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MySQL classic 50 practice questions and the most detailed analysis of the whole network

2022-08-02 06:57:00 【hungry very hungry】

MySQL练习

文章目录

50道经典SQLExercises so most detailed analytic

数据表介绍

  • 1.Learn to watch Student(SId,Sname,Sage,Ssex)

    • SId Learning Numbers

    • Sname Learn the name

    • Sage A year

    • Ssex Learn the gender

  • 2.课程表 Course(CId,Cname,TId)

    • CId 课程编号

    • Cname 课程名称

    • TId 教师编号

  • 3.教师表 Teacher(TId,Tname)

    • TId 教师编号

    • Tname 教师姓名

  • 4.成绩表 SC(SId,CId,score)

    • SId Learning Numbers

    • CId 课程编号

    • score 分数

建表语句

  • Learn to watch Student

    create table Student(
    SId varchar(10),
    Sname varchar(10),
    Sage datetime,
    Ssex varchar(10)
);

  • 课程表 Course

    create table Course(
    CId varchar(10),
    Cname nvarchar(10),
    TId varchar(10)
);

  • 教师表 Teacher

    create table Teacher(
    TId varchar(10),
    Tname varchar(10)
);

  • 成绩表 SC

    create table SC(
    SId varchar(10),
    CId varchar(10),
    score decimal(18,1)
);

插入数据

注意这里插入数据的时候,It may contain hidden characters,Couldn't show the data manually inserted again can you type again

  • Learn to watch Student

    -- 学生表 Student
-- 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');

  • 课程表 Course

    -- 科?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

    -- 教师表 Teacher 
insert into Teacher values('01' , '张三'); 
insert into Teacher values('02' , '李四'); 
insert into Teacher values('03' , '王五');

  • 成绩表 SC

    -- 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);

练习题目

1.查询" 01 “课程比” 02 "课程成绩高的学生的信息及课程分数

分析：
1、找出有01Achievement of student achievement information
2、找出有02Achievement of student achievement information
3、通过SIdWill take two table aliast1、t2进行左连接
4、加上满足01‘语文’ > 02’数学’的条件

  • 找出有01Achievement of student achievement information

    SELECT * FROM SC WHERE CId=‘01’;

  • 找出有02Achievement of student achievement information

    SELECT * FROM SC WHERE CId=‘02’;

  • 通过SIdWill take two table aliast1、t2进行左连接

    SELECT t1.SId,
    t1.CId,
    t1.score as ‘语文’,
    t2.score as ‘数学’
    FROM (
    SELECT
    SId,
    CId,
    score
    FROM SC
    WHERE CId=‘01’
    ) t1
    LEFT JOIN
    (SELECT
    SId,
    CId,
    score
    FROM SC
    WHERE CId=‘02’
    ) t2
    ON t1.SId=t2.SId;

  • 加上满足01‘语文’ > 02’数学’的条件

    SELECT t1.SId,
    t1.CId,
    t1.score as ‘语文’,
    t2.score as ‘数学’
    FROM (
    SELECT
    SId,
    CId,
    score
    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;

  • Finally the above table as a child tablett1Will we want to query table relate,Take out want to query field

    SELECT tt1.SId
    ,tt2.Sname
    ,tt3.CId
    ,tt3.score
    FROM (
    SELECT t1.SId
    FROM(
    SELECT SId
    ,CId
    ,score
    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 tt3 ON tt1.SId = tt3.SId;

2.查询同时存在" 01 “课程和” 02 "课程的情况
分析：
     满足条件的SC表中：
       1、Select course number is01的全部信息 AS命名为 t1
       2、Select course number is02的全部信息 AS命名为 t2
       3、使用joinConnection to take out the exist01课程和02课程的SId

  • Select course number is01的全部信息

    SELECT SId FROM SC WHERE CId = ‘01’;

  • Select course number is02的全部信息

    SELECT SId FROM SC WHERE CId = ‘02’;

  • 使用joinConnection to take out the exist01课程和02课程的SId

    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;

3.查询存在" 01 “课程但可能不存在” 02 "课程的情况(不存在时显示为 null )
分析：
     满足条件的SC表中：
       1、Select course number is01的全部信息 AS命名为 t1
       2、Select course number is02的全部信息 AS命名为 t2
       3、左连接

  • Select course number is01的全部信息

    SELECT SId,CId,score FROM SC WHERE CId = ‘01’;

  • Select course number is02的全部信息

    SELECT SId,CId,score FROM SC WHERE CId = ‘02’;

  • 左连接

    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,t2.CId AS t2CId
    ,t2.score AS t2Score
    FROM(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE CId = ‘01’
    ) t1 LEFT JOIN(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE CId = ‘02’
    ) t2
    ON t1.SId = t2.SId;

4.查询不存在" 01 “课程但存在” 02 "课程的情况
分析：
     满足条件的SC表中：
       1、Select course number is01的全部信息 AS命名为 t1
       2、Select course number is02的全部信息 AS命名为 t2
       3、右连接

