**微信搜索公众号【C you again】，回复“SQL”下载无水印PDF版本，方便收藏** ## 1 数据表 #### 1.1 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 #### 1.2 课程表 Course（CId，Cname，TId） > CId 课程编号，Cname 课程名称，TId 教师编号 #### 1.3 教师表 Teacher（TId，Tname） > TId 教师编号，Tname 教师姓名 #### 1.4 成绩表 SC（SId，CId，score） > SId 学生编号，CId 课程编号，score 分数 ## 2 sql语句 #### 2.1 学生表 SQL ```sql 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'); ``` #### 2.2 科目表 SQL ```sql 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'); ``` #### 2.3 教师表 SQL ```sql 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'); ``` #### 2.4 成绩表 SQL ```sql 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 ); ``` ## 3 练习题 #### 3.1 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数。 因为需要全部的学生信息，则需要在 SC 表中得到符合条件的 SId 后与 Student 表进行 join，可以 left join 也可以 right join。 ```sql # right join SELECT * FROM Student RIGHT JOIN ( SELECT t1.SId, class1, class2 FROM ( SELECT SId, score AS class1 FROM sc WHERE sc.CId = '01' ) t1, ( SELECT SId, score AS class2 FROM sc WHERE sc.CId = '02' ) t2 WHERE t1.SId = t2.SId AND t1.class1 > t2.class2 ) r ON Student.SId = r.SId; ``` ```sql # left join SELECT * FROM ( SELECT t1.SId, class1, class2 FROM ( SELECT SId, score AS class1 FROM sc WHERE sc.CId = '01' ) t1, ( SELECT SId, score AS class2 FROM sc WHERE sc.CId = '02' ) t2 WHERE t1.SId = t2.SId AND t1.class1 > t2.class2 ) r LEFT JOIN Student ON Student.SId = r.SId; ``` #### 3.2 查询同时存在" 01 "课程和" 02 "课程的情况 ```sql SELECT * FROM ( SELECT * FROM sc WHERE sc.CId = '01' ) t1, ( SELECT * FROM sc WHERE sc.CId = '02' ) t2 WHERE t1.SId = t2.SId; ``` #### 3.3 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) 这一道需要使用 join 的情况了，" 02 " 课程可能不存在，即为 left join 的右侧或 right join 的左侧即可。 ```sql SELECT * FROM ( SELECT * FROM sc WHERE sc.CId = '01' ) t1 LEFT JOIN ( SELECT * FROM sc WHERE sc.CId = '02' ) t2 ON t1.SId = t2.SId; ``` ```sql SELECT * FROM ( SELECT * FROM sc WHERE sc.CId = '02' ) t2 RIGHT JOIN ( SELECT * FROM sc WHERE sc.CId = '01' ) t1 ON t1.SId = t2.SId; ``` #### 3.4 查询不存在" 01 "课程但存在" 02 "课程的情况 ```sql SELECT * FROM sc WHERE sc.SId NOT IN ( SELECT SId FROM sc WHERE sc.CId = '01' ) AND sc.CId = '02'; ``` #### 3.5 查询平均成绩大于等于 60 分的同学的学生编号、学生姓名、平均成绩。 这里只需根据学生 ID 把成绩分组，对分组中的 score 求平均值，最后在选取结果中 AVG 大于 60 的即可。 注意，必须要给计算得到的 AVG 结果一个 alias（AS ss），得到学生信息的时候既可以用 join 也可以用一般的联合搜索。 ```sql SELECT student.SId, sname, ss FROM student, ( SELECT SId, AVG(score) AS ss FROM sc GROUP BY SId HAVING AVG(score) > 60 ) r WHERE student.sid = r.sid; ``` ```sql SELECT Student.SId, Student.Sname, r.ss FROM Student RIGHT JOIN ( SELECT SId, AVG(score) AS ss FROM sc GROUP BY SId HAVING AVG(score) > 60 ) r ON Student.SId = r.SId; ``` ```sql SELECT s.SId, ss, Sname FROM ( SELECT SId, AVG(score) AS ss FROM sc GROUP BY SId HAVING AVG(score) > 60 ) r LEFT JOIN ( SELECT Student.SId, Student.Sname FROM Student ) s ON s.SId = r.SId; ``` #### 3.6 查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和联合查询没选课的学生 ```sql SELECT student.sid, student.sname, r.coursenumber, r.scoresum FROM student, ( SELECT sc.sid, sum(sc.score) AS