Generating random number in a sequence using VBscript

Question

I'm new to vbscript.

I've got this so far

Function random_number_generator (lowerLim, upperLim)
    Randomize
    randNumber = Int(((upperLim-lowerLim+1)* Rnd())+ lowerLim)
    random_number_generator = randNumber
End Function

But how do I make it to have a sequence for generating let's say 9 digit social security number?

for example if the first social security is 200030886 next would be 200030887 and so on.

A 9-digit number is just any number between 0 and 999999999, so all you need to do is adjust your upper and lower limits accordingly. — Simon MᶜKenzie, Dec 23 '14 at 04:34
Can you post what you need the output to look like. Please include a couple of examples — Spock, Dec 23 '14 at 04:34

score 2 · Accepted Answer · answered Dec 23 '14 at 04:55

I'm not sure why you have a random element to your question. I seems like you just want a counter?

Dim start, ends
start = 200030886
ends = 200040000

for k = start to ends
    MsgBox k
next

If you need a random number formatted like the numbers above, you can try this.

MsgBox random_number_generator(1, 10000)

Function random_number_generator (lowerLim, upperLim)
    Randomize
    randNumber = Int(((upperLim-lowerLim+1)* Rnd())+ lowerLim)
    random_number_generator = "2" & RIGHT("000000000" & randNumber, 8)
End Function

Hope that helps

score -2 · Answer 2 · edited Jun 06 '16 at 13:13

-2

Dim RandNumber

RandNumber = RandomNumber(1,10000)

Print RandNumber

edited Jun 06 '16 at 13:13

Paul Roub

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answered Jun 06 '16 at 13:05

user6430578

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There is no RandomNumber function in VBScript. – Ekkehard.Horner Jun 06 '16 at 13:43

Generating random number in a sequence using VBscript

2 Answers2