经典Mysql入门必刷50题及全网最新最详细的笔记记录

文章目录

- Mysql50题
- - 练习题1
  - 练习题2
  - 练习题3
  - 练习题4
  - 练习题5
  - 练习题6
  - 练习题7
  - 练习题10
  - 练习题11
  - 练习题12
  - 练习题13
  - 练习题14
  - 练习题15
  - 练习题16
  - 练习题17
  - 练习题18
  - 练习题19
  - - - 19.按各科平均成绩进行排序，并显示排名
  - 练习题20
  - 练习题21
  - 练习题22
  - 练习题24
  - 练习题25
  - 练习题26.
  - 练习题27
  - 练习题28
  - 练习题29
  - 练习题30
  - 练习题31
  - 练习题32
  - 练习题33
  - 练习题34
  - 练习题35
  - 练习题36
  - 练习题37
  - 练习题38
  - 练习题39
  - 练习题40
  - 练习题41
  - 练习题42
  - 练习题43
  - 练习题44
  - 练习题45
  - 练习题46
  - 练习题47
  - 练习题48
  - 练习题49
  - 练习题50

Mysql50题

#建表
create table Student(
    SId varchar(10),
    Sname varchar(10),
    Sage datetime,
    Ssex varchar(10)
); 
#课程表
create table Course(
    CId varchar(10),
    Cname nvarchar(10),
    TId varchar(10)
);
# 老师表
create table Teacher(
    TId varchar(10),
    Tname varchar(10)
);
#成绩表
create table SC(
    SId varchar(10),
    CId varchar(10),
    score decimal(18,1)
);
-- 插入对应的数据
-- 学生表 Student
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' , '1990-12-20' , '男'); 
insert into Student values('04' , '李云' , '1990-12-06' , '男'); 
insert into Student values('05' , '周梅' , '1991-12-01' , '女'); 
insert into Student values('06' , '吴兰' , '1992-01-01' , '女'); 
insert into Student values('07' , '郑竹' , '1989-01-01' , '女'); 
insert into Student values('09' , '张三' , '2017-12-20' , '女'); 
insert into Student values('10' , '李四' , '2017-12-25' , '女'); 
insert into Student values('11' , '李四' , '2012-06-06' , '女'); 
insert into Student values('12' , '赵六' , '2013-06-13' , '女'); 
insert into Student values('13' , '孙七' , '2014-06-01' , '女'); 

-- 科⽬表 Course 
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 Teacher 
insert into Teacher values('01' , '张三'),('02' , '李四'),('03' , '王五'); 

-- 成绩表 SC 
insert into SC values('01' , '01' , 80); 
insert into SC values('01' , '02' , 90); 
insert into SC values('01' , '03' , 99); 
insert into SC values('02' , '01' , 70); 
insert into SC values('02' , '02' , 60); 
insert into SC values('02' , '03' , 80); 
insert into SC values('03' , '01' , 80); 
insert into SC values('03' , '02' , 80); 
insert into SC values('03' , '03' , 80); 
insert into SC values('04' , '01' , 50); 
insert into SC values('04' , '02' , 30); 
insert into SC values('04' , '03' , 20); 
insert into SC values('05' , '01' , 76); 
insert into SC values('05' , '02' , 87); 
insert into SC values('06' , '01' , 31); 
insert into SC values('06' , '03' , 34); 
insert into SC values('07' , '02' , 89); 
insert into SC values('07' , '03' , 98);

建表的层次图
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练习题1

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课程的成绩的学生信息
select * from SC where CId = '1';

在这里插入图片描述

-- 2.查询出该学生课程1和课程2的信息,且她选择的课程1的分数大于课程2的分数的学生ID
		select t1.SId FROM( select * from SC where CId = '01') t1 LEFT JOIN
			(select SId,Cid,score from SC where CId = '02') t2 on t1.SId = t2.SId
			where t1.score > t2.score

在这里插入图片描述

-- 3.查询对应的学生信息和课程分数
select tt1.SId,
		tt2.Sname,
		tt3.CId,
		tt3.score
		from(
        	select t1.SId FROM( select * from SC where CId = '01') t1 LEFT JOIN
			(select SId,Cid,score from SC where CId = '02') t2 on t1.SId = t2.SId
			where t1.score > t2.score
        )tt1
        join student tt2 on tt1.Sid = tt2.SId
        join SC tt1 on tt1.SId = tt3.SId

