AJJAX PHP Login form

JavaScript
1

Hi all,

I have a login form i want to be ajaxed only issue is im getting login failed this is my ajax and html form its as below


<script language="javascript">
//  Developed by Roshan Bhattarai 
//  Visit http://roshanbh.com.np for this script and more.
//  This notice MUST stay intact for legal use

$(document).ready(function()
{
	$("#login_form").submit(function()
	{
		//remove all the class add the messagebox classes and start fading
		$("#msgbox").removeClass().addClass('messagebox').text('Validating....').fadeIn(1000);
		//check the username exists or not from ajax
		$.post("alogin.php",{ user_name:$('#username').val(),password:$('#password').val()
	} ,function(data)
        {
		  if(data=='yes') //if correct login detail
		  {
		  	$("#msgbox").fadeTo(200,0.1,function()  //start fading the messagebox
			{ 
			  //add message and change the class of the box and start fading
			  $(this).html('Logging in.....').addClass('messageboxok').fadeTo(900,1,
              function()
			  { 
			  	 //redirect to secure page
				 document.location='secure.php';
			  });
			  
			});
		  }
		  else 
		  {
		  	$("#msgbox").fadeTo(200,0.1,function() //start fading the messagebox
			{ 
			  //add message and change the class of the box and start fading
			  $(this).html('Your login detail sucks...').addClass('messageboxerror').fadeTo(900,1);
			});		
          }
				
        });
 		return false; //not to post the  form physically
	});
	//now call the ajax also focus move from 
	$("#password").blur(function()
	{
		$("#login_form").trigger('submit');
	});
});
</script>
<!-- Show Message for AJAX response -->
<form method="post" action="" id="login_form">
<div align="center">
<div style="color:#FFF;">
   User Name : <input name="username" type="text" id="username" value="" maxlength="20" />
</div>
<div style="margin-top:5px;color:#FFF;">
   Password :
    &nbsp;&nbsp;
    <input name="password" type="password" id="password" value="" maxlength="20" />  
</div>
<div class="buttondiv" style="color:#FFF;">
    <input name="Submit" type="submit"  id="submit" value="Login" style="margin-left:-10px; height:23px" style="color:#FFF;"  />
    <span id="msgbox" style="display:none"  style="color:#FFF;"></span>
</div>
</div>

below is my php page which as follows also nothing fancy just the usual login form as it shud work when i click on login but it deosnt when using ajax it is this


				$password = mysqli_real_escape_string($con,stripslashes(trim($_POST['password'])));
				$cQuery="SELECT * FROM admin WHERE username='".$username."' AND password='".$password."'";
				echo $cQuery;
				$con;
				$rs=mysqli_query($con,$cQuery);
				if(!$rs)
				{
					echo "Unable to excute the query:".mysqli_error($con);
				}
				else
				{
					$count=$rs->num_rows;
					if($count>0)
					{
						if($data['password']=$password)
						{
							echo "yes";
							//now set the session from here if needed
							$_SESSION['u_name']=$username; 
						}
						else
						{
							echo "Password does not match";	
						}
						
					}
					else
					{
						echo "Username doesnt exist";
					}
				}

Why do i get the text login stinks what am i doing wrong when i know for sure my php code and login info are correct?

i was using a tutorial and this is what the ajax/jquery tutorial said to do not sure what to do now though

any help would be good what am i doing wrong?

Thanks,William

2

What i really want to know is why your mixing MySQLi OOP code with procedural code eg.

mysqli_query
$rs->num_row

To me this looks like where all the problems start.

3

well it seems to work ive done it many times b4 anyway ive adjusted the login code its


include("dbconnect.php");
$username = mysqli_real_escape_string($con,stripslashes(trim($_POST['username'])));
				$password = mysqli_real_escape_string($con,stripslashes(trim($_POST['password'])));
				$cQuery="SELECT * FROM admin WHERE username='".$username."' AND password='".$password."'";
				echo $cQuery;
				$con;
				$rs=mysqli_query($con,$cQuery);
				if(!$rs)
				{
					return "Unable to excute the query:".mysqli_error($con);
				}
				else
				{
					$count=mysqli_num_rows($rs);
					//$count=$rs->num_rows;
					echo $count;
					if($count>0)
					{
						
						while($data=mysqli_fetch_assoc($rs))
						{
							if($data['password']=$password)
							{
								return "yes";
								//now set the session from here if needed
								$_SESSION['user']=$username; 
							}
							else
							{
								return "Password does not match";	
							}
						}
					}
					else
					{
						return "Username doesnt exist";
					}
				}

But how can i fix my issue i am having??

4

In your javascript code before

if (data=='yes')

alert the message for data and see what response you get as it will help you debug were the error is. Also a good thing to try is returning the vital PHP variables back with a string to see what the row counts are etc.