经典sql题练习50题
-- 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
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
-- 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<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
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; -- 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); -- 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
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
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
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 a.c_id from course a join teacher b on a.t_id = b.t_id 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 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 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
--@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)
--@k1051785839的写法
select *
from student
where s_id not in(
select s_id from score t1
group by s_id having count(*) =(select count(distinct c_id) from course))
-- 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
--@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_id -- 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
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
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
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
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
-- 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) t3)
-- 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 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 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 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
-- 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<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 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
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
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 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 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 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 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 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 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; 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;
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
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;
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
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 -- 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
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
-- 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
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
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
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 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
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
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 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
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)
https://blog.csdn.net/mrbcy/article/details/68965271
https://blog.csdn.net/fashion2014/article/details/78826299
经典sql题练习50题的更多相关文章
- 经典SQL语句基础50题
很全面的sql语句大全.都是很基础性的,今天特意整理了下.大家互相学习.大家有好的都可以分享出来, 分享也是一种快乐. --创建数据库 create database SQL50 --打开SQL50 ...
- 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, ...
- MySQL经典练习题及答案,常用SQL语句练习50题
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 ...
- [转载]sql语句练习50题
Student(Sid,Sname,Sage,Ssex) 学生表 Course(Cid,Cname,Tid) 课程表 SC(Sid,Cid,score) 成绩表 Teacher(Tid,Tname) ...
- sql语句练习50题
Student(Sid,Sname,Sage,Ssex) 学生表 Course(Cid,Cname,Tid) 课程表 SC(Sid,Cid,score) 成绩表 Teacher(Tid,Tname) ...
- sql语句练习50题(Mysql版)
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) – ...
- -sql语句练习50题(Mysql学习练习版)
–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 ...
- sql语句练习50题(Mysql版) 围观
表名和字段 –.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 –.课程表 Course(c_id,c_name,t_id) ...
- 最新JAVA编程题全集(50题及答案)
[程序1]题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少? //这是一个菲波拉契数列问题 pu ...
随机推荐
- 2019牛客多校第五场G-subsequence 1 DP
G-subsequence 1 题意 给你两个字符串\(s.t\),问\(s\)中有多少个子序列能大于\(t\). 思路 令\(len1\)为\(s\)的子序列的长度,\(lent\)为\(t\)的长 ...
- 判断系统是否安装了flash插件
方法1: uses comobj; procedure TForm1.Button1Click(Sender: TObject); var v:variant; begin v:=CreateOleO ...
- 继续我们的学习。这次鸟哥讲的是LVM。。。磁盘管理 最后链接文章没有看
LVM...让我理解就是一个将好多分区磁盘帮到一起的玩意,类似于烙大饼...然后再切 新建了一个虚拟机,然后又挂了一个5G的硬盘,然后分出了5块空间,挂载到了虚拟机上.这些步骤很简单 fdisk ...
- String、StringBuuffer、StringBuilder三者的区别
string String 字符串常量(final修饰,不可被继承,线程不安全),String是常量,当创建之后即不能更改,可以给多个引用共享,在做大量字符串拼接的时候效率低.(可以通过StringB ...
- OpenGL 学习总结
最终呈现画出三角形的一个方式: public void draw(float[] mvpMatrix) { // Add program to OpenGL ES environment GLES20 ...
- 2019牛客多校第四场 I题 后缀自动机_后缀数组_求两个串de公共子串的种类数
目录 求若干个串的公共子串个数相关变形题 对一个串建后缀自动机,另一个串在上面跑同时计数 广义后缀自动机 后缀数组 其他:POJ 3415 求两个串长度至少为k的公共子串数量 @(牛客多校第四场 I题 ...
- 网页打开qq
网页打开qq 打开qq方法tencent://message/?uin=”+“541239271”+“&Menu=yes http://wpa.qq.com/msgrd?V=1&uin ...
- webservice 应用
一直以来,dashboard就会面临一个非常难堪的问题.就是刷新速度太慢了.它要连接query 来获取数据.而query每刷一次都需要时间.这是无可避免的结果.尽管它也是结果集,可还是比较慢.最近实践 ...
- Module not found: Error: Can't resolve '@babel/runtime/helpers/classCallCheck' and Module not found: Error: Can't resolve '@babel/runtime/helpers/defineProperty'
These two mistakes are really just one mistake, This is because the following file @babel/runtime ca ...
- leetcode python翻转字符串里的单词
# Leetcode 151 翻转字符串里的单词### 题目描述给定一个字符串,逐个翻转字符串中的每个单词. **示例1:** 输入: "the sky is blue" 输出: ...