Solved

Randomly generating number

Posted on 2003-12-10
16
320 Views
Last Modified: 2010-05-01
Hi, in my application, when a user clicks the add button, the text fileds are activated for the user to input client information, like client ID, name, address etc. The application also connects to a database. What i need to do is randomly generate a three digit clientID, and populate it to the client id textbox when the add button is pressed, but i also need to make sure that the ID generated should not already exist in the database... any ideas...  
0
Comment
Question by:intox1221
  • 9
  • 7
16 Comments
 

Author Comment

by:intox1221
ID: 9912216
Also im using .NET environment...
0
 
LVL 42

Expert Comment

by:frodoman
ID: 9912284
intox1221,

This is going to require generating a random number, then searching the database for it.  If it's found, you'll have to repeat the process.  The more users you get the more times you'll have to repeat this process.

I can help if you're certain this is what you want to do, but I'd take a different approach if possible.  Instead of a random number, why not just SELECT MAX(userid) +1 to increment the userid by 1 each time?

Frodoman
0
 
LVL 42

Expert Comment

by:frodoman
ID: 9912309
If you do go w/ the random approach, here's the random number code:

MyInteger = CInt(Int(1000*Rnd()))

0
 

Author Comment

by:intox1221
ID: 9912453
i declared myinteger as int32 and then equated it with txtClient.text but the application pukes it...
something im missing ?

0
 
LVL 42

Expert Comment

by:frodoman
ID: 9912485
Did you use tostring?

txtClient.text = YourInt32.ToString
0
 

Author Comment

by:intox1221
ID: 9912525
i did..

MyInteger = Cint (txtclient.text)

Also i need to have option strict on which disallows implicit conversions.
0
 
LVL 42

Expert Comment

by:frodoman
ID: 9912600
I assume it's a type conversion problem?  What is displayed in txtClient before the error?  What error is shown?

You might have extra spaces so also try MyInteger = CType(trim(txtclient.text),Int32)
0
 

Author Comment

by:intox1221
ID: 9912828
Still pukin it... this is what i got....

        Dim myinteger As Int32

        myinteger = CInt(Int(1000 * Rnd()))

myinteger = CType(Trim(txtClient.Text), Int32)
0
How your wiki can always stay up-to-date

Quip doubles as a “living” wiki and a project management tool that evolves with your organization. As you finish projects in Quip, the work remains, easily accessible to all team members, new and old.
- Increase transparency
- Onboard new hires faster
- Access from mobile/offline

 
LVL 42

Expert Comment

by:frodoman
ID: 9912886
Where is the error and what is the description of the error?  
0
 

Author Comment

by:intox1221
ID: 9912929
thats the error message.

An unhandled exception of type 'System.InvalidCastException' occurred in microsoft.visualbasic.dll

Additional information: Cast from string "" to type 'Integer' is not valid.
0
 
LVL 42

Expert Comment

by:frodoman
ID: 9913018
This means that txtClient is empty when you reach the code:

   myinteger = CType(Trim(txtClient.Text), Int32)

The block below will avoid the error and use a '0' if the textbox is empty, but you should probably examine your code for why it is empty.

If IsNumeric(txtClient.Text) Then
   myinteger = CType(Trim(txtClient.Text), Int32)
Else
   myinteger = 0
End If

What exactly are you trying to accomplish with this code?  Seems like you create the random number and you'd want to put it INTO your textbox instead of pulling something OUT of it...

  Dim myinteger As Int32
  myinteger = CInt(Int(1000 * Rnd()))
  txtClient.Text = CType(myinteger,String)
0
 

Author Comment

by:intox1221
ID: 9913334
thats exactly what i want. My textbox is empty when the add button is pressed and i want fill it with the randomly generated code... what the add button does is it activates the textfileds, as they are read only or deactivated before that. Then upon pressing the save button, is when i want the data should be sent to the database (that i have already done). So all i wanna do is populate the text field with the randomly generated code.
0
 
LVL 42

Accepted Solution

by:
frodoman earned 450 total points
ID: 9913408
Okay, so when the add button is pressed you want to do this:

  Dim myinteger As Int32
  myinteger = CInt(Int(1000 * Rnd()))
  txtClient.Text = CType(myinteger,String)

This creates a random number and places it into the txtClient textbox.  Is this not working?
0
 

Author Comment

by:intox1221
ID: 9913669
Sweeeeet. Works like a charm. But it also generates numbers like 54, 14... can i make it atleast 3 digits... or 014 instead of 14... ??????
0
 

Author Comment

by:intox1221
ID: 9913731
actually that doesnt make much of a difference. nevermind that...
0
 

Author Comment

by:intox1221
ID: 9913769
thanks for ur help on this one. I have another question to the follow up of my current question. So i am posting it as a new question, thanks for ur help. The topic should be "Randomly generating number take 2"
0

Featured Post

Do You Know the 4 Main Threat Actor Types?

Do you know the main threat actor types? Most attackers fall into one of four categories, each with their own favored tactics, techniques, and procedures.

Join & Write a Comment

Introduction I needed to skip over some file processing within a For...Next loop in some old production code and wished that VB (classic) had a statement that would drop down to the end of the current iteration, bypassing the statements that were c…
If you have ever used Microsoft Word then you know that it has a good spell checker and it may have occurred to you that the ability to check spelling might be a nice piece of functionality to add to certain applications of yours. Well the code that…
Get people started with the process of using Access VBA to control Outlook using automation, Microsoft Access can control other applications. An example is the ability to programmatically talk to Microsoft Outlook. Using automation, an Access applic…
Show developers how to use a criteria form to limit the data that appears on an Access report. It is a common requirement that users can specify the criteria for a report at runtime. The easiest way to accomplish this is using a criteria form that a…

705 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

15 Experts available now in Live!

Get 1:1 Help Now