  • Select course number is01的全部信息

    SELECT SId,CId,score FROM SC WHERE CId = ‘01’;

  • Select course number is02的全部信息

    SELECT SId,CId,score FROM SC WHERE CId = ‘02’;

  • 右连接

    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,t2.CId AS t2CId
    ,t2.score AS t2Score
    FROM(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE CId = ‘01’
    ) t1 RIGHT JOIN(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE CId = ‘02’
    ) t2
    ON t1.SId = t2.SId;

5.Average query in equal to 60 分的同学的学生编号和学生姓名和平均成绩

分析：
1、To find out the average scores greater than60分的SId,并用ROUND(X,D)保留两位小数
2、Then average as tablet1与Student表t2Make the connection the results

  • To find out the average scores greater than60分的SId,并用ROUND(X,D)保留两位小数

    SELECT SId
    ,ROUND(AVG(score),2)AS avg_score
    FROM SC
    GROUP BY SId
    HAVING avg_score>=60;

  • Then average as tablet1与Student表t2Make the connection the results

    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
    )t1 JOIN Student t2
    ON t1.SId = t2.SId;

6.查询在 SC 表存在成绩的学生信息

分析：
1、首先DISTINCTSC表的SId的数据
2、In the table as the tablet1与Student表t2Join query student information

  • 首先DISTINCTSC表的SId的数据

    SELECT
    DISTINCT SId
    FROM SC;

  • In the table as the tablet1与Student表t2Join query student information

    SELECT t1.SId
    ,t2.Sname
    FROM(
    SELECT
    DISTINCT SId
    FROM SC
    )t1 JOIN Student t2
    ON t1.SId = t2.SId;

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

分析：
1、首先统计出SC表SIdThe total number of course of study and overall
2、In the table as the tablet1与Student表t2Do the right connection query out all classmates learn code、Learn the name、选课总数、所有课程的总成绩,没成绩的显示为 null

  • 首先统计出SC表SIdThe total number of course of study and overall

    SELECT SId
    ,COUNT(CId) AS cnt
    ,SUM(score) AS sum_score
    FROM SC
    GROUP BY SId;

  • In the table as the tablet1与Student表t2Do the right connection query out all classmates learn code、Learn the name、选课总数、所有课程的总成绩,没成绩的显示为 null

    SELECT t2.SId
    ,t2.Sname
    ,t1.cnt AS ‘选课总数’
    ,t1.sum_score AS ‘总成绩’
    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;

8.查询「李」姓老师的数量

分析：直接使用COUNT(*)统计LIKE模糊查询查询「李」The number of name is t

  • 直接使用COUNT(*)统计LIKE模糊查询查询「李」The number of name is t

    SELECT COUNT(*)
    FROM Teacher
    WHERE Teacher.Tname
    LIKE (‘李%’);

9.查询学过「张三」老师授课的同学的信息

分析：
1、First of all query zhang teacherTId
2、Then the query zhang teacher teaching informationCId
3、Query studied zhang teacher taught students againSId
4、Finally will be studied zhang teacher taught studentsSId作为表t1和Student表t2Join query student information

  • First of all query zhang teacherTId

    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’;

  • Then the query zhang teacher teaching informationCId

    SELECT CId
    FROM Course
    WHERE TId =(
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    );

  • Query studied zhang teacher taught students againSId

    SELECT SId
    FROM SC
    WHERE CId = (
    SELECT CId
    FROM Course
    WHERE TId =(
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    )
    );

  • Finally will be studied zhang teacher taught studentsSId作为表t1和Student表t2Join query student information

    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;

10.查询没有学全所有课程的同学的信息

分析：
1、Query all course information before
2、Then all course information as the tablet1与Student表t2Students join queries in information
3、The number of queries all course information
4、The final table astt1Table to the course information filtering is less than the query out did not learn all of the total course of students of information

  • Query all course information before

    SELECT SId
    ,CId
    FROM SC;

  • Then all course information as the tablet1与Student表t2Students join queries in information

    SELECT t1.SId
    ,t1.SId
    ,t2.Sname
    FROM(
    SELECT SId
    ,CId
    FROM SC
    ) t1 JOIN Student t2
    ON t1.SId = t2.SId;

  • The number of queries all course information

    SELECT count(*)
    FROM Course;

  • The final table astt1Table to the course information filtering is less than the query out did not learn all of the total course of students of information

    SELECT tt1.SId
    ,tt1.SName
    ,count(tt1.CId) as cnt
    FROM (
    SELECT t1.SId
    ,t1.CId
    ,t2.SName
    FROM SC t1 JOIN Student t2
    ON t1.SId = t2.SID
    ) tt1 GROUP BY tt1.SId,tt1.SName
    Having cnt < (SELECT count(*) FROM Course);

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、Query student id, first of all, for01Students learn the course number
2、Then queries at least one course with01Students learned the same student number
3、The above-mentioned results astt2与Student表tt2进行左连接查询,找到学生信息

