SQL练习50题

1 创建表

学生表:

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

mysql> select * from Student;
+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   |
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |
| 03   | 孙风   | 1990-12-20 00:00:00 | 男   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |
| 05   | 周梅   | 1991-12-01 00:00:00 | 女   |
| 06   | 吴兰   | 1992-01-01 00:00:00 | 女   |
| 07   | 郑竹   | 1989-01-01 00:00:00 | 女   |
| 09   | 张三   | 2017-12-20 00:00:00 | 女   |
| 10   | 李四   | 2017-12-25 00:00:00 | 女   |
| 11   | 李四   | 2012-06-06 00:00:00 | 女   |
| 12   | 赵六   | 2013-06-13 00:00:00 | 女   |
| 13   | 孙七   | 2014-06-01 00:00:00 | 女   |
+------+--------+---------------------+------+
 

科目表:

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

mysql> select * from Course;
+------+--------+------+
| CId  | Cname  | TId  |
+------+--------+------+
| 01   | 语文   | 02   |
| 02   | 数学   | 01   |
| 03   | 英语   | 03   |
+------+--------+------+
3 rows in set (0.00 sec)

教师表:

create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

mysql> select * from Teacher;
+------+--------+
| TId  | Tname  |
+------+--------+
| 01   | 张三   |
| 02   | 李四   |
| 03   | 王五   |
+------+--------+
 

成绩表:

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

mysql> select * from SC;
+------+------+-------+
| SId  | CId  | score |
+------+------+-------+
| 01   | 01   |  80.0 |
| 01   | 02   |  90.0 |
| 01   | 03   |  99.0 |
| 02   | 01   |  70.0 |
| 02   | 02   |  60.0 |
| 02   | 03   |  80.0 |
| 03   | 01   |  80.0 |
| 03   | 02   |  80.0 |
| 03   | 03   |  80.0 |
| 04   | 01   |  50.0 |
| 04   | 02   |  30.0 |
| 04   | 03   |  20.0 |
| 05   | 01   |  76.0 |
| 05   | 02   |  87.0 |
| 06   | 01   |  31.0 |
| 06   | 03   |  34.0 |
| 07   | 02   |  89.0 |
| 07   | 03   |  98.0 |
+------+------+-------+

2 题目练习:

  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 s.*, sc1.score as score1, sc2.score as score2
from student s
join sc sc1 on s.sid = sc1.sid and sc1.cid = '01'
join sc sc2 on s.sid = sc2.sid and sc2.cid = '02'
where sc1.score > sc2.score;

 2.查询同时存在" 01 "课程和" 02 "课程的情况

SELECT s.*
FROM Student s
JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01'
JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02';

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 )

SELECT s.*
FROM Student s
JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02'
WHERE NOT EXISTS (
    SELECT 1 FROM SC WHERE SId = s.SId AND CId = '01'
);


4  查询不存在" 01 "课程但存在" 02 "课程的情况

SELECT s.*
FROM Student s
JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02'
WHERE NOT EXISTS (
    SELECT 1 FROM SC WHERE SId = s.SId AND CId = '01'
);



5 查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

SELECT s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING AVG(sc.score) >= 60;
查询在 SC 表存在成绩的学生信息
SELECT DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId;



6 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 )

SELECT s.SId, s.Sname, 
       COUNT(sc.CId) AS course_count,
       SUM(sc.score) AS total_score
FROM Student s
LEFT JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname;



7 查有成绩的学生信息

SELECT DISTINCT s.*
FROM Student s
WHERE EXISTS (SELECT 1 FROM SC WHERE SId = s.SId);


 8 查询「李」姓老师的数量

SELECT COUNT(*) AS li_teachers
FROM Teacher
WHERE Tname LIKE '李%';


9 查询学过「张三」老师授课的同学的信息


 

SELECT DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三';


10 查询没有学全所有课程的同学的信息


 

SELECT s.*
FROM Student s
WHERE s.SId NOT IN (
    SELECT SId 
    FROM SC 
    GROUP BY SId 
    HAVING COUNT(DISTINCT CId) = (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


SELECT DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId IN (
    SELECT CId FROM SC WHERE SId = '01'
) AND s.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


 

