MYSQL题目练习专用

原文引用地址，鼓励看原文

1.表名称和字段

 –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

2.插入测试数据

use test
 drop table sc,student,teacher
 
 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);

 准备条件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否则此种情况下会报错:
 -- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo.
 -- 原因：
 -- MySQL 5.7.5和up实现了对功能依赖的检测。如果启用了only_full_group_by SQL模式(在默认情况下是这样)，
 -- 那么MySQL就会拒绝选择列表、条件或顺序列表引用的查询，这些查询将引用组中未命名的非聚合列，而不是在功能上依赖于它们。
 -- (在5.7.5之前，MySQL没有检测到功能依赖项，only_full_group_by在默认情况下是不启用的。关于前5.7.5行为的描述，请参阅MySQL 5.6参考手册。)
 -- 执行以下个命令，可以查看 sql_mode 的内容。
 -------------------------------下边是重点

SHOW SESSION VARIABLES;
 SHOW GLOBAL VARIABLES;
 select @@sql_mode;
 -- 更改
 set global sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
 set session sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';

3.练习题目

我这是5题一小节

 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 st.*,c.c_name,sc.s_score FROM student st join score sc on st.s_id = sc.s_id join course c on sc.c_id = c.c_id where (SELECT s_score FROM score c_id="01") > (SELECT s_score FROM score c_id="02")
 01,02---course,score
 s_score---score
 student.*---student
 
 --答案：
 SELECT st.*,sc1.s_score,sc2.s_score from student st 
 LEFT JOIN score sc1 ON sc1.s_id = st.s_id AND sc1.c_id = "01" 
 LEFT JOIN score sc2 ON sc2.s_id = st.s_id AND sc2.c_id = "02" where sc1.s_score > sc2.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
 --模仿着第一题写出来的
 select st.*,sc1.s_score as '数学',sc2.s_score '语文' from student st 
 LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01' 
 LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02'
 where sc1.s_score < sc2.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***
 --自己憋出来的答案
 select st.s_id,st.s_name,SUM(sc.s_score),ROUND(AVG(sc.s_score),3) from (student st) 
 JOIN (score sc) ON sc.s_id = st.s_id GROUP BY st.s_id HAVING AVG(sc.s_score) >= 60
 
 --答案
 select st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) cjScore from student st
 left join score sc on sc.s_id=st.s_id
 group by st.s_id having AVG(sc.s_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 st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) '平均成绩' from student st 
 LEFT JOIN score sc ON sc.s_id = st.s_id GROUP BY st.s_id HAVING AVG(sc.s_score) < 60 or AVG(sc.s_score) is NULL
 
 --答案
 select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score)) end ) cjScore from student st
 left join score sc on sc.s_id=st.s_id
 group by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL
 
 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 st.s_id,st.s_name,COUNT(DISTINCT c.c_id),SUM(sc.s_score) from student st 
 LEFT JOIN score sc ON sc.s_id = st.s_id 
 LEFT JOIN course c ON c.c_id = sc.c_id GROUP BY st.s_id
 
 --答案
 select st.s_id,st.s_name,count(c.c_id),( case when SUM(sc.s_score) is null or sum(sc.s_score)="" then 0 else SUM(sc.s_score) end) from student st
 left join score sc on sc.s_id =st.s_id 
 left join course c on c.c_id=sc.c_id
 group by st.s_id

 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 st.* from student st 
 join score sc on sc.s_id = st.s_id
 join course c on c.c_id = sc.c_id 
 join teacher t on t.t_id = c.t_id
 where t_name = "张三"
 
 --答案
 select st.* from student st 
 left join score sc on sc.s_id=st.s_id
 left join course c on c.c_id=sc.c_id
 left join teacher t on t.t_id=c.t_id
  where t.t_name="张三"
 
 8、查询没学过"张三"老师授课的同学的信息 *
  -- 张三老师教的课
 --琢磨结果
 select * from student where s_id not in 
 (select st.s_id from student st 
 join score sc on sc.s_id = st.s_id 
 JOIN course c on c.c_id = sc.c_id 
 join teacher t on t.t_id = c.t_id WHERE t.t_name = "张三")
 
 --答案
  select st.* from student st where st.s_id not in(
   select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where  t.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 st.* from student st where st.s_id in (SELECT sc1.s_id from score sc1 where sc1.c_id = "01" and sc1.s_id in (SELECT sc2.s_id from score sc2 where sc2.c_id = "02"))
 
