How to generate alphanumeric unique random number in

Introduction: In this article i am going to explain how to create or generate unique random alphanumeric number or password or string in using both C# and VB language.

In previous articles i explained How to Encrypt and decrypt username and password and store in sql server database using and Jquery to show hide password characters in textbox on checkbox check uncheck and Create change password form in and Create crystal reports in visual studio 2010 using and How to bind ListBox with Sql Server Database in 

Description: Sometimes it is required to generate/create unique random number while working on applications. E.g. to create random password or to concatenate random number with the name of the file to upload through FileUpload control to make that file unique in the uploaded folder etc.

Implementation: Place a Button Control on design page(.aspx)
<asp:Button ID="btnRandomNumber" runat="server" Text="Generate"
            onclick="btnRandomNumber_Click" />

Asp.Net C# Code to create/generate unique random number in

    protected void btnRandomNumber_Click(object sender, EventArgs e)
        string num = GenerateRandomNumber(5);
public static string GenerateRandomString(int Length)
            string _allowedChars = "#@$&*abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789";
            Random randNum = new Random();
            char[] chars = new char[Length];

            for (int i = 0; i < Length; i++)
                chars[i] = _allowedChars[Convert.ToInt32((_allowedChars.Length - 1) * randNum.NextDouble())];
            return new string(chars);

Asp.Net VB Code to create/generate unique random number in

Protected Sub btnRandomNumber_Click(sender As Object, e As EventArgs) Handles btnRandomNumber.Click
        Dim num As String = GenerateRandomNumber(5)
    End Sub

    Public Shared Function GenerateRandomString(Length As Integer) As String
        Dim _allowedChars As String = "#@$&*abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789"
        Dim randNum As New Random()
        Dim chars As Char() = New Char(Length - 1) {}

        For i As Integer = 0 To Length - 1
            chars(i) = _allowedChars(Convert.ToInt32((_allowedChars.Length - 1) * randNum.NextDouble()))
        Return New String(chars)
    End Function

 Now over to you:
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Click here for comments
September 13, 2013 ×

I am getting error near NextDouble() ?

September 13, 2013 ×

What error you are getting? and what parameter you are passing to GenerateRandomNumber method?

Kadir Ansari
January 25, 2014 ×

Thanks Sir,
Nice Article.........

January 26, 2014 ×

your welcome Kadir Ansari..stay connected and keep reading for more useful updates like this.:).


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