  • Query student id, first of all, for01Students learn the course number

    SELECT CId
    FROM SC
    WHERE SId = ‘01’;

  • Then queries at least one course with01Students learned the same student number

    SELECT t2.SId
    ,COUNT(t2.SId)
    FROM(
    SELECT CId
    FROM SC
    WHERE SId = ‘01’
    )t1 JOIN SC t2
    ON t1.CId = t2.CId
    GROUP BY t2.SId;

  • The above-mentioned results astt2与Student表tt2进行左连接查询,找到学生信息

    SELECT tt2.SId
    ,tt2.Sname
    FROM(
    SELECT t2.SId
    ,COUNT(t2.SId)
    FROM(
    SELECT CId
    FROM SC
    WHERE SId = ‘01’
    )t1 JOIN SC t2
    ON t1.CId = t2.CId
    GROUP BY t2.SId
    ) tt1 LEFT JOIN Student tt2
    ON tt1.SId = tt2.SId;

解法2

分析：
1、Query student id, first of all, for01Students learn the course number
2、Then of course, is not01All the students' course information filtering
3、最后将SC表作为t1表与Student表t2表进行关联,找到学生信息

  • Query student id, first of all, for01Students learn the course number

    SELECT CId
    FROM SC
    WHERE SId = ‘01’;

  • Then of course, is not01All the students' course information filtering

    SELECT DISTINCT SId
    FROM SC
    WHERE SId != ‘01’
    AND CId IN(
    SELECT CId
    FROM SC
    WHERE SId = ‘01’
    );

  • 最后将SC表作为t1表与Student表t2表进行关联,找到学生信息

    SELECT DISTINCT t1.SId
    ,t2.Sname
    FROM
    SC t1 JOIN Student t2
    ON t1.SId = t2.SId
    WHERE t1.SId != ‘01’
    AND CId IN(
    SELECT CId
    FROM SC
    WHERE SId = ‘01’
    );

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、And then query out student id is not as01的课程信息
3、The above two tables respectively ast1、t2进行关联,得出01Students learn all course information and all the other students information
4、然后将SC表作为t1表与Student表t2表进行关联、表tt1Group of student number for01The student to carry on the screen

  • 首先查询出学号为01的课程信息

    SELECT SId
    ,CId
    FROM SC
    WHERE SId=‘01’;

  • And then query out student id is not as01的课程信息

    SELECT SId
    ,CId
    FROM SC
    WHERE SId!=‘01’;

  • The above two tables respectively ast1、t2进行关联,得出01Students learn all course information and all the other students information

    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;

  • 然后将SC表作为t1表与Student表t2表进行关联、表tt1Group of student number for01The student to carry on the screen

    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’);

  • The final table asttt1与Student表ttt2进行关联,找到学生信息

    SELECT ttt1.SId
    ,ttt2.Sname
    FROM(
    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’)
    ) ttt1 JOIN Student ttt2
    ON ttt1.SId = ttt2.SId;

13.查询没学过"张三"老师讲授的任一门课程的学生姓名

1、First query out zhang teacher teach
2、And then query out zhang teacher teach any number of course
3、To query the zhang teacher taught any of course number correspondingSId
4、Haunted by the last query learned"张三"老师讲授的任一门课程的学生姓名

  • First query out zhang teacher teach

    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’;

  • And then query out zhang teacher teach any number of course

    SELECT CId
    FROM Course
    WHERE TId = (
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    );

  • To query the zhang teacher taught any of course number correspondingSId

    SELECT SId
    FROM SC
    WHERE CId in(
    SELECT CId
    FROM Course
    WHERE TId = (
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’)
    );

  • Haunted by the last query learned"张三"老师讲授的任一门课程的学生姓名

    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’)
    )
    );

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、Failed in the first query classmate student id and course number
2、再以SIdGroups will be more than two doors and failed the course student number and average scores query out
3、Finally the table as tablett1与Student表tt2做关联,Take out to query information

  • Failed in the first query classmate student id and course number

    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE score < 60;

  • 再以SIdGroups will be more than two doors and failed the course student number and average scores query out

    SELECT t1.SId
    ,COUNT(t1.CId) AS cnt
    ,AVG(t1.score) AS avg_score
    FROM(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE score < 60) t1
    GROUP BY SId HAVING cnt >=2;

  • Finally the table as tablett1与Student表tt2做关联,Take out to query information

    SELECT tt2.SId,tt2.Sname,tt1.avg_score FROM
    (SELECT t1.SId
    ,COUNT(t1.CId) AS cnt
    ,AVG(t1.score) AS avg_score
    FROM(
    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE score < 60) t1
    GROUP BY SId HAVING cnt >=2) tt1 JOIN Student tt2
    ON tt1.SId = tt2.SId;