练习题2

2.查询同时存在01课程和02课程的信息

-- 同时具有o1和02代表我们需要进行表连接
# 最终代码
SELECT t1.SId 
FROM(
    SELECT SId 
    FROM SC 
    WHERE CId='01'
)AS t1 JOIN (
    SELECT SId 
    FROM SC 
    WHERE CId='01'
)AS t2
ON t1.SId = t2.SId;

查询的是01课程的全部信息

select * from SC where CId = '01';

在这里插入图片描述

select * from SC where CId = '02';

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查询两者都有的，利用内连接

练习题3

3…查询存在" 01 “课程但可能不存在” 02 "课程信息的情况(不存在时显示为 null )

select  t1.SId,
				t1.CId,
				t1.score,
				t2.CId as t2CId,
				t2.score as t2Score
from 
(SELECT * from SC where CId = '01' ) as t1
left join
(SELECT * from SC where CId = '02') as t2
on t1.SId = t2.SId;

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练习题4

查询可能不存在01课程但是02课程存在的情况

-- 利用右连接可以筛选出02有的01没有
SELECT t1.SId,
				t1.CId,
				t1.score,
				t2.CId as t2Cid,
				t2.score as t2Score
from
(select * from SC where CId = '01') as t1
RIGHT JOIN (SELECT SId,CId,score FROM SC WHERE CId = '02') as t2
ON t1.SId = t2.SId;

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练习题5

查询平均成绩大于等于60分的同学编号和学生姓名和平均成绩

# 先查询平均成绩大于60分的同学有哪些
SELECT SId,
			ROUND(AVG(score),2) AS avg_score
			from
			SC group by SId
			HAVING avg_score >=60;

在这里插入图片描述

# 进行表连接
select t1.SId,
t2.Sname,
t1.avg_score
FROM(
SELECT SId,
			ROUND(AVG(score),2) AS avg_score
			from
			SC group by SId
			HAVING avg_score >=60
) as t1 JOIN Student as t2 ON t1.SId = t2.SId;

在这里插入图片描述

练习题6

查询在SC表里面学生的姓名和ID

select t1.SId,t2.Sname 
from(
select DISTINCT SId from sc
) t1 JOIN student t2 ON
t1.SId = t2.SId;

在这里插入图片描述

练习题7

7…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，没成绩的显示为 null

-- 首先就是先查询对应的学生编号和选课总数和总成绩
select SId,count(CId) as cnt,SUM(score) as sum_socre from SC group by SId

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-- 题目说查询学生姓名，没成绩的显示为Null,那么就是右连接，这样可以知道没成绩的是哪些
select t2.SId,
				t2.Sname,
				t1.cnt as '选课总数',
				t1.sum_score '总成绩数'
				from
				(select SId,COUNT(CId) as cnt,SUM(score) as sum_score
				FROM sc
				GROUP BY SId) t1
				right join student t2
				on t1.SId = t2.SId

在这里插入图片描述

8 查询李姓老师的数量

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9.查询学过张三老师上课学生的信息

-- 查询到张三老师的课程号
select CId from course where TId = (select TId from teacher WHERE Tname='张三');

-- 查询上张三的学生的学生编号
select SId from SC where CId = (select CId from course where TId = (select TId from teacher WHERE Tname='张三'))

-- 查询对应的学生信息
SELECT t2.SId
,t2.Sname 
FROM(
  SELECT SId 
  FROM SC 
  WHERE CId = (
      SELECT CId 
      FROM Course 
      WHERE TId =(
            SELECT TId 
                FROM Teacher 
                WHERE Tname = '张三'
      )
))t1 JOIN Student t2
ON t1.SId = t2.SId;

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练习题10

查询没有学全所有课程的同学

分析：1.查询学生所学课程的信息

2.将学生课程表与学生表关联查出对应的学生信息

3.进行筛选出小于全课程数量的学生

# 查询出对应的学生编号和学生信息
SELECT t1.SId,
				t1.CId,
				t2.Sname
				from(
				SELECT SId,CId
				from sc				
				) as t1 join student t2 
				on t1.SId = t2.SId;

