Vijjendra's Blog

Encode a password using MD5 in C#

vijjendra — Mon, 09 Mar 2009 15:31:00 +0000

Many developers save the password in normal form in Database which is not secure for the user because every one who have the access of the database view your password it is very harmful for the user.
To encode the password use the using System.Security.Cryptography namespace;
Following is the method to encode the pwd in MD5 format.

using System;
using System.Collections;
using System.Configuration;
using System.Data;
using System.Linq;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.HtmlControls;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Xml.Linq;
using System.Collections.Generic;
using System.Security.Cryptography;
// method to encode password
public string EncodePassword(string userpassword)
{
Byte[] originalPwdBytes;
Byte[] encodedPwdBytes;
MD5 md5;
//Instantiate MD5CryptoServiceProvider, get bytes for user’s original password and encode password in MD5 format.
md5 = new MD5CryptoServiceProvider();
originalPwdBytes= ASCIIEncoding.Default.GetBytes(userpassword);
encodedPwdBytes= md5.ComputeHash(originalPwdBytes);
//Convert encoded user password in ‘readable” format.
return BitConverter.ToString(encodedPwdBytes);
}
My orginal post is here
Hope it will helpful …

Dynamically Create TextBox in .Net

vijjendra — Fri, 06 Mar 2009 18:58:34 +0000

Here we discuss to generate the textox at runtime. To create the textbox at run time for that first we take a server control Panel where we place the runtime generated textbox.Here we use a textbox and button,textbox to define the number of textbox generated at runtime ant generate the textbox on the button click.
following is the code for .aspx side:

code for code behind file:
protected void btnAdd_Click(object sender, EventArgs e)
{
for (int i = 0; i < Convert.ToInt32(txtadd.Text); i++)
{
TextBox txt = new TextBox();
txt.Width = 300;
txt.Height = 50;
txt.ID = “txtbox” + i;
txt.TextMode = TextBoxMode.MultiLine;
pnlText.Controls.Add(txt);
}
}
On the button click we define the properties of the text box which u want for runtime generated textboxes and add these textbox in the panel where we display the textbox.
This is simplay for enter the value in the given textbox and click on the button and then textboxes is display in the panel.

Trap enter key in asp.net

vijjendra — Fri, 06 Mar 2009 18:30:47 +0000

One famous problem is to call selected button on the enter key.Suppose we have registration page and want when I fill all information and press enter key the submit button call.Another situation is on the login page when we press enter key login button call.In these cases we need to trap the enter key.this can we do in two ways one for javascript and another from the C#.In C# code we also write the javascript code.
There is many cases like user want to trap the key for whole page i.e on the page when user press enter key the target button(which we want) Call,another way is when user on the inputcontrol(like text box) then press enter key then target bbutton call.
Below is The c# code to trap the enter key for all page

// Traps the enter key for all page and call the click event for the specified button control.
// defaultButton is the button to click when enter is pressed

public static void SetPageDefaultButton(Page page, WebControl defaultButton)
 {
   if (!page.Request.Browser.Browser.Equals("Firefox"))
    {
      // if browser is not fire fox.
      page.Form.DefaultButton = defaultButton.UniqueID;
    }
 else
 {
StringBuilder jscript = new StringBuilder();
jscript.Append("");
page.Form.Attributes.Add("onkeypress", "javascript:return PageDefaultButton(event,\'" + defaultButton.ClientID + "\')");
page.ClientScript.RegisterClientScriptBlock(string.Empty.GetType(), "PageDefaultButton", jscript.ToString());
 }
 }

By this function if user press enter key then always call defaultButton. call this function on the page load.
If we want when user on the specific input control and then press the enter key the specified button call.
Below is the c# code for that:
//Traps the enter key for a control and call the click event for the specified button control.
//The control to trap enter key after that input control.
//The client ID of the button to click when enter is pressed

 public static void SetDefaultButton(WebControl inputControl, string buttonClientId)
{
    StringBuilder jscript = new StringBuilder();
    jscript.Append("");
   inputControl.Attributes.Add("onkeypress", "return TrapEnterKey(event,\'" + buttonClientId + "\')");
   inputControl.Page.ClientScript.RegisterClientScriptBlock(string.Empty.GetType(), "TrapEnterKey", jscript.ToString());
        }

cheers!