 --答案
 select st.* from student st 
 inner join score sc on sc.s_id = st.s_id
 inner join course c on c.c_id=sc.c_id and c.c_id="01"
 where st.s_id in (
 select st2.s_id from student st2 
 inner join score sc2 on sc2.s_id = st2.s_id
 inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02"
 )
 --高手
 SELECT st.*
 FROM student st
 INNER JOIN score sc ON sc.`s_id`=st.`s_id`
 GROUP BY st.`s_id`
 HAVING SUM(IF(sc.`c_id`="01" OR sc.`c_id`="02" ,1,0))>1
 
 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 st.* from student st 
 WHERE st.s_id in (select sc1.s_id from score sc1 where sc1.c_id = "01" and sc1.s_id not in (select sc2.s_id from score sc2 where sc2.c_id = "02"))
 
 --答案
 select st.* from student st 
 inner join score sc on sc.s_id = st.s_id
 inner join course c on c.c_id=sc.c_id and c.c_id="01"
 where st.s_id not in (
 select st2.s_id from student st2 
 inner join score sc2 on sc2.s_id = st2.s_id
 inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02"
 )

11、查询没有学全所有课程的同学的信息
 select st.* from student st
 LEFT JOIN score sc on st.s_id = sc.s_id
 LEFT JOIN course c on c.c_id = sc.c_id 
 GROUP BY st.s_id HAVING COUNT(c.c_id) < (SELECT COUNT(1) from course)
 
 --答案--居然和我的类似（我服了你了~）
 select st.* from Student st
 left join Score S
 on st.s_id = S.s_id
 group by st.s_id
 having count(c_id)<(select count(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 DISTINCT st.* from student st
 LEFT JOIN score sc on sc.s_id = st.s_id
 where sc.c_id in (select sc.c_id from score sc where sc.s_id = '01')
 
 --答案(比较臃肿)
 select distinct st.* from student st 
 left join score sc on sc.s_id=st.s_id
 where sc.c_id in (
 select sc2.c_id from student st2
 left join score sc2 on sc2.s_id=st2.s_id
 where st2.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
 -- CONCAT_WS(separator,str1,str2,...)的解法，有点臃肿
 select st.*,CONCAT_WS(':',sc1.c_id,sc2.c_id,sc3.c_id) as pigpigpigpigpigpig from student st 
 LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01' 
 LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02' 
 LEFT JOIN score sc3 on sc3.s_id = st.s_id and sc3.c_id = '03'
 GROUP BY st.s_id HAVING pigpigpigpigpigpig in
 (select CONCAT_WS(':',sc1.c_id,sc2.c_id,sc3.c_id) as xxx from student st 
 LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01' 
 LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02' 
 LEFT JOIN score sc3 on sc3.s_id = st.s_id and sc3.c_id = '03'  where st.s_id = '01')
 
 select s_id from score GROUP BY s_id
 select s_id,GROUP_CONCAT(c_id SEPARATOR ';')from score GROUP BY s_id HAVING s_id = '01' -- 重要的一步，，意思：把s_id对应的c_id用分隔符':'连接起来成为新的字段 
--我的答案 
select st.* from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
GROUP BY st.s_id HAVING GROUP_CONCAT(sc.c_id SEPARATOR ':') = (select GROUP_CONCAT(c_id SEPARATOR ':') from score GROUP BY s_id HAVING s_id = '01') 
 
--答案 
select  st.* from student st  
left join score sc on sc.s_id=st.s_id 
group by st.s_id 
having group_concat(sc.c_id) =  
( 
select  group_concat(sc2.c_id) from student st2 
left join score sc2 on sc2.s_id=st2.s_id 
where st2.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 
select DISTINCT st.s_name from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id = sc.c_id  
LEFT JOIN teacher t on t.t_id = c.t_id  
where c.c_id in (select c.c_id from course c join teacher t on t.t_id = c.t_id where t.t_name = '张三') 
--答案 
select st.s_name from student st  
where st.s_id not in ( 
select sc.s_id from score sc  
inner join course c on c.c_id=sc.c_id 
inner join teacher t on t.t_id=c.t_id and t.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     --重要，因为我没做出来 
select st.s_id,st.s_name,AVG(sc.s_score) as '平均成绩' from student st 
LEFT JOIN score sc on sc.s_id = st.s_id  
where sc.s_id in (select sc.s_id from score sc where sc.s_score < 60 GROUP BY sc.s_id HAVING COUNT(sc.s_id) >= 2) 
GROUP BY st.s_id 
--答案 
select st.s_id,st.s_name,avg(sc.s_score) from student st 
left join score sc on sc.s_id=st.s_id 
where sc.s_id in ( 
select sc.s_id from score sc  
where sc.s_score<60 or sc.s_score is NULL 
group by sc.s_id having COUNT(sc.s_id)>=2 
) 
group by st.s_id