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# 2.查询出count(cid) < 3的；这里利用了group by 后面接having过滤的小知识
SELECT tt1.SId,
				tt1.Sname,
				count(tt1.CId) as cnt
				FROM(
				select t1.SId,
								t1.CId,
								t2.Sname
								FROM(
								SELECT SId,CId
								FROM sc
								) t1 JOIN student t2
								ON t1.SId = t2.SId
				)tt1 
				GROUP BY tt1.SId,tt1.Sname
				HAVING cnt < (SELECT count(*) from course);

练习题11

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

分析：1.查询学号为01学生的所有课程

		2. 查询至少有一门课在01课程里面的学生
		2. 查询除了01学号的有哪些

在这里插入图片描述

# 1.查询学号为01学生的所有课程
SELECT CId from sc WHERE SId = '01';

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-- 2.查询至少有一门同学的课程在01课程表的人
SELECT distinct SId
				from sc
				WHERE CId IN(SELECT CId from sc WHERE SId = '01');

在这里插入图片描述

# 除了01
SELECT DISTINCT tt1.SId,tt2.Sname
				FROM(SELECT SId
							FROM sc 
							WHERE CId IN(SELECT CId from sc WHERE SId = '01')) tt1
							JOIN student tt2
							ON tt1.SId = tt2.SId
							WHERE tt1.SId !='01';

练习题12

12.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

分析：

1、首先查询出学号为01的课程信息
2、然后查询出学号不为01的课程信息
3、再将上面两个表分别作为t1、t2进行关联，得出01同学学习的所有课程信息和其他同学的所有课程信息
4、然后将SC表作为t1表与Student表t2表进行关联、表tt1分组对学号为01的学生进行筛选

SELECT t2.SId
       ,t1.CId AS t1CId
       ,t2.CId AS t2CId 
FROM(
  SELECT SId
         ,CId 
  FROM SC 
  WHERE SId='01'
) t1 JOIN(
   SELECT SId
         ,CId 
   FROM SC 
   WHERE SId!='01') t2
   ON t1.CId = t2.CId;

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2.将上述查询的信息与student表进行连接，通过SId进行分组，CId个数进行比对

SELECT tt1.SId
       ,COUNT(tt1.t1CId) AS t1Cnt
       ,COUNT(tt1.t2CId) as t2Cnt 
FROM(
  SELECT t2.SId
         ,t1.CId AS t1CId
         ,t2.CId AS t2CId 
  FROM(
  SELECT SId
         ,CId 
  FROM SC 
  WHERE SId='01'
) t1 JOIN(
   SELECT SId
         ,CId 
   FROM SC 
   WHERE SId!='01') t2
   ON t1.CId = t2.CId) tt1 GROUP BY tt1.SId
   HAVING t1Cnt AND t2Cnt = (SELECT COUNT(*) 
                             FROM SC 
                             WHERE SId = '01');

练习题13

13.查询没学过"张三"老师讲授的任一门课程的学生姓名
分析

1.查询张三老师的课程ID

2.查询张三老师课程上的学生的SId

3.查询student表里面的学生SId不在张三课上的人的信息

-- 查询张三老师的CId
select CId FROM course
						WHERE TId = (
						SELECT TId from teacher where Tname = '张三'
						)

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-- 查询上张三课的学生
SELECT SId
				from sc where CId in(
				select CId FROM course
						WHERE TId = (
						SELECT TId from teacher where Tname = '张三'）
						)

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-- 查询对应学生的信息
SELECT SId
      ,Sname 
FROM Student
WHERE SId 
NOT IN (
    SELECT SId 
    FROM SC 
    WHERE CId in(
      SELECT CId 
      FROM Course 
      WHERE TId = (
        SELECT TId 
        FROM Teacher 
        WHERE Tname = '张三')
    )
);

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练习题14

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

分析

1.查询不及格同学的SId

2.查询两门以上不及格同学的平均分和SId

3.将上诉表和student连接，查询对应的信息

-- 查询两门不及格的信息
select SId,
				AVG(score) avg_score
				FROM sc
				WHERE score <'60'
				GROUP BY SId
				HAVING COUNT(score) >=2;

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-- 查询对应学生的信息和成绩
SELECT  tt1.SId,
				tt1.avg_score,
				t2.Sname
				from(
				select SId,
				AVG(score) avg_score
				FROM sc
				WHERE score <'60'
				GROUP BY SId
				HAVING COUNT(score) >=2)
				tt1 JOIN student t2
				on tt1.SId = t2.SId;