15.检索" 01 "课程分数小于 60,按分数降序排列的学生信息

分析：
1、首先先将SC表t1与Student表t2进行关联
2、The last searching01课程分数小于60,按分数降序排列的学生信息

  • 首先先将SC表t1与Student表t2进行关联

    SELECT t2.SId
    ,t2.Sname
    ,t1.score
    FROM
    SC t1 JOIN Student t2
    ON t1.SId = t2.SId;

  • The last searching01课程分数小于60,按分数降序排列的学生信息

    SELECT t2.SId
    ,t2.Sname
    ,t1.score
    FROM
    SC t1 JOIN Student t2
    ON t1.SId = t2.SId
    AND t1.CId = ‘01’ AND t1.score < 60
    ORDER BY t1.score DESC;

16.According to the average scores from low to show all students of all course grades and gpa

分析：
1、First of all find out first’01’语文、'02’数学、'03’English courses, the corresponding result
2、And then find out all course grade point average
3、Because of the language、数学、All three courses to the average English group
4、Finally the table after grouping astt1Table respectively with’01’语文、'02’数学、'03’English left connection,Find out all the average score of each course students,Then ascending according to average

  • First of all find out first’01’语文、'02’数学、'03’English courses, the corresponding result

    SELECT SId,score FROM SC WHERE SId = ‘01’;

SELECT SId,score FROM SC WHERE SId = '02';


SELECT SId,score FROM SC WHERE SId = '03';

  • And then find out all course grade point average

    SELECT SId
    ,AVG(score) as avg_socre
    FROM SC
    GROUP BY SId;

  • Because of the language、数学、All three courses to the average English group

    SELECT t1.SId
    ,AVG(t1.score) as avg_score
    FROM
    SC t1 GROUP BY t1.SId;

  • Finally the table after grouping astt1Table respectively with’01’语文、'02’数学、'03’English left connection,Find out all the average score of each course students,Then ascending according to average

    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;

17.The query subjects the points、最低分和平均分： 以如下形式显示：课程 ID,课程 name,The most points,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

  • 首先先将SC表和CourseCorrelation and grouped

    SELECT t1.CId
    ,t2.Cname
    FROM
    SC t1 JOIN Course t2
    ON t1.CId = t2.CId
    GROUP BY t1.CId,t2.Cname;

  • And then calculate the highest respectively、最低分、平均分、及格率、中等率、优良率、优秀率,And the query results in descending order number,若人数相同,按课程号升序排列

    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;

18.按各科平均成绩进行排序,并显示排名

分析：
1、First query the various branches of average and sorting
2、Finally define a [email protected] the table as at1Ranking table calculation,查询出结果

  • First query the various branches of average and sorting

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

  • Finally define a [email protected] the table as at1Ranking table calculation,查询出结果

    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;

19.按各科平均成绩进行排序,并显示排名,重复时不保留名次空缺

分析：
1、First query the various branches of average and sorting
2、Finally define a [email protected] the table as at1Ranking table calculation,查询出结果

  • First query the various branches of average and sorting

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

  • Finally define a [email protected] the table as at1Ranking table calculation,查询出结果

    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;

Three common ranking

row_number、dense_rank、rank在MySQL 5.7中的实现
对SC中的学生scoreIn the overall ranking

ROW_NUMBER

1 2 3 4 5 6 7 No repeat ranking,依次递增

SET @i := 0;
SELECT  t1.SId
        ,t1.CId
        ,t1.score
        ,@i := @i + 1 as row_number
from (
    SELECT  SId
            ,CId
            ,score
    from SC
    order by score desc
) t1;

DENSE_RANK

1 2 3 3 3 4 5 6 7 Have a repeat for ranking,The ranking is also continuous

SET @i := 0;
SET @p := 0;
SET @q := 0;
SELECT  t1.SId
        ,t1.CId
        ,t1.score
        ,@p := t1.score
        ,if(@[email protected],@i,@i := @i+1) as dense_rank
        ,@q :[email protected]
from (
    SELECT  SId
            ,CId
            ,score
    from SC
    order by score desc
) t1;

RANK

1 2 3 3 3 6 7 8 Have a repeat for ranking,The ranking of discontinuity

SET @i := 0;
SET @j := 0;
SET @p := 0;
SET @q := 0;
SELECT  t1.SId
        ,t1.CId
        ,t1.score
        ,@j := @j + 1
        ,@p := t1.score
        ,if(@[email protected],@i,@i := @j) as rank
        ,@q :[email protected]
from (
    SELECT  SId
            ,CId
            ,score
    from SC
    order by score desc
) t1;

To group ranked

ROW_NUMBER

The top three query each course grade one of the best 1 2 3 4 5 6 7 No repeat ranking,依次递增

SET @i := 0;
SET @p := 0;
SET @q := 0;
SELECT  tt1.SId
        ,tt2.SName
        ,tt1.CId
        ,tt1.score
        ,tt1.rn
from (
    select  t1.SId
            ,t1.CId
            ,t1.score
            ,@p := t1.CId
            ,if(@[email protected],@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.rn<=3 and tt1.SId = tt2.SId
order by tt1.CId,tt1.rn;