scoresum, count(sc.cid) AS coursenumber FROM sc GROUP BY sc.sid ) r WHERE student.sid = r.sid; ``` 如要显示没选课的学生(显示为 NULL)，需要使用 join: ```sql SELECT s.sid, s.sname, r.coursenumber, r.scoresum FROM ( SELECT student.sid, student.sname FROM student ) s LEFT JOIN ( SELECT sc.sid, sum(sc.score) AS scoresum, count(sc.cid) AS coursenumber FROM sc GROUP BY sc.sid ) r ON s.sid = r.sid; ``` #### 3.7 查有成绩的学生信息 这一题涉及到 in 和 exists 的用法，在这种小表中，两种方法的效率都差不多，但是请参考SQL查询中 in 和 exists 的区别分析，当表 2 的记录数量非常大的时候，选用 exists 比 in 要高效很多。exists 用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值 True 或 False。 结论：IN()适合B表比A表数据小的情况 结论：EXISTS()适合B表比A表数据大的情况 ```sql SELECT * FROM student WHERE EXISTS ( SELECT sc.sid FROM sc WHERE student.sid = sc.sid ); ``` ```sql SELECT * FROM student WHERE student.sid IN ( SELECT sc.sid FROM sc ); ``` #### 3.8 查询「李」姓老师的数量 ```sql SELECT count(*) FROM teacher WHERE tname LIKE '李%'; ``` #### 3.9 查询学过「张三」老师授课的同学的信息，多表联合查询 ```sql SELECT student.* FROM student, teacher, course, sc WHERE student.sid = sc.sid AND course.cid = sc.cid AND course.tid = teacher.tid AND tname = '张三'; ``` #### 3.10 查询没有学全所有课程的同学的信息。 因为有学生什么课都没有选，反向思考，先查询选了所有课的学生，再选择这些人之外的学生。 ```sql SELECT * FROM student WHERE student.sid NOT IN ( SELECT sc.sid FROM sc GROUP BY sc.sid HAVING count(sc.cid) = ( SELECT count(cid) FROM course ) ); ``` #### 3.11 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 这个用联合查询也可以，但是逻辑不清楚，我觉得较为清楚的逻辑是这样的：从sc 表查询 01 同学的所有选课 cid -- 从 sc 表查询所有同学的 sid 如果其 cid 在前面的结果中 -- 从 student 表查询所有学生信息如果sid在前面的结果中 ```sql SELECT * FROM student WHERE student.sid IN ( SELECT sc.sid FROM sc WHERE sc.cid IN ( SELECT sc.cid FROM sc WHERE sc.sid = '01' ) ); ``` #### 3.12 查询没学过"张三"老师讲授的任一门课程的学生姓名 仍然还是嵌套，三层嵌套， 或者多表联合查询 ```sql SELECT * FROM student WHERE student.sid NOT IN ( SELECT sc.sid FROM sc WHERE sc.cid IN ( SELECT course.cid FROM course WHERE course.tid IN ( SELECT teacher.tid FROM teacher WHERE tname = '张三' ) ) ); ``` ```sql SELECT * FROM student WHERE student.sid NOT IN ( SELECT sc.sid FROM sc, course, teacher WHERE sc.cid = course.cid AND course.tid = teacher.tid AND teacher.tname = '张三' ); ``` #### 3.13 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 从 SC 表中选取 score 小于 60 的，并 group by sid，having count 大于 1 ```sql 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; ``` #### 3.14 检索" 01 "课程分数小于 60 ，按分数降序排列的学生信息 双表联合查询，在查询最后可以设置排序方式，语法为ORDERBY*****DESC\ASC; ```sql SELECT student.*, sc.score FROM student, sc WHERE student.sid = sc.sid AND sc.score < 60 AND cid = '01' ORDER BY sc.score DESC; ``` #### 3.15 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 ```sql SELECT * FROM sc LEFT JOIN ( SELECT sid, avg(score) AS avscore FROM sc GROUP BY sid ) r ON sc.sid = r.sid ORDER BY avscore DESC; ``` #### 3.16 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 ```sql SELECT sc.CId, max(sc.score) AS 最高分, min(sc.score) AS 最低分 , AVG(sc.score) AS 平均分, count(*) AS 选修人数 , sum(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / count(*) AS 及格率 , sum(CASE WHEN sc.score >= 70 AND sc.score < 80 THEN 1 ELSE 0 END) / count(*) AS 中等率 , sum(CASE WHEN sc.score >= 80 AND sc.score < 90 THEN 1 ELSE 0 END) / count(*) AS 优良率 , sum(CASE WHEN sc.score >= 90 THEN 1 ELSE 0 END) / count(*) AS 优秀率 FROM sc GROUP BY sc.CId ORDER BY count(*) DESC, sc.CId ASC ``` #### 3.17 按各科成绩进行排序，并显示排名， Score重复时保留名次空缺 这一道题可以用变量，但也有更为简单的方法，即自交（左交） 用 sc 中的 score 和自己进行对比，来计算“比当前分数高的分数有几个”。 ```sql SELECT a.cid, a.sid, a.score , count(b.score) + 1 AS rank FROM sc a LEFT JOIN sc b ON a.score < b.score AND a.cid = b.cid GROUP BY a.cid, a.sid, a.score ORDER BY a.cid, rank ASC; ``` #### 3.18 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 这里主要学习一下使用变量。在SQL里面变量用@来标识。 ```sql SET @crank = 0; SELECT q.sid, total, @crank := @crank + 1 AS rank FROM ( SELECT sc.sid, sum(sc.score) AS total FROM sc GROUP BY sc.sid ORDER BY total DESC ) q; ``` #### 3.19 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 group by 以后的查询结果无法使用别名，所以不要想着先单表 group by 计算出结果再从第二张表里添上课程信息，而应该先将两张表 join 在一起得到所有想要的属性再对这张总表进行统计计算。这里就不算百分比了，道理相同。 注意一下，用 case when 返回 1 以后的统计不是用 count 而是 sum ```sql SELECT course.cname, course.cid , sum(CASE WHEN sc.score <= 100 AND sc.score > 85 THEN 1 ELSE 0 END) AS "[100-85]" , sum(CASE WHEN sc.score <= 85 AND sc.score > 70 THEN 1 ELSE 0 END) AS "[85-70]" , sum(CASE WHEN sc.score <= 70 AND sc.score > 60 THEN 1 ELSE 0 END) AS "[70-60]" , sum(CASE WHEN sc.score <= 60 AND sc.score > 0 THEN 1 ELSE 0 END) AS "[60-0]" FROM sc LEFT JOIN course ON sc.cid = course.cid GROUP BY sc.cid; ``` #### 3.20 查询各科成绩前三名的记录 mysql 不能 group by了以后取 limit，所以不要想着讨巧了。思路有两种，第一种比较暴力，计算比自己分数大的记录有几条，如果小于 3 就 select，因为对前三名来说不会有 3 个及以上的分数比自己大了，最后再对所有 select 到的结果按照分数和课程编号排名即可。 ```sql SELECT * FROM sc WHERE ( SELECT count(*) FROM sc a WHERE sc.cid = a.cid AND sc.score < a.score ) < 3 ORDER BY cid ASC, sc.score DESC; ``` 第二种比较灵巧一些，用自身左交，但是有点难以理解。先用自己交自己，条件为a.cid=b.cidanda.score 1; ``` 嵌套查询列出同名的全部学生的信息 ```sql SELECT * FROM student WHERE sname IN ( SELECT sname FROM student GROUP BY sname HAVING count(*) > 1 ); ``` #### 3.26 查询 1990 年出生的学生名单 ```sql SELECT * FROM student WHERE YEAR(student.Sage) = 1990; ``` #### 3.27 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 ```sql SELECT sc.cid, course.cname, AVG(SC.SCORE) AS average FROM sc, course WHERE sc.cid = course.cid GROUP BY sc.cid ORDER BY average DESC, cid ASC; ``` #### 3.28 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 having 也可以用来截取结果表，在这里就先得到平均成绩总表，再截取 AVG 大于 85 的即可 ```sql SELECT student.sid, student.sname, AVG(sc.score) AS aver FROM student, sc WHERE student.sid = sc.sid GROUP BY sc.sid HAVING aver > 85; ``` #### 3.29 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数 ```sql SELECT student.sname, sc.score FROM student, sc, course WHERE student.sid = sc.sid AND course.cid = sc.cid AND course.cname = '数学' AND sc.score < 60; ``` #### 3.30 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） ```sql SELECT student.sname, cid, score FROM student LEFT JOIN sc ON student.sid = sc.sid; ``` #### 3.31 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 ```sql 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; ``` #### 3.32 