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练习题15

– 检索01课程学生分数低于60分，按分数降序排列的同学信息

分析：

1.查询出对应的学生的编号、姓名、和分数

2.查询出对应的的学生课程为01且分数为60的学生信息

SELECT t1.SId,t2.Sname,t1.score 
			FROM sc t1
			JOIN student t2
			on t1.SId = t2.SId

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SELECT t1.SId,t2.Sname,t1.score 
			FROM sc t1
			JOIN student t2
			on t1.SId = t2.SId
			WHERE t1.CId = '01' AND t1.score <'60'
			ORDER BY t1.score DESC;

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练习题16

16.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

分析：

1.查出对应的SId,score的平均分在sc表并以SId为分组

2.指定对应的CId ,并查出对应的SId和Score从sc

3.将表1和表2进行左连接

SELECT tt1.SId,
				tt1.avg_score AS '平均分',
				tt2.score AS '语文',
				tt3.score as '数学',
				tt4.score AS '英语'
FROM (
		SELECT t1.SId,AVG(t1.score) as avg_score from sc t1 GROUP BY t1.SId
)tt1
LEFT JOIN(SELECT SId,score from sc WHERE CId = '01') tt2 ON tt1.SId = tt2.SId
LEFT JOIN(SELECT SId,score FROM sc WHERE CId = '02') tt3 on tt1.SId = tt3.SId
LEFT JOIN(SELECT SId,score FROM sc where CId = '03') tt4 ON tt1.SId = tt4.SId
ORDER BY tt1.avg_score DESC;

练习题17

– 17.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

分析：

1.将我们的CID和我们的SC表进行连接，查询出对应的CId和Cname

2.然后分别计算出最高分、最低分、平均分、及格率、中等率、优良率、优秀率，并将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  t1.CId,t2.Cname 
				FROM sc t1 JOIN
				course t2 ON t1.CId = t2.CId;

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SELECT t1.CId,
				t2.Cname,
				MAX(score) AS '最高分',
				MIN(score) AS '最低分',
				AVG(score) AS '平均分'
	      ,CONCAT(ROUND(SUM(IF(score >= 60,1,0))*100/COUNT(score),2),"%") AS '及格率' 
       ,CONCAT(ROUND(SUM(IF(score >= 70,1,0))*100/COUNT(score),2),"%") AS '中等率' 
       ,CONCAT(ROUND(SUM(IF(score >= 80,1,0))*100/COUNT(score),2),"%") AS '优良率' 
       ,CONCAT(ROUND(SUM(IF(score >= 90,1,0))*100/COUNT(score),2),"%") AS '优秀率' 
				,COUNT(score) as '人数'
				FROM sc t1 JOIN
				course t2 ON t1.CId = t2.CId
				GROUP BY t1.CId,t2.Cname;
				ORDER BY '人数' DESC,CId ASC

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练习题18

按各科平均成绩进行排序，并显示排名

引入变量的知识： @i := 0;

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在这里插入图片描述

-- 1.先查询出各科平均成绩
select CId,
		AVG(score) avg_score
		from sc
		group by CId
		order by avg_score desc;

-- 2.引入变量，进行排名
set @i := 0;
select t1.CId,
		t1.avg_score,
		@i := @i+1 as '排名'
		FROM(select CId,
		AVG(score) avg_score
		from sc
		group by CId
		order by avg_score desc) t1;

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练习题19

19.按各科平均成绩进行排序，并显示排名

分析：

首先查出各科平均成绩，用group by CId就行
最后定义一个变量@，将其上表作为t1表进行排名计算，查询出结果

首先查询出各科的平均分并进行排序

SELECT CId
       ,AVG(score) avg_score 
FROM SC 
GROUP BY CId 
ORDER BY avg_score DESC;

然后定义一个变量@i 然后将上表作为t1表进行排名计算

SET @i :=0;-- 定义一个变量
SELECT t1.CId
       ,t1.avg_score
       ,@i := @i + 1 AS '排名' 
FROM(
   SELECT CId
          ,AVG(score) avg_score 
FROM SC 
GROUP BY CId 
ORDER BY avg_score DESC
) t1;

