根據流傳着各種版本的Mysql 50題，做一份自己的解答跟進；
已經是第二遍再次回顧這部分題目，如有錯誤歡迎糾正～
另外十分感謝@啓明星的指引所提供的原文，附鏈接如下，題目摘自博主原文，答案爲個人所寫。

附第二彈題庫鏈接：
SQL題庫–45題版
——————————————————————————————————————————
另附個人期間所使用工具
sqlfiddle在線工具
網頁版方便個人在閒暇時間的小題目練習

————————————————
版權聲明：本文爲CSDN博主「啓明星的指引」的原創文章，遵循 CC 4.0 BY-SA 版權協議，轉載請附上原文出處鏈接及本聲明。
原文鏈接：https://blog.csdn.net/fashion2014/article/details/78826299
——————————————————————————————————————————

正文

建表

表名和字段
–1.學生表
Student(s_id,s_name,s_birth,s_sex) --學生編號,學生姓名, 出生年月,學生性別
–2.課程表
Course(c_id,c_name,t_id) – --課程編號, 課程名稱, 教師編號
–3.教師表
Teacher(t_id,t_name) --教師編號,教師姓名
–4.成績表
Score(s_id,c_id,s_score) --學生編號,課程編號,分數

--建表
--學生表
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`)
);
--課程表
CREATE TABLE `Course`(
	`c_id`  VARCHAR(20),
	`c_name` VARCHAR(20) NOT NULL DEFAULT '',
	`t_id` VARCHAR(20) NOT NULL,
	PRIMARY KEY(`c_id`)
);
--教師表
CREATE TABLE `Teacher`(
	`t_id` VARCHAR(20),
	`t_name` VARCHAR(20) NOT NULL DEFAULT '',
	PRIMARY KEY(`t_id`)
);
--成績表
CREATE TABLE `Score`(
	`s_id` VARCHAR(20),
	`c_id`  VARCHAR(20),
	`s_score` INT(3),
	PRIMARY KEY(`s_id`,`c_id`)
);
--插入學生表測試數據
insert into Student values('01' , '趙雷' , '1990-01-01' , '男');
insert into Student values('02' , '錢電' , '1990-12-21' , '男');
insert into Student values('03' , '孫風' , '1990-05-20' , '男');
insert into Student values('04' , '李雲' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吳蘭' , '1992-03-01' , '女');
insert into Student values('07' , '鄭竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--課程表測試數據
insert into Course values('01' , '語文' , '02');
insert into Course values('02' , '數學' , '01');
insert into Course values('03' , '英語' , '03');

--教師表測試數據
insert into Teacher values('01' , '張三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成績表測試數據
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);

題目及解答

-- 1、查詢"01"課程比"02"課程成績高的學生的信息及課程分數
    select a.* , b.s_score from Student  as a
    join Score as b on a.s_id = b.s_id
    join Score as c on b.s_id = c.s_id
    where b.c_id = '01' and c.c_id = '02' and b.s_score > c.s_score
        
 

-- 2、查詢"01"課程比"02"課程成績低的學生的信息及課程分數    
   select a.* ,b.s_score from student as a 
   join score as b on a.s_id=b.s_id
   join score as c on b.s_id=c.s_id
   where b.c_id = '01' and c.c_id = '02' and b.s_score < c.s_score
       		 
       		 
-- 3、查詢平均成績大於等於60分的同學的學生編號和學生姓名和平均成績
  select a.s_id, a.s_name , AVG(b.s_score) as avg_score from student as a 
  join score as b on a.s_id = b.s_id
  GROUP BY  a.s_id , a.s_name HAVING  avg_score >=60
       

-- 4、查詢平均成績小於60分的同學的學生編號和學生姓名和平均成績
-- (包括有成績的和無成績的)
  select a.s_id ,a.s_name ,AVG(b.s_score)as avg_score from student as a
 join score as b on a.s_id = b.s_id
 GROUP BY a.s_id, a.s_name HAVING avg_score <60 
 union
 select a.s_id,a.s_name,0 as avg_score from student as a 
 where a.s_id not in (select distinct s_id from score)
        
	        
-- 5、查詢所有同學的學生編號、學生姓名、選課總數、所有課程的總成績
 select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) 
 as sum_score from  student as a 
 left join score b on a.s_id=b.s_id
 GROUP BY a.s_id,a.s_name;


