15.7 Imagine a simple database storing information for students' grades. Design what this database might look like and provide a SQL query to return a list of the honor roll students (top 10%), sorted by their grade point average.
在一个简化的数据库中我们有三个表，Students表，Courses表和CourseEnrollment表如下：
TABLE Students
+-------------+-------------+ | Students | +-------------+-------------+ | StudentID | int(11) | | StudentName | varchar(30) | | Address | varchar(50) | +-------------+-------------+
TABLE Courses
+-------------+-------------+ | Courses | +-------------+-------------+ | CourseID | int(11) | | CourseName | varchar(30) | | ProfessorID | int(11) | +-------------+-------------+
TABLE CourseEnrollment
+-----------+---------+ | CourseEnrollment | +-----------+---------+ | CourseID | int(11) | | StudentID | int(11) | | Grade | float | | Term | int(11) | +-----------+---------+
使用SQL Server的Top .. Percent 函数可以如下代码：
-- SQL Server (Incorrect Code) SELECT TOP 10 PERCENT AVG(CourseEnrollment.Grade) AS GPA, CourseEnrollment.StudentID FROM CourseEnrollment GROUP BY CourseEnrollment.StudentID ORDER BY AVG(CourseEnrollment.Grade);
由于我使用的是MySQL，本来可以用Limit关键字来做，但是MySQL的Limit关键字后面只能为常量，不能为变量，所以只能换一种写法，所以代码如下：
SET @limit = (SELECT 0.1 * COUNT(*) FROM CourseEnrollment); SELECT Grade FROM ( SELECT *, @rownum := @rownum + 1 AS rank FROM CourseEnrollment, (SELECT @rownum := 0) init ORDER BY Grade DESC ) d WHERE rank <= @limit ;
但是上面的写法确实返回了top 10%的行，但是假如我们有100个学生，前15个学生都是4.0的GPA，而上面的方法只能返回10个学生，而我们需要返回那15个都是4.0的学生，所以为了实现这个，我们可以这么做：
-- SQL Server DECLARE @GPACutOff FLOAT; SET @GPACutOff = (SELECT MIN(GPA) AS 'GPAMin' FROM ( SELECT TOP 10 PERCENT AVG(CourseEnrollment.Grade) AS GPA, FROM CourseEnrollment GROUP BY CourseEnrollment.StudentID ORDER BY GPA DESC) Grades); SELECT StudentName, GPA FROM ( SELECT AVG(CourseEnrollment.Grade) AS GPA, CourseEnrollment.StudentID FROM CourseEnrollment GROUP BY CourseEnrollment.StudentID HAVING AVG(CourseEnrollment.Grade) >= @GPACutOff) Honors INNER JOIN Students ON Honors.StudentID = Students.StudentID;
上面的方法先定义了一个GPACutOff变量，算出了前10%的GPA，然后在后面的代码中遍历所有的GPA，返回所有大于等于GPACutOff的行。而是用MySQL可以写出实现同样功能的代码，参考了我之前的博客Department Top Three Salaries，Department Highest Salary和Second Highest Salary，用更简洁的方式如下：
-- MySQL SELECT c.Grade FROM CourseEnrollment c WHERE (SELECT COUNT(DISTINCT Grade) FROM CourseEnrollment WHERE Grade > c.Grade) < (SELECT 0.1 * COUNT(*) FROM CourseEnrollment) ORDER BY c.Grade DESC;
本文转自博客园Grandyang的博客，原文链接：学生成绩[CareerCup] 15.7 Student Grade ，如需转载请自行联系原博主。