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练习题20

20.查询学生的总成绩，并进行排名，总分重复保留名次空缺（122345）

分析

1.先查询总成绩，以SId进行分组，并进行排序

2.最后将上表作为t1表然后定义变量@i@j进行排名，@p@q用来控制总分重复时保留名次空缺

SELECT SId
       ,SUM(score) 
AS sum_score 
FROM SC 
GROUP BY SId 
ORDER BY sum_score DESC;

-- 定义变量用来排名的
set @i :=0;
set @j :=0;
set @p :=0;
set @q :=0;
SELECT t1.SId,
				t1.sum_score AS '总分',
				@j :=@j+1,
				@p := t1.sum_score,
				IF(@p=@q,@j,@i :=@j) AS '排名',
				@q := @p
				FROM(
				SELECT SId,SUM(score) as sum_score
				FROM sc 
				GROUP BY SId
				ORDER BY sum_score DESC
				)t1;

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练习题21

– 21.查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

set @i :=0;
set @p :=0;
set @q :=0;
SELECT t1.SId
			,t1.sum_score as '总分'
			,@p := t1.sum_score
			,IF(@p=@q,@i,@i :=@i+1) -- 如果分数相等就可以等于i否则就进行加一
			,@q := @p
			FROM(
				SELECT SId,SUM(score) as sum_score
				FROM sc 
				GROUP BY SId
				ORDER BY sum_score DESC
			) t1;

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练习题22

– 22.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），[60-0）及所占百分比

分析

查询我们对应的人数的话是sum求和：1、0来记录人数，百分比需要用到concat，sum/concat
再将信息表与课程表关联

select
	t1.CId as '课程编号',t2.Cname as '课程名字',
	sum(case when score between 85 and 100 then 1 else 0 end) as '[100-85]',
	concat(round(sum(case when score between 85 and 100 then 1 else 0 end) / count(*) * 100, 2), '%') as '百分比',
	sum(case when score between 70 and 85 then 1 else 0 end) as '[85-70]',
	concat(round(sum(case when score between 70 and 85 then 1 else 0 end) / count(*) * 100, 2), '%') as '百分比',
	sum(case when score between 60 and 70 then 1 else 0 end) as '[70-60]',
	concat(round(sum(case when score between 60 and 70 then 1 else 0 end) / count(*) * 100, 2), '%') as '百分比',
	sum(case when score between 0 and 60 then 1 else 0 end) as '[0-60]',
	concat(round(sum(case when score between 0 and 60 then 1 else 0 end) / count(*) * 100, 2), '%') as '百分比'
from sc t1
inner join course t2
on t1.CId = t2.CId
group by t1.CId,t2.Cname;

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– 23.查询各科成绩前三名的记录

分析

1.查看各科成绩信息，进行降序排名，找出前三

2.进行学生表，课程表相连接

SELECT tt3.CId,tt3.Cname,tt1.SId,tt2.Sname,tt1.score
FROM(
SELECT * FROM
sc t1
where(SELECT count(*) from sc t2
			where t1.CId = t2.CId
			and t1.score < t2.score
) < 3
) tt1
INNER JOIN student tt2 on tt1.SId = tt2.SId
INNER JOIN course tt3 on tt1.CId = tt3.CId
ORDER BY tt1.CId,tt1.score DESC

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练习题24

– 24.查询每门课程选修的学生人数

SELECT CId,COUNT(SId) as '人数'
FROM sc
GROUP BY CId;

练习题25

查询出对应学生所学课程的数量

SELECT t1.SId,
				t2.Sname
				FROM(
				SELECT SId,
				count(CId) as num
				FROM sc
				GROUP BY SId HAVING num = 2
				) t1 JOIN student t2
				ON t1.SId = t2.SId

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练习题26.

查询男生，女生各自的人数

SELECT Ssex
,COUNT(1) 
FROM Student 
GROUP BY Ssex;

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练习题27

查询名字中含有风字的学生信息

SELECT * FROM Student WHERE Sname LIKE '%风%';

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练习题28

28.查询同名同性学生名单，并统计同名同性人数

SELECT Sname
       ,Ssex
       ,COUNT(1) 
AS cnt 
FROM Student 
GROUP BY Sname,Ssex 
HAVING cnt > 1;

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练习题29

查询1999 年出生的学生名单

SELECT * FROM Student 
WHERE Sage 
LIKE '1990%';

