2009-07-27 108 views
3

我想测试一种情况,但我需要添加一行虚拟数据来测试假设。因此,根据this mySQL manual page for SELECT INTO,我的查询是正确的:SELECT INTO not working

SELECT INTO courses.sections_rfip 
    (SectionID, CourseID, SectionNumber, Term, Credits, CutOffDate, StartDate, EndDate, LastDateToWithDraw, ContinuousIntake, AcceptsRegistration, Fee, Instructor, SectionDescription, RegistrationRestrictions, MeetingTime, Notes, Active, Created, SetInactive) 
    SELECT 3, 
     s.CourseID, 
     s.SectionNumber, 
     s.Term, 
     s.Credits, 
     s.CutOffDate, 
     s.StartDate, 
     s.EndDate, 
     s.LastDateToWithDraw, 
     s.ContinuousIntake, 
     s.AcceptsRegistration, 
     s.Fee, 
     s.Instructor, 
     s.SectionDescription, 
     s.RegistrationRestrictions, 
     s.MeetingTime, 
     s.Notes, 
     s.Active, 
     s.Created, 
     s.SetInactive 
    FROM courses.sections_rfip s 
    WHERE s.sectionid = 1 

但我得到以下错误信息:

“您的SQL语法错误;检查对应于你的MySQL服务器版本的手册为正确的语法使用近“变成courses.sections_rfip (SectionID,CourseID,SectionNumber,期限,信用,”第1" 行

所以有什么东西烂在INTO,这不明摆着我为什么 - 帮助?

回答