1
查詢定義:誰擁有的85或更高的藝術品位
選擇學生
和
誰也有85或更好的等級中的任何計算機課程
QUERY(在3個部分解釋是關於MySQLWorkBench執行單個查詢):
MySQL的複雜的內部聯接查詢收到錯誤
select Students.StudFirstName,Student_Schedules.Grade
from
(Select Distinct Students.StudentID,Students.StudFirstName,Student_Schedules.Grade
from (((Students
Inner Join Student_Schedules
On Student_Schedules.StudentID = Students.StudentID)
Inner Join Classes
On Classes.ClassID = Student_Schedules.ClassID)
Inner Join Subjects
On Subjects.SubjectID = Classes.SubjectID)
Inner Join Categories
On Subjects.CategoryID = Categories.CategoryID
where Categories.CategoryDescription = 'Art' and Student_Schedules.Grade >= 85)
As Stud_Art
上面的代碼中提取誰在藝術具有85或更好的成績優異
Inner Join
(Select Distinct Students.StudentID,Students.StudFirstName,Student_Schedules.Grade
from (((Students
Inner Join Student_Schedules
On Student_Schedules.StudentID = Students.StudentID)
Inner Join Classes
On Classes.ClassID = Student_Schedules.ClassID)
Inner Join Subjects
On Subjects.SubjectID = Classes.SubjectID)
Inner Join Categories
On Subjects.CategoryID = Categories.CategoryID
WHERE Categories.CategoryDescription LIKE '%Computer%' AND Student_Schedules.Grade >= 85)
As Stud_CS
上面的代碼中提取誰在電腦
On Stud_CS.StudentID = Stud_Art.StudentID;
有85或更好的成績優異 以上代碼匹配Art和Computer的學生ID
錯誤出自MySQLWorkBench:
錯誤代碼:在「字段列表」 1054年未知列「Students.StudFirstName」
上面的查詢可以用子查詢技術來解決,但我想了解是怎麼回事可能使用Inner Join技術
是你right.I已經糾正了第一part.Then我也忘了,包括Students.StudentID兩個選擇查詢 – Robin 2013-03-07 15:41:48
很樂意幫忙:-) – 2013-03-07 15:44:58