mysql题51Cto博客

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mob6454cc6e8f43 2024-09-17 15:16:01

文章标签 mysql题51Cto博客 Code Sage 升序 文章分类 MySQL 数据库

网上流传的50道sql练习，发现其答案部分有问题，故在此记录下个人用mysql 的实现。

数据表介绍

--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

--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 学生编号,CId 课程编号,score 分数

学生表 Student

mysql题51Cto博客_Code

mysql题51Cto博客_Code_02

-- ----------------------------
-- Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `SId` varchar(10) DEFAULT NULL,
  `Sname` varchar(10) DEFAULT NULL,
  `Sage` datetime DEFAULT NULL,
  `Ssex` varchar(10) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES ('01', '赵雷', '1990-05-18 00:00:00', '男');
INSERT INTO `student` VALUES ('02', '钱电', '1990-05-24 00:00:00', '男');
INSERT INTO `student` VALUES ('03', '孙风', '1990-05-20 00:00:00', '男');
INSERT INTO `student` VALUES ('04', '李云', '1990-05-25 00:00:00', '男');
INSERT INTO `student` VALUES ('05', '周梅', '1991-12-01 00:00:00', '女');
INSERT INTO `student` VALUES ('06', '吴兰', '1992-01-01 00:00:00', '女');
INSERT INTO `student` VALUES ('07', '郑竹', '1989-01-01 00:00:00', '女');
INSERT INTO `student` VALUES ('09', '张三', '2017-12-20 00:00:00', '女');
INSERT INTO `student` VALUES ('10', '李四', '2017-12-25 00:00:00', '女');
INSERT INTO `student` VALUES ('11', '李四', '2012-06-06 00:00:00', '女');
INSERT INTO `student` VALUES ('12', '赵六', '2013-06-13 00:00:00', '女');
INSERT INTO `student` VALUES ('13', '孙七', '2014-06-01 00:00:00', '女');

View Code

科目表 Course

mysql题51Cto博客_Code

mysql题51Cto博客_Code_02

-- ----------------------------
-- Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `CId` varchar(10) DEFAULT NULL,
  `Cname` varchar(10) DEFAULT NULL,
  `TId` varchar(10) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of 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');

View Code

教师表 Teacher

mysql题51Cto博客_Code

mysql题51Cto博客_Code_02

-- ----------------------------
-- Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `TId` varchar(10) DEFAULT NULL,
  `Tname` varchar(10) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '张三');
INSERT INTO `teacher` VALUES ('02', '李四');
INSERT INTO `teacher` VALUES ('03', '王五');

View Code

成绩表 SC

mysql题51Cto博客_Code

mysql题51Cto博客_Code_02

-- ----------------------------
-- Table structure for `sc`
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc` (
  `SId` varchar(10) DEFAULT NULL,
  `CId` varchar(10) DEFAULT NULL,
  `score` decimal(18,1) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES ('01', '02', '90.0');
INSERT INTO `sc` VALUES ('01', '01', '80.0');
INSERT INTO `sc` VALUES ('01', '03', '99.0');
INSERT INTO `sc` VALUES ('02', '02', '60.0');
INSERT INTO `sc` VALUES ('02', '01', '70.0');
INSERT INTO `sc` VALUES ('02', '03', '80.0');
INSERT INTO `sc` VALUES ('03', '01', '80.0');
INSERT INTO `sc` VALUES ('03', '02', '80.0');
INSERT INTO `sc` VALUES ('03', '03', '80.0');
INSERT INTO `sc` VALUES ('04', '01', '50.0');
INSERT INTO `sc` VALUES ('04', '02', '30.0');
INSERT INTO `sc` VALUES ('04', '03', '20.0');
INSERT INTO `sc` VALUES ('05', '01', '76.0');
INSERT INTO `sc` VALUES ('05', '02', '87.0');
INSERT INTO `sc` VALUES ('06', '01', '31.0');
INSERT INTO `sc` VALUES ('06', '03', '34.0');
INSERT INTO `sc` VALUES ('07', '02', '90.0');
INSERT INTO `sc` VALUES ('07', '03', '98.0');

View Code

下面是题目和mysql实现

-- 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.*, class1, class2 FROM student st JOIN 
(SELECT * FROM 
(SELECT SId s1, score class1 FROM sc WHERE CId = '01') c1, 
(SELECT SId s2, score class2 FROM sc WHERE CId = '02') c2
WHERE s1 = s2 AND class1 > class2) c 
ON st.SId = c.s1;

-- 1.1 查询同时存在" 01 "课程和" 02 "课程的情况
SELECT * FROM
((SELECT * FROM sc WHERE CId = '01') t1
JOIN
(SELECT * FROM sc WHERE CId = '02') t2
ON
t1.Sid = t2.Sid);

-- 1.2 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 * FROM
((SELECT * FROM sc WHERE CId = '01') t1
left JOIN
(SELECT * FROM sc WHERE CId = '02') t2
ON
t1.Sid = t2.Sid);

-- 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
select * from sc 
where sc.CId = '02' AND 
sc.SId NOT IN
(SELECT SId FROM sc WHERE sc.CId = '01');