DENSE_RANK

The top three query each course grade one of the best 1 2 3 3 3 4 5 6 7 Have a repeat for ranking,The ranking is also continuous

SET @i := 0;
SET @p := 0;
SET @q := 0;
SET @j := 0;
SET @k := 0;
SELECT  tt1.SId
        ,tt2.SName
        ,tt1.CId
        ,tt1.score
        ,tt1.rn
from (
    select  t1.SId
            ,t1.CId
            ,t1.score
            ,@p := t1.CId
            ,@j := t1.score
            ,if(@[email protected],if(@[email protected],@i,@i := @i + 1),@i :=1) as rn
            ,@q := @p
            ,@k := @j
    from (
        select  SId
                ,CId
                ,score
        from SC
        order by CId,score DESC
    ) t1
) tt1 join Student tt2 on tt1.rn<=3 and tt1.SId = tt2.SId
order by tt1.CId,tt1.rn;

RANK

The top three query each course grade one of the best 1 2 3 3 3 6 7 8 Have a repeat for ranking,The ranking of discontinuity

SET @i := 0;
SET @p := 0;
SET @q := 0;
SET @j := 0;
SET @k := 0;
SET @m := 1;
SELECT  tt1.SId
        ,tt2.SName
        ,tt1.CId
        ,tt1.score
        ,tt1.rn
from (
    select  t1.SId
            ,t1.CId
            ,t1.score
            ,@p := t1.CId
            ,@j := t1.score
            ,if(@[email protected],@m := @m + 1,@m := 1)
            ,if(@[email protected],if(@[email protected],@i,@i := @m),@i :=1) as rn
            ,@q := @p
            ,@k := @j
    from (
        select  SId
                ,CId
                ,score
        from SC
        order by CId,score DESC
    ) t1
) tt1 join Student tt2 on tt1.rn<=3 and tt1.SId = tt2.SId
order by tt1.CId,tt1.rn;
20.查询学生的总成绩,并进行排名,总分重复时保留名次空缺

122345
分析：
1、First query first grade,以SId进行分组,并进行排序
2、The final table ast1Table and then define [email protected]、@j进行排名、@p、@qWhen used to control total repeat 保留名次空缺

  • First query first grade,以SId进行分组,并进行排序

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

*The final table ast1Table and then define [email protected]、@j进行排名、@p、@qWhen used to control total repeat 保留名次空缺

SET @i :=0;
SET @j :=0;
SET @p :=0;
SET @q :=0;
SELECT t1.SId
       ,t1.sum_score AS '总分'
       ,@j :[email protected] + 1
       ,@p :=t1.sum_score
       ,IF(@[email protected],@j,@i :[email protected]) AS '排名'
       ,@q :[email protected] 
FROM(
  SELECT SId
         ,SUM(score) 
  AS sum_score 
  FROM SC 
  GROUP BY SId 
  ORDER BY sum_score DESC
) t1;

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

122456

  • First query first grade,以SId进行分组,并进行排序

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

*The final table ast1Table and then define [email protected]进行排名、@p、@qWhen used to control the total score repeated don't keep vacancy place

SET @i :=0;
SET @p :=0;
SET @q :=0;
SELECT  t1.SId
        ,t1.sum_score AS '总分'
        ,@p := t1.sum_score 
        ,if(@[email protected],@i,@i := @i+1) as '排名'
        ,@q :[email protected]
FROM(
  SELECT SId
         ,SUM(score) 
  AS sum_score 
  FROM SC 
  GROUP BY SId 
  ORDER BY sum_score DESC
) t1;

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

分析：
1、首先将SC表t1与Course表t2进行关联,并以CId进行分组
2、Calculated for each block final score,显示出来

  • 首先将SC表t1与Course表t2进行关联,并以CId进行分组

    SELECT t1.CId
    ,t2.Cname
    FROM
    SC t1 JOIN Course t2
    ON t1.CId = t2.CId
    GROUP BY t1.CId,t2.Cname;

  • Calculated for each block final score,显示出来

    SELECT t1.CId
    ,t2.Cname
    ,CONCAT(ROUND(SUM(IF(score<=100 AND score>85,1,0)),2),“%”) AS ‘[100-85)’
    ,CONCAT(ROUND(SUM(IF(score<=85 AND score>70,1,0)),2),“%”) AS ‘[85-70)’
    ,CONCAT(ROUND(SUM(IF(score<=70 AND score>60,1,0)),2),“%”) AS ‘[70-60)’
    ,CONCAT(ROUND(SUM(IF(score<=60 AND score>0,1,0)),2),“%”) AS ‘[60-0)’
    FROM
    SC t1 JOIN Course t2
    ON t1.CId = t2.CId
    GROUP BY t1.CId,t2.Cname;

23.查询各科成绩前三名的记录

分析：
1、First of all find out two subjects and sorting
2、Then the table as at1表定义@i控制排序,@q、@pRepetitive control keep vacancy
3、The final table astt1表与Student表tt2Correlation to remove the record of the top three results in all the subjects