16、检索"01"课程分数小于60，按分数降序排列的学生信息 
select st.* from student st join score sc on sc.s_id = st.s_id WHERE sc.c_id = '01' and sc.s_score < 60 ORDER BY sc.s_score DESC 
--答案 
select st.*,sc.s_score from student st  
inner join score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score<60 
order by sc.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 
-- 可加round,case when then else end 使显示更完美 
select st.s_id as '学生编号',st.s_name,sc1.s_score as '语文',sc2.s_score as '数学',sc3.s_score as '外语',ROUND(AVG(sc4.s_score),2) as '平均成绩' from student st 
LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01'  
LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02'  
LEFT JOIN score sc3 on sc3.s_id = st.s_id and sc3.c_id = '03' 
LEFT JOIN score sc4 on sc4.s_id = st.s_id 
GROUP BY st.s_id ORDER BY SUM(sc4.s_score) DESC 
 
--答案 
select st.s_id,st.s_name,avg(sc4.s_score) "平均分",sc.s_score "语文",sc2.s_score "数学",sc3.s_score "英语" from student st 
left join score sc  on sc.s_id=st.s_id  and sc.c_id="01" 
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id="02" 
left join score sc3 on sc3.s_id=st.s_id and sc3.c_id="03" 
left join score sc4 on sc4.s_id=st.s_id 
group by st.s_id  
order by SUM(sc4.s_score) 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 
--我的答案有误 
select c.c_id as '课程ID',c.c_name as '课程name',MAX(sc.s_score) as '最高分',MIN(sc.s_score) as '最低分',AVG(sc.s_score) as '平均分', 
(select COUNT(s_id) from score where s_score >=60 and c_id = c.c_id)/(select COUNT(s_id) from score where c_id = c.c_id) as '及格率', 
COUNT(sc.s_score >=70 and sc.s_score < 80)/COUNT(sc.s_score) as '中等率',COUNT(sc.s_score >=80 and sc.s_score < 90)/COUNT(sc.s_score) as '优良率',COUNT(sc.s_score >= 90)/COUNT(sc.s_score) as '优秀率' from score sc  
LEFT JOIN course c on c.c_id = sc.c_id GROUP BY c.c_id 
 
--答案 
select c.c_id,c.c_name,max(sc.s_score) "最高分",MIN(sc2.s_score) "最低分",avg(sc3.s_score) "平均分"  
,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "及格率" 
,((select count(s_id) from score where s_score>=70 and s_score<80 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "中等率" 
,((select count(s_id) from score where s_score>=80 and s_score<90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "优良率" 
,((select count(s_id) from score where s_score>=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "优秀率" 
from course c 
left join score sc on sc.c_id=c.c_id  
left join score sc2 on sc2.c_id=c.c_id  
left join score sc3 on sc3.c_id=c.c_id  
group by c.c_id 
 
19、按各科成绩进行排序，并显示排名(实现不完全) ***多看多练 
-- mysql没有rank函数 
select c1.s_id,c1.c_id,c1.c_name,@score := c1.s_score,@x := @x + 1 from  
(SELECT sc.s_id,c.c_name,c.c_id,sc.s_score from course c  
join score sc on sc.c_id = c.c_id  
where c.c_id = '01' ORDER BY sc.s_score DESC) as c1, 
(select @x := 0) y 
UNION ALL  
(select c2.s_id,c2.c_id,c2.c_name,@score := c2.s_score,@q := @q + 1 from  
(SELECT sc.*,c.c_name FROM course c  
LEFT JOIN score sc on sc.c_id = c.c_id where c.c_id = '02' ORDER BY sc.s_score DESC) as c2, 
(select @q := 0) w) 
UNION ALL 
(select c3.s_id,c3.c_id,c3.c_name,@score := c3.s_score,@e := @e + 1 from  
(select sc.*,c.c_name from course c  
LEFT JOIN score sc on sc.c_id = c.c_id where c.c_id = '03' ORDER BY sc.s_score desc)as c3, 
(select @e := 0) r) 
 