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练习题30

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

分析：

查询出课程的平均成绩
利用order by自带升序，让后多个字段代表相同时按第二个排列

-- 查询出平均成绩
SELECT  SC.CId
        ,Course.Cname
        ,AVG(SC.score) AS avg_score
FROM SC 
JOIN Course 
ON SC.CId = Course.CId 
GROUP BY SC.CId,Course.Cname;

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-- 加入排序字段
SELECT  SC.CId
        ,Course.Cname
        ,AVG(SC.score) AS avg_score
FROM SC 
JOIN Course 
ON SC.CId = Course.CId 
GROUP BY SC.CId,Course.Cname
ORDER BY avg_score DESC,SC.CId;

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练习题31

31.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

分析：

1.查询出成绩大于等于平均成绩的课程信息

2.然后将上表与student表进行关联

-- 首先查询出成绩大于等于85的平均成绩的课程信息
SELECT SId
       ,ROUND(AVG(score),2) AS avg_score 
FROM SC 
GROUP BY SId
HAVING avg_score>=85;

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SELECT t1.SId
       ,t2.Sname
       ,t1.avg_score 
FROM(
   SELECT SId
          ,ROUND(AVG(score),2) AS avg_score 
   FROM SC 
   GROUP BY SId
   HAVING avg_score>=85
) t1 JOIN Student t2
ON t1.SId = t2.SId;

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练习题32

32.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

分析：

1、首先查询出课程名为数学的CId
2、然后再将上表作为字表查询出课程名为数学，且分数低于60的课程信息
3、最后将上表作为表t1与Student表t2做关联取出学生信息

SELECT t2.Sname
       ,t1.Score 
FROM(
  SELECT SId
         ,Score 
FROM SC 
WHERE CId =(
      SELECT  CId
      FROM Course 
      WHERE CName = '数学'
)AND score<60
) t1 JOIN Student t2
ON t1.SId = t2.SId;

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练习题33

.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

分析：

1.查出对应学生学号的各科成绩分数情况

2.与student进行右连接就可以，取出学生姓名

-- 1.查询学生各科成绩
SELECT  SId
        ,SUM(CASE CId WHEN '01' THEN score ELSE 0 END) AS '语文成绩'
        ,SUM(CASE CId WHEN '02' THEN score ELSE 0 END) AS '数学成绩'
        ,SUM(CASE CId WHEN '03' THEN score ELSE 0 END) AS '英语成绩'
FROM SC
GROUP BY SId;

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-- 进行连接
SELECT t1.Sname
       ,t2.语文成绩
       ,t2.数学成绩
       ,t2.英语成绩 
FROM
Student t1 LEFT JOIN (
   SELECT  SId
           ,SUM(CASE CId WHEN '01' THEN score ELSE 0 END) AS '语文成绩'
        ,SUM(CASE CId WHEN '02' THEN score ELSE 0 END) AS '数学成绩'
        ,SUM(CASE CId WHEN '03' THEN score ELSE 0 END) AS '英语成绩'
        FROM SC
        GROUP BY SId
) t2
ON t1.SId = t2.SId;

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练习题34

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

分析：

1、首先查询出成绩在70分以上的课程信息
2、然后再将上表作为t1表与Student表t2关联，再与Course表t3做关联取出任何一门课程成绩在 70 分以上的姓名、课程名称和分数信息

-- 查询出成绩在70分的课程成绩
SELECT  SId
        ,CId
        ,score
FROM SC
WHERE score>70;

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-- 然后再将上表作为t1表与Student表t2关联，再与Course表t3做关联取出任何一门课程成绩在 70 分以上的姓名、课程名称和分数信息
SELECT t2.Sname
       ,t3.Cname
       ,t1.Score 
FROM(
   SELECT  SId
           ,CId
           ,score
   FROM SC
   WHERE score>70
) t1 JOIN Student t2
ON t1.SId = t2. SId
JOIN Course t3
ON t1.CId = t3.CId;

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练习题35

查询不及格的课程学生姓名及课程名

SELECT t2.Sname,
				t3.Cname,
				t1.CId,
				t1.score
				FROM(
				select SId,CId,score
				FROM sc where score <60
				)t1 JOIN student t2
				ON t1.SId = t2.SId
				JOIN course t3
				ON t1.CId = t3.CId;

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练习题36

– 36.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

分析

1.查询出课程为01且分数在80的人的SId

2.进行表连接查询

SELECT SId
FROM sc
WHERE CId = '01' AND score >=80;

