的学生信息
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
--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
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
-- 47、查询本周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))-- 48、查询下周过生日的学生
-- 48、查询下周过生日的学生
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
-- 49、查询本月过生日的学生
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
-- 50、查询下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)重要参考：
sql语句练习50题(Mysql版)：https://blog.csdn.net/fashion2014/article/details/78826299
W3School在线教程：https://www.w3school.com.cn/index.html
经典SQL练习题(MySQL版)（有运行结果）：https://blog.csdn.net/paul0127/article/details/82529216