--验证： 
select COUNT(1) as '各科课程数量' from score GROUP BY c_id 
 
--答案 
-- 加@score是为了防止用union all 后打乱了顺序 
select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c  
left join score sc on sc.c_id=c.c_id 
where c.c_id="01" order by sc.s_score desc) c1 , 
(select @i:=0) a 
union all  
select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c  
left join score sc on sc.c_id=c.c_id 
where c.c_id="02" order by sc.s_score desc) c2 , 
(select @ii:=0) aa  
union all 
select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c  
left join score sc on sc.c_id=c.c_id 
where c.c_id="03" order by sc.s_score desc) c3; 
set @iii=0; 
 
20、查询学生的总成绩并进行排名 --改进 
select s1.s_id,s1.total,@i := @i + 1 as '排名' from  
(select st.s_id,(case when SUM(sc.s_score) is null then 0 else SUM(sc.s_score) end) as total from student st LEFT JOIN score sc on sc.s_id = st.s_id  
GROUP BY st.s_id ORDER BY SUM(sc.s_score) DESC) as s1, 
(select @i := 0) j 
 
--答案 
select st.s_id,st.s_name 
,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end) 
from student st 
left join score sc on sc.s_id=st.s_id 
group by st.s_id order by sum(sc.s_score) desc

21、查询不同老师所教不同课程平均分从高到低显示  
select t.t_name,t.t_id,c.c_name,ROUND(AVG(sc.s_score),2) as '平均成绩' from score sc 
LEFT JOIN course c on c.c_id = sc.c_id  
LEFT JOIN teacher t on t.t_id = c.t_id  
GROUP BY t.t_id ORDER BY AVG(sc.s_score) DESC 
 
--答案 
select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t  
left join course c on c.t_id=t.t_id  
left join score sc on sc.c_id =c.c_id 
group by t.t_id 
order by avg(sc.s_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 st.*,kkk.* FROM 
(select c1.s_id,c1.c_name,@score := c1.s_score,@x := @x + 1 as Rank from  
(select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
where c.c_id = '01' ORDER BY sc.s_score DESC LIMIT 1,2) as c1, 
(SELECT @x := 1) y  
UNION ALL  
(select c2.s_id,c2.c_name,@score := c2.s_score,@xx := @xx + 1 as Rank from  
(select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
where c.c_id = '02' ORDER BY sc.s_score DESC LIMIT 1,2) as c2, 
(SELECT @xx := 1) yy)  
UNION ALL  
(select c3.s_id,c3.c_name,@score := c3.s_score,@xxx := @xxx + 1 as Rank from  
(select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
where c.c_id = '03' ORDER BY sc.s_score DESC LIMIT 1,2) as c3, 
(SELECT @xxx := 1) yyy)) as kkk 
LEFT JOIN student st on st.s_id = kkk.s_id 
 
--答案 
select a.* from ( 
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id =sc.c_id and c.c_id="01" 
order by sc.s_score desc LIMIT 1,2 ) a 
union all 
select b.* from ( 
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id =sc.c_id and c.c_id="02" 
order by sc.s_score desc LIMIT 1,2) b 
union all 
select c.* from ( 
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id =sc.c_id and c.c_id="03" 
order by sc.s_score desc LIMIT 1,2) c 
 
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 c.c_id,c.c_name, 
((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score <= 100 and sc.s_score >= 85)/(select COUNT(1) from score sc where c.c_id = sc.c_id)) "100-85", 
((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score < 85 and sc.s_score >= 70)/(select COUNT(1) from score sc where c.c_id = sc.c_id)) "85-70", 
((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score < 70 and sc.s_score >= 60)/(select COUNT(1) from score sc where c.c_id = sc.c_id)) "70-60", 
((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score < 60 and sc.s_score >= 0)/(select COUNT(1) from score sc where c.c_id = sc.c_id)) "60-0"  
from course c ORDER BY c.c_id ASC 
 