SELECT t1.SId,
				t2.Sname
				FROM(
				SELECT SId
				FROM sc
				WHERE CId = '01' AND score >=80
				)t1 JOIN student t2
				ON t1.SId = t2.SId;

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练习题37

求每门课程的学生人数

SELECT t1.CId,
				t2.Cname,
				t1.人数
				FROM(
				SELECT CId,
			COUNT(SId) AS '人数'
			from sc GROUP BY CId
				)t1 JOIN course t2
				ON t1.CId = t2.CId;

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练习题38

– 38.成绩不重复，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

分析：

1、查询张三老师的CId;

2.查询此CId的学生信息

SELECT CId 
FROM course
WHERE TId = (SELECT TId from teacher WHERE Tname = '张三');

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SELECT SId,score FROM sc 
WHERE CId =(SELECT CId 
FROM course
WHERE TId = (SELECT TId from teacher WHERE Tname = '张三'))
ORDER BY score DESC
LIMIT 1

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-- 查询出对应学生的信息
SELECT t1.SId AS '学号',
				t2.Sname as '姓名',
				t1.CId,
				t1.score as '分数'
				FROM(
				SELECT SId,score,CId FROM sc 
				WHERE CId =(SELECT CId 
				FROM course
				WHERE TId = (SELECT TId from teacher WHERE Tname = '张三'))
				ORDER BY score DESC
				LIMIT 1
				)t1 JOIN student t2
				ON t1.SId = t2.SId;

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练习题39

– 39.成绩有重复的情况下，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

 -- 1.查询出对应的SId,CId,score等的信息
 SELECT SId
       ,CId
       ,score 
FROM SC
WHERE CId IN (
   SELECT CId 
   FROM Course 
   WHERE TId = (
     SELECT TId 
     FROM Teacher 
     WHERE Tname = '张三'
   )
) ORDER BY score DESC;

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-- 与其他表进行相连接
SET @i := 0;
SET @p := 0;
SET @q := 0;
SELECT tt1.SId,
				tt1.CId,
				tt1.score,
				tt2.Sname
				FROM(
				SELECT  t1.SId
        ,t1.CId
        ,t1.score
        ,@p := t1.score
        ,IF(@p=@q,@i,@i := @i+1) AS `dense_rank`
        ,@q :=@p
				FROM (
				SELECT SId
         ,CId
         ,score 
				FROM SC
				WHERE CId IN 
			(
					SELECT CId 
					FROM Course 
					WHERE TId = (
					SELECT TId 
					FROM Teacher 
					WHERE Tname = '张三'
   )
) ORDER BY score DESC) t1
				)tt1 JOIN student tt2
				ON tt1.`dense_rank` = 1
				AND tt1.SId = tt2.SId;

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练习题40

– 40.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

分析：

1.查询sc中分数一样但是cid不一样的学生SId,以及CId,score

2.将上述表查询出来的进行联结

SELECT 
				t1.SId,
				t1.CId,
				t2.score
				FROM sc t1
				JOIN sc t2
				on t1.score = t2.score
				AND t1.CId != t2.CId;

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SELECT t3.Sname,
				t1.CId,
				t2.score
				FROM sc t1
				JOIN sc t2
				on t1.score = t2.score
				AND t1.CId != t2.CId
				JOIN student t3
				ON t1.SId = t3.SId;

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练习题41

查询出每门课程成绩最好的前两名

分析：首先这里提到前两门也就代表了我们需要进行排序，可以利用变量也可以利用窗口函数，窗口函数要求mysql是在8.0以上的版本

1.查询出每门课程成绩的排名

2.进行表连接，取排名对应的前两名

方法一

# 将成绩按降序排名
SELECT  SId,
				CId,
				score
				FROM sc 
				ORDER BY CId,score DESC

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-- 定义变量，求出排名情况
set @i :=0;
set @p :=0;
set @q :=0;
SELECT 	t1.SId,
				t1.CId,
				t1.score,
				@p := t1.CId,
				IF(@p=@q,@i := @i+1,@i :=1) as rn,
				@q :=@p
				FROM(
				SELECT  SId,
				CId,
				score
				FROM sc 
				ORDER BY CId,score DESC
				) t1