-- 6、查詢"李"姓老師的數量 
 select count(t_name) from teacher 
 where t_name like '李%'


-- 7、查詢學過"張三"老師授課的同學的信息 
	  select * from student  
 		 where s_id in (select s_id from score 
        		where c_id in (select c_id from course
              		where t_id in (select t_id from teacher
                     		 where t_name = '張三')))

-- 8、查詢沒學過"張三"老師授課的同學的信息
  select * from student 
	where s_id in (select s_id from score 
   		where c_id not in (select c_id from course
      	   where t_id in (select t_id from teacher 
     		 where t_name = '張三')))

-- 9、查詢學過編號爲"01"並且也學過編號爲"02"的課程的同學的信息
select * from student as a
join score as b on a.s_id=b.s_id and b.c_id = '01'
join score as c on c.s_id=b.s_id and c.c_id = '02'


-- 10、查詢學過編號爲"01"但是沒有學過編號爲"02"的課程的同學的信息
  		
select * from student 
  where s_id in (select s_id from score where c_id = '01')
    and s_id not in (select s_id from score where c_id ='02')


-- 11、查詢沒有學全所有課程的同學的信息 
select a.* from student as a left join score as b on a.s_id=b.s_id
group by a.s_id having  count(b.c_id)  < (select count (*) from course)
--提示：使用left join保留student 表中一條數據，join會缺失（無分數）


-- 12、查詢至少有一門課與學號爲"01"的同學所學相同的同學的信息 
select a.* from student as a 
join score as b on a.s_id = b.s_id
where b.c_id in (select c_id from score where s_id ='01')
group by a.s_id


-- 13、查詢和"01"號的同學學習的課程完全相同的其他同學的信息 
-- 第二遍開始思路仍然是模糊的哈哈
select * from student 
where s_id in 
(select s_id from score group by s_id HAVING COUNT(s_id)=
-- 下面的語句是找到'01'同學學習的課程數
 (select count (c_id) from score where s_id='01'))
 
 
 and s_id not in
-- 下面的語句是找到學過‘01’同學沒學過的課程，有哪些同學。並排除他們
 (select s_id from score where c_id in (
-- 下面的語句是找到‘01’同學沒學過的課程
   select distinct c_id from score
 where c_id not in (
-- 下面的語句是找出‘01’同學學習的課程
   select c_id from score where s_id='01')
   group by s_id)) 
   
   
-- 下面的條件是排除01同學   
     and s_id not in ('01')


-- 14、查詢沒學過"張三"老師講授的任一門課程的學生姓名 
select s_name from student 
where s_id  not in (select s_id from score 
  where c_id in (select c_id from course 
     where t_id in (select t_id from teacher
        where t_name = '張三')))


-- 15、查詢兩門及其以上不及格課程的同學的學號，姓名及其平均成績 

--個人解法1，錯誤方法，暫未發現錯誤，歡迎糾正～
    select a.s_id,a.s_name,avg(b.s_score) as avg_score from student as a
    join score as b on a.s_id = b.s_id 
    group by a.s_id 
    having (select count 
           (case when (b.s_score < 60 ) then 1 else 0 end )from score)>=2
           
    
   
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、檢索"01"課程分數小於60，按分數降序排列的學生信息
select * from student 
where s_id in 
(select s_id from score 
 where c_id = '01' and s_score <60
 order by s_score desc)


-- 17、按平均成績從高到低顯示所有學生的所有課程的成績以及平均成績
select a.s_id,
(select s_score from score where a.s_id=s_id and c_id='01') as 語文,
(select s_score from score where a.s_id=s_id and c_id='02') as 數學,
(select s_score from score where a.s_id=s_id and c_id='03') as 英語,
 avg(a.s_score) as avg_score from score as a 
group by a.s_id
order by avg_score desc


-- 18.查詢各科成績最高分、最低分和平均分：以如下形式顯示：
-- 課程ID，課程name，最高分，最低分，平均分，及格率，中等率，優良率，優秀率
-- 及格爲>=60，中等爲：70-80，優良爲：80-90，優秀爲：>=90
--原解法使用round四捨五入。也可以使用cast優化顯示
select a.c_id,b.c_name,max(a.s_score),min(a.s_score),avg(a.s_score) as avg_score,
(sum (case when a.s_score >=60 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 及格率,
(sum (case when 70<a.s_score and a.s_score<80 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 中等率,
(sum (case when 80<a.s_score and a.s_score<90 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 優良率,
(sum (case when 90<= a.s_score then 1 else 0 end)/ sum (case when a.s_score then 1 else 0 end)) as 優秀率
from score as a join course as b on a.c_id=b.c_id
group by a.c_id

