分享整理的12条sql语句连同数据

WBOY
풀어 주다: 2016-06-07 18:07:11
원래의
1008명이 탐색했습니다.

原本sql写得也不好,近几年数据库用得少,sql更是荒废了,最近复习sql,网上例 子很多,但都只是提供sql例子,没有配套的数据用来自己练习和调试

俺觉得自 己试着写写sql,调试调试还是有帮助的,读人家sql例子好像读懂了,自己写就未 必思路正确,调试得通,写得简洁。

跟着网上流行的学生选课表的例子复习了一下:

这篇文字在网上被转载烂了,里面有些sql适合用在应用系统里,有些“报表”的感 觉更重些,主要是想复习前者。前20条大体还挺好,后30条明显偏报表风格了,而 且后面选例良莠不齐,选了12个例子做练习,(其实很多语法,case, any/all, union之类的都没包括),用mysql数据库,并共享自己造出来的数据。关于这12条 sql, 修正了原文中有纰漏的地方。

sql是基本技能,若能写得好也挺精彩的,还在继续练习。绝不提倡努力写复杂sql 解决业务问题。应用系统里如果存在很复杂的sql,往往揭示了业务逻辑向下泄露 到sql层的问题,不利于维护和扩展,虽然这样确实常能提高运行效率。具体情况 自行取舍。
下面的例子都是比较通用的sql, 其实针对特定的数据库,需要学的也挺多,比如 oracle db的decode函数, rowid, rownum, connect by 虽然不通用,但是很实用。

数据可以在这里下载,只是用作练习,没做任何外键关联:

整理的sql在下面:
Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表

1. 选出每门功课都及格的学号
select distinct `s#` from sc where `s#` not in (select `s#` from sc where score 2. 查询“1”课程比“2”课程成绩高的所有学生的学号;
SELECT c01.`s#` from (select `s#`, `score` from sc where `c#`=1) c01,
(select `s#`, `score` from sc where `c#`=2) c02
where c01.`s#` = c02.`s#` and c01.score > c02.score
3. 查询平均成绩大于60分的同学的学号和平均成绩;
select `s#`, avg(score) from sc group by `s#` having avg(score) > 60
4. 查询所有同学的学号、姓名、选课数、总成绩;
select student.`s#`, student.`Sname`, count(`c#`), sum(score) from student left outer join sc on student.`s#` = sc.`s#` group by student.`s#`, sc.`s#`

5.查询没学过“叶平”老师课的同学的学号、姓名;
select student.`s#`, student.`Sname` from student where student.`s#` not in (select distinct(sc.`s#`) from teacher, course, sc where Tname='叶平' and teacher.`t#` = course.`t#` and sc.`c#`= course.`c#` )
6. 查询学过“001”并且也学过编号“002”课程的同学的学号、姓名
select student.`s#`, student.sname from student, sc where student.`s#` = sc.`s#` and sc.`c#` = 1 and exists (select * from sc sc_2 where sc_2.`c#`=2 and sc.`s#`=sc_2.`s#`)
7. 查询学过“叶平”老师所教的所有课的同学的学号、姓名 (巧妙)
select `s#`, sname from student where `s#` in
(select `s#` from sc, teacher, course where tname='叶平' and teacher.`t#`=course.`t#` and course.`c#`= sc.`c#` group by `s#` having count(sc.`c#`)=
(select count(`c#`) from teacher, course where tname='叶 平' and teacher.`t#`=course.`t#`) )

8. 查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名 (有代表性)
select `s#`, sname from (select student.`s#`, student.sname, score, (select score from sc sc_2 where student.`s#`=sc_2.`s#` and sc_2.`c#`=2) score2 from student , sc where
sc.`s#`=student.`s#` and sc.`c#`=1) s_2 where score2 9.查询没有学全所有课的同学的学号、姓名
select student.`S#`, Sname from student, sc where student.`s#` = sc.`s#` group by `s#`, sname having count(`c#`)
10. 查询至少有一门课与学号为“002”的同学所学相同的同学的学号和姓名;
select distinct(sc.`s#`), sname from student, sc where student.`s#`=sc.`s#` and `c#` in (select `c#` from sc where `s#`=002)
11. 把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;

update sc inner join
(select sc2.`c#`, avg(sc2.score) score from sc sc2, teacher, course where
sc2.`c#`=course.`c#` and tname='叶平' and teacher.`t#` = course.`t#` and course.`c#`=sc2.`c#` group by course.`c#`) sc3 on sc.`c#`=sc3.`c#` set sc.score=sc3.score
12. 查询2号的同学学习的课程他都学了的同学的学号;(注意理解:where语句的 第一个条件过滤掉不满足c#的记录,再group by,就比较清晰)
select `S#` from SC where `C#` in (select `C#` from SC where `S#`=2)
group by `S#` having count(*)=(select count(*) from SC where `S#`=2);

作者 人在江湖
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