What I would do:

* The COURSES table:
 - Probably it's good to choose a different name for COURSES. (Course 
registration details or maybe better name can be invented).
 - Add a unique/primary key field for each entry in the table.
 - If possible - would be easier to work if you have EXPIRATION_DATE there.

* In STUDENT table:
 - Add a column referencing the unique/primary key which I just suggested 
for COURSES. (Create a foreign key from this column to the table with 
course details)

Having these it would be trivial to join both tables.

On Wednesday, November 14, 2012 8:50:47 PM UTC+2, SQL-Help wrote:
>
> I have a STUDENT table that has 
> Student_id 
> Student_name 
> Course_id 
> Course_date 
>
> So my records look like 
>
> 123 Kathy Smith  A103    12/01/2011 
> 123 Kathy Smith  A102    12/01/2011 
> 124 Chris Jones   A103    06/01/2011 
> 124 Chris Jones   A102    06/01/2011 
>
> The I have another table COURSE 
> Course_id 
> Course_name 
> Course_fee 
> Effective_date 
>
> A103 Accounting1  200.00 04/01/2011 
> A103 Accounting1  210.00 06/01/2011 
> A103 Accounting1  220.00 10/01/2011 
> A102 Writing101    150.00  04/01/2011 
>
>
> what I need is 
>
> Student_id, Student_name, Course_id, Course_name, course_date and 
> Course_fee 
>
> (based on the date a particular student signed up for the course, 
> course fee should be determined) 
>
> so for 
>
> 123 Kathy smith   A103  Accounting1  12/01/2011    220.00 
> 124 Chris Jones   A103  Accounting1   06/01/2011   210.00 
> 123 Kathy Smith  A102  Writing102     12/01/2011   150.00 
> 124 Chris Jones    A102 Writing 102    04/01/2011   150.00 
>
>
>
> This is how I tried and it doesn't work 
>
> select a.student_id, a.course_id, b.course_name, a.course_date, 
> b.course_fee 
> from student a left outer join on Course b where a.course_id = 
> b.course_id 
> and b.effective_Date = (select effective_Date from course where 
> effective_date <= a.course_date) 
>
> This is not for my school homework..working as a junior developer (so 
> I created a scenario similar to my issue at work and am asking for 
> help). 
> Thank you 
>

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