Sign In
Create Account
Follow Us on Twitter
Subscribe on Youtube
2
Hello to all friends
I write a program for Web-based e-Learning.
In this system there are three types of user: administrator, student, professor
I do not know for now identify the three classes of membership. Net or do I use the sql for each table created each time the user types information into tables that do recognize
Please tell me what a way to identify users is appropriate.
My English is not good. Excuse me please:closedeyes::closedeyes:
Tanks alot
4 replies to this topic
Back to top|
|
|
#2
Posted 16 August 2009 - 04:18 AM
WingedPanther
-
- Moderators
-
- 16,822 posts
A spammer's worst nightmare
- Location:Upstate, South Carolina
- Programming Language:C, C++, PL/SQL, Delphi/Object Pascal, Pascal, Transact-SQL, Others
- Learning:Java, C#, PHP, JavaScript, Lisp, Fortran, Haskell, Others
You can have three tables if you want, but all of them are a "User" in the end and will need a common table to store login credentials and indicate the type of user (1,2, or 3 in user-type, for example).
#3
Posted 16 August 2009 - 10:27 PM
zkazemi
-
- Members
-
- 7 posts
Newbie
WingedPanther said:
You can have three tables if you want, but all of them are a "User" in the end and will need a common table to store login credentials and indicate the type of user (1,2, or 3 in user-type, for example).
Tanks alot WingedPanther:thumbup:
#4
Guest_Jordan_*
Posted 17 August 2009 - 05:58 AM
Guest_Jordan_*
Guest_Jordan_*
-
- Guests
You can use an ENUM column to indicate users:
MySQL :: MySQL 5.0 Reference Manual :: 10.4.4 The ENUM Type
CREATE TABLE users (
type ENUM('administrator', 'student', 'professor')
);
MySQL :: MySQL 5.0 Reference Manual :: 10.4.4 The ENUM Type
#5
Posted 04 March 2010 - 07:56 AM
gokuajmes
-
- Members
-
- 518 posts
Programming God
well u have a ready to use login controls, try dropping the loginbox and other controls from your toolbox then configure them to use the sql membership also make Roles to for each person ie admin ,student and the professor
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
