how to generate random number when you click on refresh button?

aswinialuri-19361
Posted by in ASP.NET category on for Beginner level | Points: 250 | Views : 6053 red flag
Rating: 4.67 out of 5  
 3 vote(s)

This article will explain how to generate a random number


To generate random number 
Add new item it solution explorer
drag and drop the text box and label

.aspx page:
design mode:
aspx
source:
<table><tr><td>RANDOM NUMBER:</td><td><asp:TextBox ID="txtrobocomp" runat="server"  Font-Names="Vivaldi" 
                        Font-Size="X-Large" Font-Bold="True" BackColor="#CC99FF" ForeColor="#CC0066" 
                        Width="148px" Height="33px"></asp:TextBox></td></tr>


code in .aspx.cs page:
 private int RandomNumber(int min, int max)
    {
        Random rd = new Random();
        return rd.Next(min, max);
    }

    private string RandomString(int size, bool lowerCase)
    {
        StringBuilder builder = new StringBuilder();
        Random rd = new Random();
        char ch;
        for (int i = 0; i < size; i++)
        {
            ch = Convert.ToChar(Convert.ToInt32(Math.Floor(26 * rd.NextDouble() + 65)));
            builder.Append(ch);
        }
        if (lowerCase)
            return builder.ToString().ToLower();
        return builder.ToString();
    }
Th output will be seen like this

random


After refreshing the page randomnumber will be generated
Place a refresh button in .aspx page
Image will shown like this

refresh
code in .aspx.cs page
in page load event
 protected void Page_Load(object sender, EventArgs e)
    {
        if (!Page.IsPostBack)
        {
            StringBuilder builder = new StringBuilder();
            builder.Append(RandomNumber(10, 99));
            builder.Append(RandomString(3, true));
            builder.Append(RandomString(2, false));
            txtrobocomp.Text = builder.ToString();
        }
    }
code in refresh button click event
 protected void ImageButton1_Click(object sender, ImageClickEventArgs e)
        {
            //if (!Page.IsPostBack)
            //{
                StringBuilder builder = new StringBuilder();
                builder.Append(RandomNumber(10, 99));
                builder.Append(RandomString(3, true));
                builder.Append(RandomString(2, false));
                txtrobocomp.Text = builder.ToString();
            //}
         }

the output will be like this
 
After clicking on refresh button the same as our captcha image
the output will be like this the random number is changed in particular text box
If you click on refresh button the random number will be generated by using the code above we can generate a random number and refresh button it will be seen like our captcha image

I hope you understand this article and will be useful.


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About the Author

aswinialuri-19361
Full Name: Aswini Aluri
Member Level: Starter
Member Status: Member
Member Since: 11/21/2012 12:44:34 PM
Country: India
Mark as Answer if it helps you Thanks&Regards Aswini Aluri
http://www.dotnetfunda.com
Aswini Aluri. Software developer

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Comments or Responses

Posted by: Shitalr on: 8/13/2013 | Points: 25
Really nice i learn something new today thanks!!
Posted by: Ses on: 9/16/2013 | Points: 25
Really nice article

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