-- 19、按各科成績進行排序，並顯示排名
-- 篩選課程號‘01’的排序
select * from  
(select t1.c_id,t1.s_score,
-- 選擇2號score表中不重複的項目，條件爲，2號score表中的分數大於1號score表並且課程號是‘01’，註釋爲rank
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
-- 大條件限制爲1號score表中課程號爲‘01’的數據
FROM score t1 where t1.c_id='01'
order by t1.s_score desc) t1

union
-- 篩選課程號‘02’的排序
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
-- 篩選課程號‘03’的排序
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

-- ADD：還有一種全局變量賦值的解法，還沒研究完畢 後續有機會補充

-- 20、查詢學生的總成績並進行排名
select s_id 
  @r := @r + 1 as i
  @s := (case when @score =a.sum_score then @s else @r 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) as a
  (select @k:=0,@i:=0,@score:=0) as s
  --sql fiddle 不支持，需mysql跑

-- 21、查詢不同老師所教不同課程平均分從高到低顯示 
select c.t_id , b.c_id , avg(a.s_score) as avg_score from score as a 
left join course as b on a.c_id = b.c_id
left join teacher as c on b.t_id = c.t_id
group by c.t_id 
order by avg_score desc

-- 22、查詢所有課程的成績第2名到第3名的學生信息及該課程成績
select a.c_id ,a.s_score from score as a
where (select count(1) from score as b 
       where b.c_id=a.c_id and b.s_score >=a.s_score) =2
   or (select count(1) from score as c
       where c.c_id=a.c_id and c.s_score >=a.s_score) =3

-- 23、統計各科成績各分數段人數：
-- 課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]及所佔百分比
-- 第二次做題的遺漏點是，忘記了篩選和join的匹配，遺失了幾行數據
select a.c_id,'85-100',百分比1,'70-85',百分比2,'60-70',百分比3,'0-60',百分比4 from score as a
 join (select c_id, sum(case when s_score>85 and s_score <=100 then 1 else 0 end) as '85-100',
 round ('85-100'/count(*),2) as 百分比1 from score group by c_id) as b on a.c_id=b.c_id  
 join (select c_id, sum(case when s_score>70 and s_score <=85 then 1 else 0 end) as '70-85',
 round ('70-85'/count(*),2) as 百分比2 from score group by c_id) as c on a.c_id=c.c_id 
 join (select c_id, sum(case when s_score>60 and s_score <=70 then 1 else 0 end) as '60-70',
 round ('60-70'/count(*),2) as 百分比3 from score group by c_id) as d on a.c_id=d.c_id 
 join (select c_id, sum(case when s_score>=0 and s_score <=60 then 1 else 0 end) as '0-60',
 round ('0-60'/count(*),2) as 百分比4 from score group by c_id) as e on a.c_id=e.c_id 
 group by a.c_id


-- 24、查詢學生平均成績及其名次 
-- 全局變量解法典型例題    
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、查詢各科成績前三名的記錄
-- 1.選出b表比a表成績大的所有組
-- 2.選出比當前id成績大的 小於三個的           
select * from score as a join score as b 
on a.s_id=b.s_id and b.s_score>a.s_score
group by a.s_id 
having count(b.s_score)<3
order by b.s_score desc


-- 26、查詢每門課程被選修的學生數
select c_id,count(s_id) from score a GROUP BY c_id
    

-- 27、查詢出只有兩門課程的全部學生的學號和姓名 
select a.s_id,b.s_name from score as a 
join student as b on a.s_id=b.s_id
group by a.s_id 
    having count(s_score)=2


-- 28、查詢男生、女生人數 
select s_sex, count(*) from student 
group by s_sex


-- 29、查詢名字中含有"風"字的學生信息
select * from student
where s_name like '%風%'
       