SELECT s.*
FROM Student s
WHERE s.SId IN (
    SELECT SId FROM SC WHERE SId != '01'
    GROUP BY SId
    HAVING GROUP_CONCAT(CId ORDER BY CId) = (
        SELECT GROUP_CONCAT(CId ORDER BY CId) 
        FROM SC 
        WHERE SId = '01'
    )
);


13 查询没学过"张三"老师讲授的任一门课程的学生姓名


 

SELECT s.*
FROM Student s
WHERE s.SId NOT IN (
    SELECT DISTINCT sc.SId
    FROM SC sc
    JOIN Course c ON sc.CId = c.CId
    JOIN Teacher t ON c.TId = t.TId
    WHERE t.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


 


SELECT s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.score < 60
GROUP BY s.SId, s.Sname
HAVING COUNT(*) >= 2;


15 检索" 01 "课程分数小于 60,按分数降序排列的学生信息


 

SELECT s.*, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId = '01' AND sc.score < 60
ORDER BY sc.score DESC;


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


 

SELECT s.SId, s.Sname, sc.CId, sc.score, 
       (SELECT AVG(score) FROM SC WHERE SId = s.SId) AS avg_score
FROM Student s
LEFT JOIN SC sc ON s.SId = sc.SId
ORDER BY avg_score DESC;


17 查询各科成绩最高分、最低分和平均分:


 

SELECT 
    c.CId, 
    c.Cname, 
    MAX(sc.score) AS max_score,
    MIN(sc.score) AS min_score,
    AVG(sc.score) AS avg_score,
    COUNT(sc.SId) AS student_count,
    CONCAT(ROUND(SUM(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS pass_rate,
    CONCAT(ROUND(SUM(CASE WHEN sc.score >= 70 AND sc.score < 80 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS medium_rate,
    CONCAT(ROUND(SUM(CASE WHEN sc.score >= 80 AND sc.score < 90 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS good_rate,
    CONCAT(ROUND(SUM(CASE WHEN sc.score >= 90 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS excellent_rate
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
ORDER BY student_count DESC, c.CId ASC;



18 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


 

SELECT 
    sc.CId, 
    sc.SId, 
    sc.score,
    RANK() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank
FROM SC sc
ORDER BY sc.CId, rank;


19 按各科成绩进行排序,并显示排名, Score 重复时保留名次空缺



 

SELECT 
    sc.CId, 
    sc.SId, 
    sc.score,
    DENSE_RANK() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank
FROM SC sc
ORDER BY sc.CId, rank;


20 按各科成绩进行排序,并显示排名, Score 重复时合并名次


 

SELECT 
    s.SId, 
    s.Sname, 
    SUM(sc.score) AS total_score,
    RANK() OVER (ORDER BY SUM(sc.score) DESC) AS rank
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
ORDER BY rank;


21 查询学生的总成绩,并进行排名,总分重复时保留名次空缺


 

SELECT 
    s.SId, 
    s.Sname, 
    SUM(sc.score) AS total_score,
    DENSE_RANK() OVER (ORDER BY SUM(sc.score) DESC) AS rank
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
ORDER BY rank;


22 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


 

SELECT 
    s.SId, 
    s.Sname, 
    SUM(sc.score) AS total_score,
    DENSE_RANK() OVER (ORDER BY SUM(sc.score) DESC) AS rank
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
ORDER BY rank;


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],[60-0] 及所占百分比

SELECT 
    c.CId,
    c.Cname,
    SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS '[85-100]',
    CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS '[85-100]%',
    SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS '[70-85]',
    CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS '[70-85]%',
    SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS '[60-70]',
    CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS '[60-70]%',
    SUM(CASE WHEN sc.score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) AS '[0-60]',
    CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) / COUNT(*) * 100, 2), '%') AS '[0-60]%'
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;


24 查询各科成绩前三名的记录

SELECT *
FROM (
    SELECT 
        sc.CId, 
        sc.SId, 
        sc.score,
        RANK() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank
    FROM SC sc
) t
WHERE t.rank <= 3;


25 查询每门课程被选修的学生数


 

SELECT c.CId, c.Cname, COUNT(DISTINCT sc.SId) AS student_count
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;


26 查询出只选修两门课程的学生学号和姓名


 

SELECT s.SId, s.Sname
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING COUNT(*) = 2;