--答案 
select c.c_id,c.c_name  
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100 and sc.s_score>80)/(select count(1) from score sc where sc.c_id=c.c_id )) "100-85" 
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85 and sc.s_score>70)/(select count(1) from score sc where sc.c_id=c.c_id )) "85-70" 
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70 and sc.s_score>60)/(select count(1) from score sc where sc.c_id=c.c_id )) "70-60" 
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60 and sc.s_score>=0)/(select count(1) from score sc where sc.c_id=c.c_id )) "60-0" 
from course c order by c.c_id 
 
24、查询学生平均成绩及其名次 
select s1.s_id,s1.s_name,@xxx := s1.kkk as '平均成绩',@i := @i + 1 from  
(select st.s_id,st.s_name,ROUND((case when AVG(sc.s_score) is null then 0 else AVG(sc.s_score) end),2) as kkk FROM student st LEFT JOIN score sc on st.s_id = sc.s_id GROUP BY st.s_id ORDER BY AVG(sc.s_score) DESC) as s1, 
(select @i := 0) j 
 
--重要的语句 
case  
when AVG(sc.s_score) is null then 0 else AVG(sc.s_score)  
end 
--答案 
set @i=0; 
select a.*,@i:=@i+1 from ( 
select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2) "平均分" from student st 
left join score sc on sc.s_id=st.s_id 
group by st.s_id order by sc.s_score desc) a 
 
25、查询各科成绩前三名的记录 
select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i := @i + 1 from  
(select st.s_id,st.s_name,sc.c_id,c.c_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id =sc.c_id  
where sc.c_id = '01' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, 
(select @i := 0) as j  
UNION ALL  
(select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i2 := @i2 + 1 from  
(select st.s_id,st.s_name,sc.c_id,c.c_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id =sc.c_id  
where sc.c_id = '02' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, 
(select @i2 := 0) as j2 )  
UNION ALL  
(select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i3 := @i3 + 1 from  
(select st.s_id,st.s_name,sc.c_id,c.c_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id =sc.c_id  
where sc.c_id = '03' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, 
(select @i3 := 0) as j3 ) 
 
--答案 
select a.* from ( 
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id='01' 
order by sc.s_score desc LIMIT 0,3) a 
union all  
select b.* from ( 
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id='02' 
order by sc.s_score desc LIMIT 0,3) b 
union all 
select c.* from ( 
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id='03' 
order by sc.s_score desc LIMIT 0,3) c

26、查询每门课程被选修的学生数 
select sc.c_id,c.c_name,COUNT(1) as '选修人数' from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY sc.c_id  
 
--验证：select COUNT(s_score) FROM score  
--答案 
select c.c_id,c.c_name,count(1) from course c  
left join score sc on sc.c_id=c.c_id 
inner join student st on st.s_id=c.c_id 
group by st.s_id 
 
27、查询出只有两门课程的全部学生的学号和姓名 
select st.s_id,st.s_name from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
GROUP BY st.s_id HAVING COUNT(1) = 2 
 
--答案 
select st.s_id,st.s_name from student st  
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id  
group by st.s_id having count(1)=2 
 
28、查询男生、女生人数 
select s_sex as '性别',COUNT(1) as '人数' from student st  
GROUP BY s_sex  
--答案 
select st.s_sex,count(1) from student st group by st.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%' 
 
--答案 
select st.* from student st where st.s_name like "%风%"; 
 
30、查询同名同性学生名单，并统计同名人数 
select st.s_name,COUNT(1) from student st GROUP BY st.s_name,st.s_sex HAVING COUNT(1) >= 2 
--答案 
select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1

31、查询1990年出生的学生名单 
select st.s_name,st.s_birth from student st where st.s_birth like '1990%' 
 
--答案 
select st.* from student st where st.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 sc.c_id,c.c_name,ROUND(AVG(sc.s_score),2) as '平均成绩' from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY sc.c_id ORDER BY AVG(sc.s_score) DESC ,sc.c_id ASC 
 
--答案 
select c.c_id,c.c_name,avg(sc.s_score) from course c 
inner join score sc on sc.c_id=c.c_id   
group by c.c_id order by avg(sc.s_score) desc,c.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 st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) as '平均成绩' FROM student st  
LEFT JOIN score sc on sc.s_id = st.s_id GROUP BY st.s_id HAVING AVG(sc.s_score) >= 85 
 