-- 30、查詢同名同性學生名單，並統計同名人數 
select *,count(*) from student as a
join student as b on a.s_name=b.s_name and a.s_sex=b.s_sex
group by a.s_name,a.s_sex 


-- 31、查詢1990年出生的學生名單
select s_name from student 
where s_birth like '1990%'
      
      
-- 32、查詢每門課程的平均成績，結果按平均成績降序排列， 平均成績相同時，按課程編號升序排列 
select c_id,round(avg(s_score),2) as avg_score from score 
group by c_id
order by avg(s_score) desc,c_id asc      


-- 33、查詢平均成績大於等於85的所有學生的學號、姓名和平均成績 
select a.s_id,a.s_name,avg(b.s_score) as avg_score from student as a 
join score as b on a.s_id=b.s_id
group by a.s_id 
having avg_score>=85
    
                
-- 34、查詢課程名稱爲"數學"，且分數低於60的學生姓名和分數 
select a.s_name ,b.s_score from student as a 
join score as b on a.s_id = b.s_id
where b.c_id in (select c_id from course 
where c_name='數學') and b.s_score<60
   
   
-- 35、查詢所有學生的課程及分數情況； 
select a.c_id ,
sum(case when b.c_name='語文' then a.s_score else 0 end) as '語文',
sum(case when b.c_name='數學' then a.s_score else 0 end) as '數學',
sum(case when b.c_name='英語' then a.s_score else 0 end) as '英語',
sum(a.s_score)as '總分'
from score as a join course as b on a.c_id=b.c_id


 -- 36、查詢任何一門課程成績在70分以上的姓名、課程名稱和分數；
 select c.s_name ,b.c_name ,a.s_score from score as a
 join course as b on a.c_id=b.c_id
 join student as c on a.s_id=c.s_id
 where a.s_score>=70


-- 37、查詢不及格的課程
select 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、查詢課程編號爲01且課程成績在80分以上的學生的學號和姓名；
select a.s_id ,a.s_name from student as a 
join score as b on a.s_id=b.s_id
where b.c_id='01' and b.s_score >80


-- 39、求每門課程的學生人數 
select a.c_id ,b.c_name,count(*) from score as a
join course as b on a.c_id=b.c_id
group by a.c_id


-- 40、查詢選修"張三"老師所授課程的學生中，成績最高的學生信息及其成績
select a.*,max(b.s_score)  from student as a
join score as b on a.s_id=b.s_id
where c_id in (select c_id from course
   where t_id in (select t_id from teacher 
        where t_name = '張三'))


-- 41、查詢不同課程成績相同的學生的學生編號、課程編號、學生成績 
select  distinct a.s_id,a.c_id,a.s_score from score as a
join score as b on a.c_id != b.c_id and a.s_score =b.s_score


-- 42、查詢每門功成績最好的前兩名 
 select a.s_id ,a.c_id ,a.s_score from score as a
 where (select count(1) from score as b 
   where b.c_id=a.c_id and b.s_score >=a.s_score ) <=2
 order by a.c_id


-- 43、統計每門課程的學生選修人數（超過5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 c_id ,count(*) as total  from score 
group by c_id
having total>5  
order by total desc ,c_id asc


-- 44、檢索至少選修兩門課程的學生學號 
select s_id,count(s_score)as tol from score
group by s_id
having tol >=2



-- 45、查詢選修了全部課程的學生信息 
select * from student 
where s_id in ( select s_id from score group by s_id having count(*) =(select count(*) from course))


-- 46、查詢各學生的年齡
-- 按照出生日期來算，當前月日 < 出生年月的月日則，年齡減一

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、查詢本週過生日的學生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)


-- 48、查詢下週過生日的學生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
        

-- 49、查詢本月過生日的學生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

	
-- 50、查詢下月過生日的學生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

Finish~~~

Mysql經典50題--第一彈

正文

建表

題目及解答

Python神經網絡模型--Sequential應用舉例

Python神經網絡模型--Sequential詳解及樣例

python--安裝TensorFlow解決Cannot uninstall‘wrapt’It is a distutils installed project

Python--時間序列一本通----實例大舞臺，有碼你就來

Python--KMeans聚類實例--航空公司客戶價值分析

Mac下配置sublime實現LaTeX

https://yachay.unat.edu.pe/blog/index.php?comment_area=format_blog&comment_component=blog&comment_co

linux以太網驅動總結