Generation 9 digits random number including leading zero

Question

Hi,

I want to generate random number, which is 9 digits including leading zero if the number is less than 9 digits, say 123 will be 000000123. I have the following code which doesn't include leading zero :

Dim RandomClass As New Random()
Dim RandomNumber = RandomClass.Next(1, 999999999)

Thanks.

Answer 1

EDIT: While I still quite like my "individual digits" approach below, there's an easier way - just give a custom number format:

C#:

Random rng = new Random();
int number = rng.Next(1, 1000000000);
string digits = number.ToString("000000000");
Console.WriteLine(digits);

VB:

Dim rng As New Random
Dim number As Integer = rng.Next(1, 1000000000)
Dim digits As String = number.ToString("000000000") 
Console.WriteLine(digits)

EDIT: As has been pointed out in the comments, a format string of D9 will also do the job:

Dim digits As String = number.ToString("D9") 

Personally I'd have to look up exactly what that would do, whereas I'm comfortable with custom number formats - but that says more about me than about the code :)

---

Rather than generating a single number between 1 and 999999999, I would just generate 9 numbers between 0 and 9. Basically you're generating a string rather than a number (as numerically 000000000 and 0 are equivalent, but you don't want the first).

So generate 9 characters '0' to '9' in a Character array, and then create a string from that.

Here's some sample C# code:

using System;

class Test
{
    static void Main(string[] args)
    {
        Random rng = new Random();
        string digits = GenerateDigits(rng, 9);
        Console.WriteLine(digits);
    }

    static string GenerateDigits(Random rng, int length)
    {
        char[] chars = new char[length];
        for (int i = 0; i < length; i++)
        {
            chars[i] = (char)(rng.Next(10) + '0');            
        }
        return new string(chars);
    }
}

... and converting it to VB:

Public Class Test

    Public Shared Sub Main()
        Dim rng As New Random
        Dim digits As String = Test.GenerateDigits(rng, 9)
        Console.WriteLine(digits)
    End Sub

    Private Shared Function GenerateDigits(ByVal rng As Random, _
                     ByVal length As Integer) As String
        Dim chArray As Char() = New Char(length  - 1) {}
        Dim i As Integer
        For i = 0 To length - 1
            chArray(i) = Convert.ToChar(rng.Next(10) + &H30)
        Next i
        Return New String(chArray)
    End Function

End Class

One point to note: this code can generate "000000000" whereas your original code had a minimum value of 1. What do you actually want the minimum to be?

Answer 2

You need to declare your variable as type string. Integer can't have leading zero. Check my sample below:

 Dim RandomClass As New Random()


    Dim randomString As String


    For i = 1 To 1000 Step 1

        randomString = RandomClass.Next(1, 999999999).ToString

        If randomString.Length < 9 Then
            randomString = randomString.PadLeft(9, "0")
        End If
        Response.Write(randomString & "<br/>")
    Next

Answer 3

Is 999999999 a legal value? If so...

'
Dim myRand As New Random
'
Private Function NineDigitRand() As String
    'return a numeric string between 000000001 and 999999999 inclusive
    Return myRand.Next(1, 1000000000).ToString("d9")
End Function