Server Side Validation on a particular Button Click

vijjendra — Thu, 26 Feb 2009 20:02:53 +0000

In ASP.NET there is a server side validation which is very powerful and easy to use for developers and automatically validates the user input. Beginner developer facing problem when they validate a particular section of the page and page has more than one button in this case they are facing problem.
suppose we have a registration page there is two button one for Save and another for Cancel and we want validation only on Save button in this case we face the problem how to restrict validation on the cancel button and it is only working on Save button.

Solution:
The solution of this problem is to use the CausesValidation Property of button control set false.
Syntax:

another way is to use the ValidationGroup property.
Syntax:

ValidationGroup property is used on Save button where we want to Validate the User Input i.e. ValidationGroup is used on that button on which button we want validation after that CausesValidation Property is not required on the Cancel button. if we did not use the ValidationGroup any of Server Validators then that user input field is not validate. so this is require, that every server validation control which is used on the page is contains the ValidationGroup property.

Generate Random Password

vijjendra — Tue, 24 Feb 2009 20:18:36 +0000

Here we discuss how to generate the random password.
Random password increase the security of website it simply provide the non easily guessable password.Here we generate a set of random digits by combining a set of alphabets and numbers and special characters. After generating the random passsword it is the combination of string,number and special characters. One of the simplest way to generate the random password via GUID(Globally Unigue ID),it is a 128 bit number producing a hexadecimal number. following is a sample code to generate random password via GUID.

public string RandomPassword(int length)
{
string randomPass= Guid.NewGuid().ToString();
randomPass= randomPass.Replace(“-”,string.Empty);
return randomPass.Substring(0,length).ToUpper() // this random password contains all character in uppercase.
//if you want ur random password contains all character in lower case then use .ToLower()
}
In this NewGuid().ToString() returns a string of 32 hexadecimals (excluding the “-”).
Another way to generate the random number is :

public string RandomPassword(int length)
{
// give character which you want in random generated password.
char[] allowCharacter = “abcdefgijkmnopqrstwxyzABCDEFGHJKLMNPQRSTWXYZ23456789*$-+?_&=!%{}/”.ToCharArray();

// create object of random
Random RandChar = new Random();
string randomPassword = “”;
for (int i = 0; i < length; i++)
{
randomPassword += allowCharacter[RandChar.Next(0, allowCharacter.Length)];
}
//.ToUpper() for your password is in Uppercase and ToLower() password is in Lower case.
return randomPassword.ToUpper();
}

Maintain Scroll Position of ASP.NET Ajax Rating Control

vijjendra — Fri, 06 Feb 2009 13:59:15 +0000

In ASP.NET’s Ajax Rating control there is a bug.
When user click on the rating control it Jumps on the top of the page because it add # with url.
If you put your rating control on the bottom of the page and you click on the rating control it jumps on the top of the page.This happens because your all image tag are in HTML’s Anchor tag and its href attribute set to #(view page source of ajax rating control).
To remove this problem you need to add following line on the page load.

protected void Page_Load(object sender, EventArgs e)
{
Ratingcontrol.Attributes.Add(“Onclick”, “return false;”);
}

This will create java script dynamically for onclick event. That java script will override that unwanted jump to top of page.

FileUpload inside the Ajax UpdatePanel

vijjendra — Fri, 06 Feb 2009 13:56:55 +0000

File upload control is not working when this control is inside the update panel.When we place the file upload control control inside the updatepanel and upload the file then FileUpload1.Hasfile is always false because file upload control is inside updatepanel then your file upload control does not post back. Because the file upload control works only when the full page postback. If you want to upload the file then you need to postback the page.So we need to create the PostBackTrigger for the button.

We place between and

Triggers allow us to force a post back.