-- 2.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 st.SId, Sname, AVG(score) 'avg' FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 60;

-- 3.查询在 SC 表存在成绩的学生信息
SELECT * FROM student WHERE SId IN
(SELECT DISTINCT SId FROM sc);

-- 4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和
SELECT st.sid, st.Sname, COUNT(sc.CId) 'nums', SUM(sc.score) 'sum' FROM
student st JOIN sc on st.SId = sc.SId
GROUP BY st.SId;

-- 4.1显示没选课的学生(显示为NULL)
SELECT st.sid, st.Sname, CASE WHEN COUNT(sc.SId) > 0 THEN COUNT(sc.SId) ELSE NULL END 'nums', SUM(sc.score) 'sum' FROM
student st left JOIN sc on st.SId = sc.SId
GROUP BY st.SId;

-- 4.2查有成绩的学生信息
SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); -- 适用于右表小
SELECT * FROM student st WHERE EXISTS (SELECT * FROM sc WHERE sc.SId = st.SId); -- 适用于右表大


-- 5.查询「李」姓老师的数量
SELECT COUNT(*) FROM teacher t WHERE t.Tname LIKE '李%';

-- 6.查询学过「张三」老师授课的同学的信息
SELECT * FROM student st WHERE st.SId IN (
SELECT SId from sc WHERE sc.CId = (
SELECT CId FROM course WHERE course.TId = (
SELECT TId FROM teacher WHERE teacher.Tname = '张三')));

-- 7.查询没有学全所有课程的同学的信息
SELECT * FROM student WHERE SId NOT IN(
SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course));

-- 8.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 * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01'));

-- 9.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
-- 解法一
SELECT * FROM student WHERE SId IN
(SELECT SId FROM sc GROUP BY SId HAVING SId <> '01' AND GROUP_CONCAT(CId ORDER BY CId) = 
(SELECT GROUP_CONCAT(CId ORDER BY CId) FROM sc WHERE SId = '01'));
-- 解法二
SELECT * FROM student WHERE SId IN
(SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01') AND SId <> '01' 
GROUP BY SId 
HAVING COUNT(*) = (SELECT COUNT(*) FROM sc WHERE SId = '01'));

-- 10.查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT student.Sname FROM student WHERE SId NOT IN
(SELECT SId FROM sc WHERE CId in 
(SELECT course.CId FROM course WHERE TId =
(SELECT teacher.TId FROM teacher WHERE tname = '张三')));

-- 11.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 st.SId, sname, avg(score) FROM student st JOIN sc ON st.sid = sc.SId GROUP BY st.SId 
HAVING COUNT(score <= 60 OR NULL) >= 2;
-- 错误解法：
select student.sid, student.sname, AVG(sc.score) from student,sc
where 
    student.sid = sc.sid and sc.score< 60
group by sc.sid 
having count(*)>1;

-- 12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息
SELECT st.*, score
    FROM student st JOIN sc 
        ON st.SId = sc.SId AND sc.CId = '01' AND sc.score <  60
            ORDER BY score DESC;

-- 13.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 * from sc JOIN 
(SELECT SId, avg(score) avg FROM sc GROUP BY SId) avgs
ON sc.SId = avgs.SId
ORDER BY avgs.avg desc;

-- 14.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
SELECT 
sc.CId '课程 ID',
count(*) '选修人数',
course.Cname '课程 name',
MAX(score) '最高分', 
MIN(score) '最低分', 
AVG(score) '平均分',
COUNT(score >= 60 OR NULL) / COUNT(*) '及格率',
COUNT(score >= 70 AND score < 80 OR NULL) / COUNT(*) '中等率',
COUNT(score >= 80 AND score < 90 OR NULL) / COUNT(*) '优良率',
COUNT(score >= 90 OR NULL) / COUNT(*) '优秀率'
FROM sc join course ON sc.CId = course.CId GROUP BY sc.CId
ORDER BY COUNT(*) desc, sc.CId;

-- 15.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 CId, SId, score , (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 rank
FROM sc o ORDER BY CId, score DESC;

-- 16.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 @rank = 0;
SELECT q.sid, q.sum, (@rank := @rank + 1) rank FROM
(SELECT sid, sum(score) sum  FROM sc GROUP BY SId ORDER BY sum) q;

-- 17.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 s.CId, c.Cname,
CONCAT(COUNT(score >= 85 OR NULL), ', ',COUNT(score >= 85 OR NULL) / COUNT(*)) '[100-85]',
COUNT(score < 85 AND score >= 70 OR NULL) '[85-70]',
COUNT(score < 70 AND score >= 60 OR NULL) '[70-60]',
COUNT(score < 60 OR NULL) '[60-0]'
FROM sc s JOIN course c ON s.CId = c.CId GROUP BY s.CId;

-- 18. 查询各科成绩前三名的记录
SELECT o.* FROM sc o HAVING 
(SELECT COUNT(*) FROM sc WHERE o.CId = CId AND o.score < score) + 1 <= 3 
ORDER BY CId, score DESC, SId;

-- 19.查询每门课程被选修的学生数
SELECT CId, COUNT(*) FROM sc GROUP BY CId;

