这个示例代码她只是计算出表3的平均成绩,而不是真正的GPA:SELECT * FROM( 选择AVG(g.Grade)作为average_grade,g.SSN 从等级g g.SSN小组)a 内部联接学生在a.ssn = s.ssn;
找出正确的公式:这是我想出的但是不正确 选择 * FROM( SELECT SUM(Grade * CreditHour)/(SUM(CreditHour)as average_grade,g.SSN 从等级g g.SSN小组)a 内部联接学生在a.ssn = s.ssn;
Need some help stuck on this problem?
答案 0 :(得分:1)
在我看来,好像你只是忘了加入课程表的成绩表。或者我错过了什么?
SELECT *
FROM
(
SELECT
SUM(g.Grade*c.CreditHour) / SUM(c.CreditHour) as average_grade,
g.SSN
FROM Grade g
INNER JOIN Course c ON c.cno = g.cno
GROUP BY g.SSN
) a
INNER JOIN Student s ON a.ssn = s.ssn;