27 查询男生、女生人数


 

SELECT Ssex, COUNT(*) AS count
FROM Student
GROUP BY Ssex;


28 查询名字中含有「风」字的学生信息


 

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


29 查询同名同性学生名单,并统计同名人数


 

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


30 查询 1990 年出生的学生名单


SELECT *
FROM Student
WHERE YEAR(Sage) = 1990;


31 查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列


 

SELECT c.CId, c.Cname, AVG(sc.score) AS avg_score
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
ORDER BY avg_score DESC, c.CId ASC;


32 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 s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING AVG(sc.score) >= 85;


33 查询课程名称为「数学」,且分数低于 60 的学生姓名和分数


 

SELECT s.Sname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
WHERE c.Cname = '数学' AND sc.score < 60;


34 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)


 

SELECT s.SId, s.Sname, c.Cname, sc.score
FROM Student s
LEFT JOIN SC sc ON s.SId = sc.SId
LEFT JOIN Course c ON sc.CId = c.CId;


35 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 DISTINCT s.Sname, c.Cname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
WHERE sc.score > 70;


36 查询不及格的课程


 

SELECT s.Sname, c.Cname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
WHERE sc.score < 60;


37 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 s.SId, s.Sname
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId = '01' AND sc.score > 80;


38 求每门课程的学生人数


 

SELECT c.CId, c.Cname, COUNT(sc.SId) AS student_count
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;


39 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

SELECT s.*, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三'
ORDER BY sc.score DESC
LIMIT 1;


40 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

SELECT s.*, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三' AND sc.score = (
    SELECT MAX(score) 
    FROM SC 
    JOIN Course ON SC.CId = Course.CId
    JOIN Teacher ON Course.TId = Teacher.TId
    WHERE Teacher.Tname = '张三'
);


41 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩


 

SELECT DISTINCT a.SId, a.CId, a.score
FROM SC a
JOIN SC b ON a.SId = b.SId AND a.CId != b.CId AND a.score = b.score;


 


42 查询每门功成绩最好的前两名


 

SELECT *
FROM (
    SELECT 
        sc.CId, 
        sc.SId, 
        sc.score,
        RANK() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank
    FROM SC sc
) t
WHERE t.rank <= 2;
————————————————

                            版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
                        
原文链接:https://blog.youkuaiyun.com/2202_75594192/article/details/147042195


43 统计每门课程的学生选修人数(超过 5 人的课程才统计)。


 

SELECT c.CId, c.Cname, COUNT(sc.SId) AS student_count
FROM Course c
JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
HAVING COUNT(sc.SId) > 5;


44 检索至少选修两门课程的学生学号

SELECT SId
FROM SC
GROUP BY SId
HAVING COUNT(DISTINCT CId) >= 2;


45 查询选修了全部课程的学生信息


 

SELECT s.*
FROM Student s
WHERE NOT EXISTS (
    SELECT c.CId FROM Course c
    WHERE NOT EXISTS (
        SELECT 1 FROM SC WHERE SC.SId = s.SId AND SC.CId = c.CId
    )
);


46 查询各学生的年龄,只按年份来算


 

SELECT SId, Sname, YEAR(CURRENT_DATE) - YEAR(Sage) AS age
FROM Student;


47 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一

SELECT 
    SId, 
    Sname, 
    CASE 
        WHEN MONTH(CURRENT_DATE) < MONTH(Sage) OR 
             (MONTH(CURRENT_DATE) = MONTH(Sage) AND DAY(CURRENT_DATE) < DAY(Sage))
        THEN YEAR(CURRENT_DATE) - YEAR(Sage) - 1
        ELSE YEAR(CURRENT_DATE) - YEAR(Sage)
    END AS age
FROM Student;


48 查询本周过生日的学生


 

SELECT *
FROM Student
WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(CURRENT_DATE);


49 查询下周过生日的学生


 

SELECT *
FROM Student
WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(CURRENT_DATE) + 1;


50 查询本月过生日的学生


 

SELECT *
FROM Student
WHERE MONTH(Sage) = MONTH(CURRENT_DATE);


51 查询下月过生日的学生


SELECT *
FROM Student
WHERE MONTH(Sage) = MONTH(CURRENT_DATE) + 1;

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