查询存在不及格的课程 可以用 group by 来取唯一，也可以用 distinct ```sql SELECT cid FROM sc WHERE score < 60 GROUP BY cid; ``` ```sql SELECT DISTINCT sc.CId FROM sc WHERE sc.score < 60; ``` #### 3.33 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 ```sql SELECT student.sid, student.sname FROM student, sc WHERE cid = '01' AND score >= 80 AND student.sid = sc.sid; ``` #### 3.34 求每门课程的学生人数 ```sql SELECT sc.CId, count(*) AS 学生人数 FROM sc GROUP BY sc.CId; ``` #### 3.35 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 用 having max() 理论上也是对的，但是下面那种按分数排序然后取 limit 1 的更直观可靠 ```sql SELECT student.*, sc.score, sc.cid FROM student, teacher, course, sc WHERE teacher.tid = course.tid AND sc.sid = student.sid AND sc.cid = course.cid AND teacher.tname = '张三' HAVING max(sc.score); ``` ```sql SELECT student.*, sc.score, sc.cid FROM student, teacher, course, sc WHERE teacher.tid = course.tid AND sc.sid = student.sid AND sc.cid = course.cid AND teacher.tname = '张三' ORDER BY score DESC LIMIT 1; ``` #### 3.36 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 为了验证这一题，先修改原始数据 UPDATE sc SET score= 90 wheresid=" 07 " and cid=" 02 "; 这样张三老师教的 02 号课就有两个学生同时获得 90 的最高分了。 这道题的思路继续上一题，我们已经查询到了符合限定条件的最高分了，这个时候只用比较这张表，找到全部 score 等于这个最高分的记录就可，看起来有点繁复。 ```sql SELECT student.*, sc.score, sc.cid FROM student, teacher, course, sc WHERE teacher.tid = course.tid AND sc.sid = student.sid AND sc.cid = course.cid AND teacher.tname = '张三' AND sc.score = ( SELECT Max(sc.score) FROM sc, student, teacher, course WHERE teacher.tid = course.tid AND sc.sid = student.sid AND sc.cid = course.cid AND teacher.tname = '张三' ); ``` #### 3.37 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 同上，在这里用了 inner join 后会有概念是重复的记录：“ 01 课与 03 课”=“ 03 课与 01课”，所以这里取唯一可以直接用 group by ```sql SELECT a.cid, a.sid, a.score FROM sc a INNER JOIN sc b ON a.sid = b.sid AND a.cid != b.cid AND a.score = b.score GROUP BY cid, sid; ``` #### 3.38 查询每门功成绩最好的前两名 同上 19 题 ```sql SELECT a.sid, a.cid, a.score FROM sc a LEFT JOIN sc b ON a.cid = b.cid AND a.score < b.score GROUP BY a.cid, a.sid HAVING count(b.cid) < 2 ORDER BY a.cid; ``` #### 3.39 统计每门课程的学生选修人数（超过 5 人的课程才统计） ```sql SELECT sc.cid, count(sid) AS cc FROM sc GROUP BY cid HAVING cc > 5; ``` #### 3.40 检索至少选修两门课程的学生学号 ```sql SELECT sid, count(cid) AS cc FROM sc GROUP BY sid HAVING cc >= 2; ``` #### 3.41 查询选修了全部课程的学生信息 ```sql SELECT student.* FROM sc, student WHERE sc.SId = student.SId GROUP BY sc.SId HAVING count(*) = ( SELECT DISTINCT count(*) FROM course ) ``` #### 3.42 查询各学生的年龄，只按年份来算 一般都用 41 题的方法精确到天，按照出生日期来算，当前月日 <出生年月的月日则，年龄减一 ```sql select student.sid as 学生编号, student.sname 学生姓名 , timestampdiff(year, student.sage, curdate()) 学生年龄 from student ``` #### 3.43 查询本周过生日的学生 * where weekofyear(student.sage)="WEEKOFYEAR(CURDATE());" 3.44 查询下周过生日的学生 + 1; 3.45 查询本月过生日的学生 month(student.sage)="MONTH(CURDATE());" 3.46 查询下月过生日的学生 ## 关于作者> **作者：** C you again，从事软件开发 努力在IT搬砖路上的技术小白 > > **公众号：** 【**[C you again](https://cyouagain.cn/)**】，分享计算机类毕业设计源码、IT技术文章、游戏源码、网页模板、程序人生等等 > > **关于转载**：欢迎转载博主文章，转载时标明出处 > > **求赞环节**：创作不易，记得 点赞评论转发 谢谢你一路支持