--答案 
select st.s_id,st.s_name,avg(sc.s_score) from student st 
left join score sc on sc.s_id=st.s_id 
group by st.s_id having avg(sc.s_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 st.s_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id = sc.c_id where c.c_name = '数学' and sc.s_score < 60 
--答案 
select st.s_id,st.s_name,sc.s_score from student st 
inner join score sc on sc.s_id=st.s_id and sc.s_score<60 
inner join course c on c.c_id=sc.c_id and c.c_name ="数学" 
 
35、查询所有学生的课程及分数情况； 
select st.s_id,st.s_name,sc.c_id,c.c_name,sc.s_score from student st 
LEFT JOIN score sc on st.s_id = sc.s_id  
LEFT JOIN course c on c.c_id = sc.c_id  
ORDER BY st.s_id,sc.c_id 
--答案 
select st.s_id,st.s_name,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
left join course c on c.c_id =sc.c_id 
order by st.s_id,c.c_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 st.s_name,c.c_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id =sc.c_id where st.s_id in  
(select sc.s_id FROM score sc where sc.s_score > 70 GROUP BY sc.s_id HAVING COUNT(sc.s_score) = 3) 
ORDER BY st.s_id ASC 
 
--答案,,我的答案和答案有点出入 
select st2.s_id,st2.s_name,c2.c_name,sc2.s_score from student st2 
left join score sc2 on sc2.s_id=st2.s_id 
left join course c2 on c2.c_id=sc2.c_id  
where st2.s_id in( 
select st.s_id from student st  
left join score sc on sc.s_id=st.s_id  
group by st.s_id having min(sc.s_score)>=70) 
order by s_id 
37、查询不及格的课程 
select st.s_id,st.s_name,c.c_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id = sc.c_id where sc.s_score <60 
--答案 
select st.s_id,c.c_name,st.s_name,sc.s_score from student st 
inner join score sc on sc.s_id=st.s_id and  sc.s_score<60 
inner join course c on c.c_id=sc.c_id  
 
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 st.s_id,st.s_name,sc.c_id,sc.s_score from student st  
JOIN score sc on sc.s_id = st.s_id where sc.c_id = '01' and sc.s_score >= 80  
 
--答案 
select st.s_id,st.s_name,sc.s_score from student st 
inner join score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score>=80 
 
39、求每门课程的学生人数 
select sc.c_id,c.c_name,COUNT(1) from score sc  
LEFT JOIN course c on c.c_id = sc.c_id GROUP BY sc.c_id 
 
--答案 
select c.c_id,c.c_name,count(1) from course c 
inner join score sc on sc.c_id=c.c_id 
group by c.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 
select st.*,c1.c_name,c1.kkk as '分数' from student st join  
(select sc.s_id,c.c_name,MAX(sc.s_score) as kkk from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
LEFT JOIN teacher t on t.t_id = c.t_id where t.t_name = '张三') as c1 on c1.s_id = st.s_id 
 
--答案 
select st.*,c.c_name,sc.s_score,t.t_name from student st 
inner join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id  
inner join teacher t on t.t_id=c.t_id and  t.t_name="张三" 
order by sc.s_score desc 
limit 0,1

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 
--理解偏题了 
select st.s_id,sc1.s_score,sc2.s_score,sc3.s_score from student st  
LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01'  
LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02'  
LEFT JOIN score sc3 on sc3.s_id = st.s_id and sc3.c_id = '03'  
where sc1.s_score != sc2.s_score and sc1.s_score != sc3.s_score and sc2.s_score != sc3.s_score 
--参考答案 
select st.s_id,st.s_name,sc.c_id,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id = sc.c_id  
where (select COUNT(1) from student st2  
LEFT JOIN score sc2 on sc2.s_id = st2.s_id  
LEFT JOIN course c2 on c2.c_id = sc2.c_id  
where sc.s_score = sc2.s_score and c.c_id != c2.c_id) > 1 
--答案 
select st.s_id,st.s_name,sc.c_id,sc.s_score from student st  
left join score sc on sc.s_id=st.s_id 
left join course c on c.c_id=sc.c_id 
where ( 
select count(1) from student st2  
left join score sc2 on sc2.s_id=st2.s_id 
left join course c2 on c2.c_id=sc2.c_id 
where sc.s_score=sc2.s_score and c.c_id!=c2.c_id  
)>1 
 