  • First of all find out two subjects and sorting

    SELECT SId
    ,CId
    ,score
    FROM SC
    ORDER BY CId,score DESC;

  • Then the table as at1表定义@i控制排序,@q、@pRepetitive control keep vacancy

    SET @i := 0;
    SET @p := 0;
    SET @q := 0;
    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,@p := t1.CId
    ,if(@[email protected],@i := @i + 1,@i :=1) as rn
    ,@q := @p
    FROM (
    SELECT SId
    ,CId
    ,score
    FROM SC
    ORDER BY CId,score DESC
    ) t1

  • The final table astt1表与Student表tt2Correlation to remove the record of the top three results in all the subjects

    SET @i := 0;
    SET @p := 0;
    SET @q := 0;
    SELECT tt1.SId
    ,tt2.Sname
    ,tt1.CId
    ,tt1.score
    ,tt1.rn
    FROM(
    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,@p := t1.CId
    ,if(@[email protected],@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.rn<=3 AND tt1.SId = tt2.SId
    ORDER BY tt1.CId,rn;

24.查询每门课程被选修的学生数

分析：
以CIdGrouping query out each course is the number of students who take

  • 以CIdGrouping query out each course is the number of students who take

    SELECT CId
    ,COUNT(CId)
    AS cnt
    FROM SC
    GROUP BY CId;

25.查询出只选修两门课程的学生学号和姓名

分析：
1、First query of students of two courses have courses and student id
2、Then the table as at1表与Studen表t2Associated to take out the student's name,找出结果

  • First query of students of two courses have courses and student id

    SELECT SId
    ,COUNT(CId) AS cnt
    FROM SC
    GROUP BY SId
    HAVING cnt = 2;

  • Then the table as at1表与Studen表t2Associated to take out the student's name,找出结果

    SELECT t1.SId
    ,t2.Sname
    FROM(
    SELECT SId
    ,COUNT(CId) AS cnt
    FROM SC
    GROUP BY SId
    HAVING cnt = 2
    ) t1 JOIN Student t2
    ON t1.SId = t2.SId;

26.查询男生、女生人数

分析：
Gender group respectively for the boys、女生的人数

  • Gender group respectively for the boys、女生的人数

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

27.查询名字中含有「风」字的学生信息

分析：
使用 LIKE Fuzzy query name contains「风」字的学生信息

  • 使用 LIKE Fuzzy query name contains「风」字的学生信息

    SELECT * FROM Student WHERE Sname LIKE ‘%风%’;

Through observation found that there is no name phoenix word student information is contained in

28.查询同名同性学生名单,并统计同名同性人数

分析：
To the student's name,性别分组,然后countThe number of the same name gay

  • To the student's name,性别分组,然后countThe number of the same name gay

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

29.查询 1990 年出生的学生名单

分析：该题有两种解法
1、使用LIKE模糊查询
2、使用BETWEEN…AND…查询

  • 使用LIKE模糊查询

    SELECT * FROM Student
    WHERE Sage
    LIKE ‘1990%’;

  • 使用BETWEEN…AND…查询

    SELECT * FROM Student
    WHERE Sage
    BETWEEN ‘1990-1-1’ AND ‘1990-12-31’;

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：
1、将SC表与CourseAssociated query table do scored an average of each course
2、Finally according to the above-mentioned will average descending order,平均成绩相同时,按课程编号升序排列

  • 将SC表与CourseAssociated query table do scored an average of each course

    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;

  • Finally according to the above-mentioned will average descending order,平均成绩相同时,按课程编号升序排列

    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;

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、First query result is greater than or equal to85The average grades of course information
2、Then the table as at1表与Studen表t2Link to take out the student information

  • First query result is greater than or equal to85The average grades of course information

    SELECT SId
    ,ROUND(AVG(score),2) AS avg_score
    FROM SC
    GROUP BY SId
    HAVING avg_score>=85;

  • Then the table as at1表与Studen表t2Link to take out the student information

    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;

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、First of all query a course called mathCId
2、Then the above-mentioned as word table query a course called mathematics,且分数低于60的课程信息
3、Finally the table as tablet1与Student表t2Link to take out the student information

  • First of all query a course called mathCId

    SELECT CId
    FROM Course
    WHERE CName = ‘数学’;

  • Then the above-mentioned as word table query a course called mathematics,且分数低于60的课程信息

    SELECT SId
    ,Score
    FROM SC
    WHERE CId =(
    SELECT CId
    FROM Course
    WHERE CName = ‘数学’
    )AND score<60;

  • Finally the table as tablet1与Student表t2Link to take out the student information

    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;

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、First query out all course grades,并以SId分组（存在学生没成绩,The situation of the course for0分）
2、然后将Student表t1With the above as a tablet2做左连接,取出学生姓名

  • First query out all course grades,并以SId分组（存在学生没成绩,The situation of the course for0分）

    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;