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-- 进行表相连接			
SELECT tt1.SId,
				tt2.Sname,
				tt1.CId,
				tt1.score,
				tt1.rn
				FROM(
				SELECT 	t1.SId,
								t1.CId,
								t1.score,
								@p := t1.CId,
								IF(@p=@q,@i := @i+1,@i :=1) as rn,
								@q :=@p
				FROM(
								SELECT  SId,
								CId,
								score
				FROM sc 
				ORDER BY CId,score DESC
				) t1
				)tt1 JOIN student tt2
				ON tt1.SId = tt2.SId AND tt1.rn <=2
				ORDER BY tt1.CId,tt1.score DESC;

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解法二（窗口函数）

-- 按课程分类进行成绩降序排名
SELECT  *,
			ROW_NUMBER() OVER (
  PARTITION BY CId
  ORDER BY score DESC
) AS rn FROM sc

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SELECT  t1.SId,
				t2.Sname,
				t1.CId,
				t1.score,
				t1.rn
				FROM(
				SELECT *,
			       ROW_NUMBER() OVER (
             PARTITION BY CId
             ORDER BY score DESC
             ) AS rn FROM sc 
				) t1 JOIN student t2 ON t1.rn <=2 AND t1.SId = t2.SId
				ORDER BY CId,score DESC;

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练习题42

统计每门课程学生的选修人数（超过5人的课程才统计）

SELECT CId,
				COUNT(SId) as '选课人数'
				FROM sc
				GROUP BY CId
				HAVING 选课人数 > 5;

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练习题43

检索至少选修2门课程的学生学号

分析：

1.直接统计课程人数，按学生学号进行分组

SELECT SId,
				COUNT(CId) AS cn
				FROM sc
				GROUP BY SId
				HAVING cn>=2;

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练习题44

查询选修了全部课程的学生信息

分析：

查询课程表中共有多少课程
查询出按学生分组计算课程数等于表1查询的课程数
进行表连接，查询出学生姓名等信息

SELECT SId,
				count(CId) as cn
				FROM sc
				GROUP BY SId
				HAVING cn = (SELECT COUNT(CId) FROM course)

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SELECT  t1.SId,
				t2.Sname
				FROM(
				SELECT SId,
				count(CId) as cn
				FROM sc
				GROUP BY SId
				HAVING cn = (SELECT COUNT(CId) FROM course)
				)t1 JOIN student t2 ON
				t1.SId = t2.SId;

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练习题45

查询各学生的年龄，只按年份来算

select year(now())-date_formate(Sage,'%Y') from student

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练习题46

46.按照出生日期来算，当前月日 < 出生年月的月日，则年龄减一

分析：

1.先查询出对应的生日日期以及当前日期

2.在进行当前日期与生日日期的比较

SELECT SId 
       ,Sname
       ,(YEAR(now()) - date_format(Sage,'%Y') ) AS age
       ,date_format(Sage,'%m-%d') AS month_day
       ,date_format(now(),'%m-%d') AS now_month_day
FROM Student;

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SELECT SId 
       ,SName
       ,CASE WHEN now_month_day<month_day THEN age-1 ELSE age END AS new_age
       ,age
FROM(
  SELECT SId 
         ,Sname
         ,(YEAR(now()) - date_format(Sage,'%Y') ) AS age
         ,date_format(Sage,'%m-%d') AS month_day
         ,date_format(now(),'%m-%d') AS now_month_day
FROM Student
) t1;

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练习题47

查询本周过生日的学生

SELECT  SId,
				Sname,
				Sage,
				WEEK(Sage)
				FROM student
				WHERE WEEK(Sage) = WEEK(NOW());

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代表本周没有人生日

练习题48

查询下周生日的学生

SELECT SId,
				Sname,
				Sage,
				WEEK(Sage)
				FROM student
				WHERE WEEK(Sage) = WEEK(DATE_ADD(NOW(),INTERVAL 1 week));

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练习题49

查询本月过生日的学生

SELECT SId,
				Sname,
				Sage,
				MONTH(Sage)
				FROM student
				WHERE MONTH(Sage) = MONTH(NOW());

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练习题50

查询下个月过生日的学生

SELECT SId,
				Sname,
				Sage,
				MONTH(Sage)
				FROM student
				WHERE WEEK(Sage) = WEEK(DATE_ADD(NOW(),INTERVAL 1 WEEK));

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