-- 20.查询出只选修两门课程的学生学号和姓名
SELECT st.SId, st.Sname
FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING COUNT(*) = 2;

-- 21.查询男生、女生人数
select ssex, count(*) from student
group by ssex;

-- 22.查询名字中含有「风」字的学生信息
SELECT * FROM student WHERE Sname LIKE '%风%';

-- 23.查询同名学生名单，并统计同名人数
SELECT st.*, (SELECT COUNT(*) FROM student WHERE Sname = st.Sname) 'same' FROM student st WHERE st.SId IN
(SELECT s1.SId FROM student s1 JOIN student s2 ON s1.SId != s2.SId AND s1.Sname = s2.Sname);

-- 24.查询 1990 年出生的学生名单
SELECT * FROM student WHERE YEAR(Sage) = 1990;

-- 25.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 CId, avg(score) avg
FROM sc GROUP BY CId ORDER BY avg DESC, CId;

-- 26.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.SId, Sname, AVG(score) avg
FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 85;

-- 27.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.Sname, s.score
FROM student st, sc s, course c
WHERE st.SId = s.SId AND s.CId = c.CId AND s.score < 60 AND c.Cname = '数学';

-- 28.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 Sname, st.SId, score 
FROM student st LEFT JOIN sc s ON st.SId = s.SId;

-- 29.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
-- 这个我没太理解，理解一是任意一门成绩均在70分以上
SELECT st.Sname, c.Cname, s.score
FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN
(SELECT sid FROM sc GROUP BY SId HAVING MIN(score) > 70);
-- 理解二是存在一门成绩在70分以上即可满足条件
SELECT st.Sname, c.Cname, s.score
FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN
(SELECT sid FROM sc GROUP BY SId HAVING MAX(score) > 70);
-- 理解三就是找出所有大于70分的得分。
select student.sname, course.cname,sc.score from student,course,sc
where sc.score>70
and student.sid = sc.sid
and sc.cid = course.cid;
SELECT * FROM sc WHERE score > 70

-- 30.查询存在不及格的课程
SELECT cid FROM sc GROUP BY CId HAVING MIN(score) < 60;
select cid from sc
where score< 60
group by cid;
SELECT DISTINCT cid FROM sc WHERE score < 60;

-- 31.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名
SELECT sid, sname FROM student WHERE sid IN
(SELECT SId FROM sc WHERE CId = '01' AND score >= 80);

-- 32.求每门课程的学生人数
SELECT cid, count(*) FROM sc GROUP BY cid;

-- 33.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 st.*, score FROM student st JOIN sc s ON st.SId = s.SId AND s.CId =
(SELECT CId FROM course WHERE TId =
(SELECT TId FROM teacher WHERE Tname = '张三')) ORDER BY score DESC LIMIT 1;

-- 34.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 st.*, score, CId FROM student st JOIN sc s ON st.SId = s.SId AND s.CId = 
(SELECT CId FROM course WHERE TId =
(SELECT TId FROM teacher WHERE Tname = '张三'))
WHERE score = (SELECT max(score) FROM sc WHERE CId = s.CId);

-- 35.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
-- 这个问题其实一开始没太明白啥意思，后来理解为某个人的几科分数是一样的，需要把这个人找出来
select  a.cid, a.sid,  a.score from sc as a
inner join 
sc as b
on a.sid = b.sid
and a.cid != b.cid
and a.score = b.score
group by cid, sid;

-- 36.查询每门功成绩最好的前两名
SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 <= 2  ORDER BY o.cid, sid;

-- 37.统计每门课程的学生选修人数（超过 5 人的课程才统计）
SELECT CId, COUNT(*) sum FROM sc GROUP BY CId HAVING sum > 5;

-- 38.检索至少选修两门课程的学生学号
SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) >= 2;

-- 39.查询选修了全部课程的学生信息
SELECT * FROM student WHERE sid IN
(SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course)); 

-- 40.查询各学生的年龄，只按年份来算
SELECT SId, Sname, Sage, YEAR(NOW()) - YEAR(Sage)
FROM student;

-- 41.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 '学生编号', Sname '学生姓名', TIMESTAMPDIFF(YEAR, Sage, NOW()) '学生年龄'
FROM student;

-- 42.查询本周过生日的学生
-- 有点复杂，需要拼接出本周的起止日期
SELECT * FROM student 
WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage)))
BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) DAY)) 
AND DATE(DATE_ADD(NOW(),INTERVAL 6 - WEEKDAY(NOW()) DAY));
SELECT * FROM student 

-- 43. 查询下周过生日的学生
-- 同42
SELECT * FROM student 
WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage)))
BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) - 7 DAY)) 
AND DATE(DATE_ADD(NOW(),INTERVAL 13 - WEEKDAY(NOW()) DAY));

-- 44.查询本月过生日的学生
SELECT * FROM student WHERE month(sage) = month(NOW())

-- 45.查询下月过生日的学生
-- 注意本月是12月的话，下一个月份是1即可
SELECT * FROM student WHERE month(Sage) = (CASE WHEN month(NOW()) = 12 THEN 1 ELSE MONTH(NOW()) + 1 END);

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