Check Digit Validation?

[lizzieah]

I have no clue how to do the following Check Digit Validation

Step 1- Add 1st and 5th Digit of a ID number, giving total of 1
Add 2nd and 4th Digit of a Id number, giving total of 2

Step 2- Add 3rd digit of Id Number to total 1 and total 2, giving total 3.

Step 3- While total 3 is greater than 9, add all the digits of total 3 together giving a new total 3. Repeat this unitl total 3 is single digit.

Step 4 - The Id number is valid when the 6th digit of the id number is same as toatl 3.

Confused? I am. Could be because im very new to the whole VB concept. Any help much appreciated
Thanks

 

[Pat Hartman]

I'm not going to write all the code for you and there are no guarentees because I'm not going to test it, but here's a start.

Public Function YourCheckSum(YourField As Long) As Boolean
Dim Sub1 As Long
Dim Sub2 As Long
Dim sub3 As Long
Dim Work3 As Long
Dim n As Int

Sub1 = Left(YourField,1) + Mid(YourField,5,1)
Sub2 = Mid(YourField,2,1) + Mid(YourField,4,1)
Sub3 = Sub1 + Sub2 + Mid(YourField,3,1)
Do Until Sub3 <10
n = Len(Sub3)
I'm too tired to work this out but you need to do the inner loop n times where n is the current length of Sub3 and at the end set the new value of Sub3 = Work3
Work3 = Mid(Sub3,n,1)

Loop
If Mid(YourField,6,1) = Sub3 Then
YourCheckSum = True
Else
YourCheckSum = False
End If
End Function

 

[neileg]

I don'tdispute the generality of Pat's approach, but a little maths can bring the task under control.

Total 1 and 2 are irrelevant as they just get totalled again. So the basis for the check digit is Digit 1+2+3+4+5

This has a maximum value of 45 (9+9+9+9+9)

The maximum value of the sum of the digits here is 12 (when the value is 39)

So there will only ever be a maximum of 2 iterations.

So the logic becomes
Work3 = Left(YourField,1) + Mid(YourField,2,1) + Mid(YourField,3,1) + Mid(YourField,4,1) + Mid(YourField,5,1)
Work3=IIF(Work3>9,Left(Work3,1)+Right(Work3,1),Work3)

 

[Pat Hartman]

Nice solution neileg. At 10:48 PM I'm not at my best and probably shouldn't have tried to code anything anyway. Wonder why lizzieah got such stupid directions? Your solution is so simple.

 

[lizzieah]

thats ok

Hey,
It doesnt matter, all helps.

 

[lizzieah]

ok
it didnt work. it brings back the 1st and 5th number.

Eg- type in 123456
brings up 15

I dont know if i said but this is a assignment and he gave us examples of what should happen.

eg ID Number 123456
1+5=6, 2+4= 6
6+6+3=15
1+5= 6
6 is the same as the 6th digit, therefore valid id number

eg ID number 950178
9+7=16, 5+1=6
16+6+0=22
2+2=4
4 is not the same as the 6th digit, invalid id number

When it is invalid i also have to add the msg box that the id is invalid.
Im trying real hard to get it, its just not coming to me.

 

[neileg]

Sorry, I was trying to be too clever!

By using the string functions, left, mid and right, the number is converted to text. We need to convert it back. So try this:
Work3 = Left(YourField,1) + Mid(YourField,2,1) + Mid(YourField,3,1) + Mid(YourField,4,1) + Mid(YourField,5,1)
ChkDgt=IIF(Value(Work3)>9,Value(Left(Work3,1)+Right(Work3,1)), Value(Work3))