  • 然后将Student表t1With the above as a tablet2做左连接,取出学生姓名

    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;

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、First query the result in70分以上的课程信息
2、Then the table as at1表与Student表t2关联,再与Course表t3Be associated remove any course grades in 70 分以上的姓名、Course name and score information

  • First query the result in70分以上的课程信息

    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE score>70;

  • Then the table as at1表与Student表t2关联,再与Course表t3Be associated remove any course grades in 70 分以上的姓名、Course name and score information

    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;

35.查询不及格的课程

分析：
1、First query the results below60分的课程信息
2、Then the table as a tablet1与Student表t2表做连接,再与Course表t3做连接,Failed to find students、课程编号、课程名称、成绩

  • First query the results below60分的课程信息

    SELECT DISTINCT CId
    ,score
    FROM SC
    WHERE score < 60;

*Then the table as a tablet1与Student表t2表做连接,再与Course表t3做连接,Failed to find students、课程编号、课程名称、成绩

SELECT  DISTINCT 
        t2.Sname
        ,t1.CId
        ,t3.CName
        ,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;

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、The query out the Numbers for the course01且课程成绩在80分以上的课程信息
2、Then the table as a tablet1与Student表t2做关联,Take out the student's name and student id

  • The query out the Numbers for the course01且课程成绩在80分以上的课程信息

    SELECT SId
    ,score
    FROM SC
    WHERE CId = ‘01’
    AND score >= 80;

  • Then the table as a tablet1与Student表t2做关联,Take out the student's name and student id

    SELECT t1.SId
    ,t2.Sname
    FROM(
    SELECT SId
    ,score
    FROM SC
    WHERE CId = ‘01’
    AND score >= 80
    ) t1 JOIN Student t2
    ON t1.SId = t2.SId;

37.求每门课程的学生人数

  • 将SC表t1与Course表t2做关联,并按CId分组,Take out the course number、课程名称、每门课程

    SELECT t1.CId
    ,t2.cname
    ,COUNT(t1.CId) AS ‘人数’
    FROM SC t1 JOIN Course t2
    ON t1.CId = t2.CId
    GROUP BY t1.CId,t2.Cname;

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、First query zhang teacher taught coursesTId
2、然后在CourseIn the table number query zhang teacher taught coursesCId
3、再在SC表查询,Grades don't repeat course information
4、And then will result descending order take out the first resultLIMIT 1,Is the highest grade
5、The final table ast1表与Student表t2连接,Remove the top student information

  • First query zhang teacher taught coursesTId

    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’;

  • 然后在CourseIn the table number query zhang teacher taught coursesCId

    SELECT CId
    FROM Course
    WHERE TId = (
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    );

  • 再在SC表查询,Grades don't repeat course information

    SELECT SId
    ,CId
    ,score
    FROM SC
    WHERE CId IN (
    SELECT CId
    FROM Course
    WHERE TId = (
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    )
    );

  • And then will result descending order take out the first resultLIMIT 1,Is the highest grade

    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
    LIMIT 1;

  • The final table ast1表与Student表t2连接,Remove the top student information

    SELECT t1.SId
    ,t2.Sname
    ,t1.CId
    ,t1.score
    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
    LIMIT 1) t1 JOIN Student t2
    ON t1.SId = t2.SId;

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

分析：
1、First query zhang teacher taught coursesTId
2、然后在CourseIn the table number query zhang teacher taught coursesCId
3、再在SC表查询,Grades don't repeat course information,并降序排列
4、The table as a tablet1再定义@i、@p、@q三个变量@i控制排序,@p、@qControl scores repeat reserved place
5、Then the table as att1表与Student表tt2做连接,Take out the highest score,Information and students

  • First query zhang teacher taught coursesTId

    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’;

  • 然后在CourseIn the table number query zhang teacher taught coursesCId

    SELECT CId
    FROM Course
    WHERE TId = (
    SELECT TId
    FROM Teacher
    WHERE Tname = ‘张三’
    );

  • 再在SC表查询,Grades don't repeat course information,并降序排列

    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;

  • The table as a tablet1再定义@i、@p、@q三个变量@i控制排序,@p、@qControl scores repeat reserved place

    SET @i := 0;
    SET @p := 0;
    SET @q := 0;
    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,@p := t1.score
    ,IF(@[email protected],@i,@i := @i+1) AS dense_rank
    ,@q :[email protected]
    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;

  • Then the table as att1表与Student表tt2做连接,Take out the highest score,Information and students

    SET @i := 0;
    SET @p := 0;
    SET @q := 0;
    SELECT tt2.Sname
    ,tt1.SId
    ,tt1.CId
    ,tt1.score
    FROM(
    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,@p := t1.score
    ,IF(@[email protected],@i,@i := @i+1) AS dense_rank
    ,@q :[email protected]
    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;

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、首先现将SC表t1与自己SC表t2进行关联,Take out the result of the same student number、课程编号、学生成绩
2、然后再与Student表t3Correlation to take out the student information