42、查询每门功成绩最好的前两名 
select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i :=@i + 1 from  
(select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '01' 
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, 
(select @i := 0) as j 
UNION ALL  
(select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i2 :=@i2 + 1 from  
(select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '02' 
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, 
(select @i2 := 0) as j2 
) 
UNION ALL  
(select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i3 :=@i3 + 1 from  
(select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st  
LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '03' 
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, 
(select @i3 := 0) as j3 
) 
--验证： 
SELECT sc.c_id,sc.s_score from score sc where sc.c_id = '01' ORDER BY sc.s_score desc LIMIT 2 
 
--答案 
select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id="01" 
order by sc.s_score desc limit 0,2) a 
union all 
select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id="02" 
order by sc.s_score desc limit 0,2) b 
union all 
select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id 
inner join course c on c.c_id=sc.c_id and c.c_id="03" 
order by sc.s_score desc limit 0,2) c 
 
--高手 
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 sc.c_id,c.c_name,COUNT(1) from score sc  
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY sc.c_id HAVING COUNT(1)> 5 ORDER BY COUNT(1) DESC,sc.c_id ASC 
 
--答案 
select sc.c_id,count(1) from score sc 
left join course c on c.c_id=sc.c_id 
group by c.c_id having count(1)>5 
order by count(1) desc,sc.c_id asc 
 
44、检索至少选修两门课程的学生学号 
select st.s_id from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
GROUP BY st.s_id HAVING COUNT(1) >= 2 
 
--答案 
select st.s_id from student st  
left join score sc on sc.s_id=st.s_id 
group by st.s_id having count(1)>=2 
 
45、查询选修了全部课程的学生信息 
--借鉴了答案的用法。在最后一句 
select st.* from student st  
LEFT JOIN score sc on sc.s_id = st.s_id  
LEFT JOIN course c on c.c_id = sc.c_id  
GROUP BY st.s_id HAVING COUNT(1) = (select COUNT(1) from course)  
 
--答案 
select st.* from student st  
left join score sc on sc.s_id=st.s_id 
group by st.s_id having count(1)=(select count(1) from course)

46、查询各学生的年龄 
--尝试用函数进行解答，方便快捷，无纰漏 
SELECT st.*,TIMESTAMPDIFF(YEAR, st.s_birth, CURDATE()) from student st 
 
--答案 
select st.*,timestampdiff(year,st.s_birth,now()) from student st 
 
47、查询本周过生日的学生 
--这个我的答案有问题 
select * from student st 
where year(st.s_birth)<year(CURDATE()) 
and month(st.s_birth)=month(CURDATE()) and  
day(st.s_birth)>day(CURDATE())-DATE_FORMAT(CURDATE()+1,'%e') and day(st.s_birth)<=day(CURDATE())+(8-DATE_FORMAT(CURDATE(),'%e')) 
 
--答案 
select st.* from student st  
where week(now())=week(date_format(st.s_birth,'%Y%m%d')) 
 
48、查询下周过生日的学生 
我参考的答案 
--答案 
select st.* from student st  
where week(now())+1=week(date_format(st.s_birth,'%Y%m%d')) 
 
49、查询本月过生日的学生 
select st.* from student st  
where month(now()) = month(DATE_FORMAT(st.s_birth,'%Y%m%d')) 
 
--答案 
select st.* from student st  
where month(now())=month(date_format(st.s_birth,'%Y%m%d')) 
 
50、查询下月过生日的学生 
--参考了答案的 
select st.* from student st  
where (month(now())+1) mod 12 = month(DATE_FORMAT(st.s_birth,'%Y%m%d')) 
 
--答案； -- 注意:当 当前月为12时,用month(now())+1为13而不是1,可用timestampadd()函数或mod取模 
select st.* from student st  
where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,'%Y%m%d')) 
-- 或 
select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,'%Y%m%d'))

 
tips：

总结：SQL就是要去熟悉表结构，然后找关系，多做题，多思考逻辑的东西，慢慢的磨，慢慢的提升，要有耐心~

posted @ 2020-10-07 01:07 清风吹拂啊狂风肆虐阅读(159) 评论(0) 编辑收藏举报

会员力量，点亮园子希望

刷新页面返回顶部

清风吹拂啊狂风肆虐

MYSQL题目练习专用

原文引用地址，鼓励看原文

1.表名称和字段

2.插入测试数据

3.练习题目

我这是5题一小节

公告