  • 首先现将SC表t1与自己SC表t2进行关联,Take out the result of the same student number、课程编号、学生成绩

    SELECT t1.CId
    ,t2.score
    FROM SC t1 JOIN SC t2
    ON t1.score = t2.score
    AND t1.CId != t2.CId;

  • 然后再与Student表t3Correlation to take out the student information

    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;

41.查询每门课程成绩最好的前两名

分析：
1、首先在SCQuery the grades for each course in the table and pressCId分数降序排列
2、定义函数@i、@p、@q,According to the results obtained by each course sorting,And will the above-mentioned ast1表,然后再作为tt1表与Student表tt2To connect to take out the best top two each course grade

  • 首先在SCQuery the grades for each course in the table and pressCId分数降序排列

    SELECT SId
    ,CId
    ,score
    FROM SC
    ORDER BY CId,score DESC;

  • 定义函数@i、@p、@q,According to the results obtained by each course sorting,And will the above-mentioned ast1表,然后再作为tt1表与Student表tt2To connect to take out the best top two each course grade

    SET @i := 0;
    SET @p := 0;
    SET @q := 0;
    SELECT tt1.SId
    ,tt2.SName
    ,tt1.CId
    ,tt1.score
    ,tt1.rn
    FROM (
    SELECT t1.SId
    ,t1.CId
    ,t1.score
    ,@p := t1.CId
    ,IF(@[email protected],@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.rn<=2 AND tt1.SId = tt2.SId
    ORDER BY tt1.CId,rn;

42.统计每门课程的学生选修人数（超过 5 人的课程才统计）.

分析：
1、以CId进行分组,统计出超过5人的课程信息

  • 以CId进行分组,统计出超过5人的课程信息

    SELECT CId
    ,COUNT(SId) AS ‘选课人数’
    FROM SC
    GROUP BY CId
    HAVING 选课人数 > 5;

43.检索至少选修两门课程的学生学号

分析：
1、以SId进行分组,Statistics of at least two courses students student id

  • 以SId进行分组,Statistics of at least two courses students student id

    SELECT SId
    ,COUNT(CID) AS ‘选课数量’
    FROM SC
    GROUP BY SId
    HAVING 选课数量 >= 2;

44.查询选修了全部课程的学生信息

分析：
1、First query the total course attributes as tablet3
2、然后将Student表t1与SC表t2进行关联,以SId、Sname进行分组,用表t3Selected students enrolled in the whole course information

  • First query the total course attributes as tablet3

    SELECT COUNT(t3.CId)
    FROM Course t3;

  • 然后将Student表t1与SC表t2进行关联,以SId、Sname进行分组,用表t3Selected students enrolled in the whole course information

    SELECT t1.SId
    ,t1.Sname
    ,COUNT(t2.CId) AS ‘选课数量’
    FROM Student t1 JOIN SC t2
    ON t2.SId = t1.SId
    GROUP BY t1.SId,t1.Sname
    HAVING COUNT(t2.CId) = (
    SELECT COUNT(t3.CId)
    FROM Course t3
    );

45.查询各学生的年龄,只按年份来算

分析：Get the current date minus the students date of birth

SELECT YEAR(now()) - date_format(Sage,'%Y') FROM Student;

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、首先先在StudentQuery the date of birth in the table、当前月日、出生年月的月日
2、Then the table as a tablet1,Calculate the current date < 出生年月的月日,则年龄减一

  • 首先先在StudentQuery the date of birth in the table、当前月日、出生年月的月日

    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;

  • Then the table as a tablet1,Calculate the current date < 出生年月的月日,则年龄减一

    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;

47.查询本周过生日的学生

分析：Query current weeks and weeks in the table are equal to

SELECT SId
       ,SName
       ,Sage
       ,WEEK(Sage)
FROM Student
WHERE WEEK(Sage) = WEEK(now());

Nobody's birthday this week so there is no data

48.查询下周过生日的学生

分析：Query the current week+1And in the table weeks whether is equal to

SELECT SId
       ,SName
       ,Sage
       ,WEEK(Sage)
FROM Student
WHERE WEEK(Sage) = WEEK(date_add(now(),INTERVAL 1 WEEK));

Nobody's birthday next week so there is no data

49.查询本月过生日的学生

分析：Query whether the current month and the month in the table number is equal to

SELECT SId
       ,SName
       ,Sage
       ,MONTH(Sage)
FROM Student
WHERE MONTH(Sage) = MONTH(now());

50.查询下月过生日的学生

分析：Query the current month+1With the number in the table are equal to

SELECT SId
       ,SName
       ,Sage
       ,MONTH(Sage)
FROM Student
WHERE MONTH(Sage) = MONTH(date_add(now(),INTERVAL 1 MONTH));

Nobody in January birthday so there is no data

week()Function to refer to the following url

https://blog.csdn.net/moakun/article/details/82528773

date_format()Function to refer to the following url

https://www.w3school.com.cn/sql/func_date_format.asp

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