Problem In Ajax

msafvati · October 18, 2009

hi my friends

i want to use Ajax in php,but when i use from it i encounter with a problem.

that is:"i send a request to server but it dont give me Response"

please see this code and guide me,

thank you

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>

<script type="text/javascript">
    function GetXmlHttpObject()
    {
    var xmlHttp=null;
    try
     {
     // Firefox, Opera 8.0+, Safari
     xmlHttp=new XMLHttpRequest();
     }
    catch (e)
     {
     //Internet Explorer
     try
      {
      xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
      }
     catch (e)
      {
      xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
     }
    return xmlHttp;
    }
    
    
    function checkUser()
    { 
      xmlHttp=GetXmlHttpObject();
      if (xmlHttp==null)
      {
        alert ("Browser does not support HTTP Request");
        return;
      }
    
      url='http://127.0.0.1/mas/index.php?user='+document.getElementById('user').value;
      //alert(url);
      xmlHttp.onreadystatechange=stateChanged;
      xmlHttp.open("GET",url,true);
      xmlHttp.send(null);

    }
    
    function stateChanged() 
    { 
    if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
     { 
        if (xmlHttp.responseText==1)    
        {    
            document.getElementById("user").style.border="1px solid #FF0000";
        }
        else{
            document.getElementById("user").style.border="1px solid #00CC00";
        }
}

  
    }

</script>
</head>

<body>
<form action="http://127.0.0.1/mas/index.php" method="get">
<table width="300" border="0">
  <tr>
    <td><label for="name">name </label></td>
    <td><input type="text" id="name" /></td>
    <td> </td>
  </tr>
  <tr>
    <td><label for="user">user </label></td>
    <td><input type="text" id="user"  /></td>
    <td style="text-align:left"><input type="button" value="check"  onclick="checkUser()"/></td>
  </tr>
  <tr>
    <td><label for="pass">password</label></td>
    <td><input type="password" id="pass" /></td>
    <td> </td>
  </tr>
  <tr>
      <td></td>
    <td colspan="2"><input type="submit" id="send" /></td>
  </tr>
</table>
</form>
</body>
</html>

and PHP:

<?php
   $con=mysql_connect('127.0.0.1','root','root');
    if($con){
        echo "Connection Successfuly \n";
    }
    $db=mysql_select_db('Ajax',$con);
    $sql="SELECT *  FROM `users` WHERE `user` ='".$_GET['user']."'";
    $r=mysql_query($sql);
    $result=mysql_num_rows($r);
    echo( $result);
?>

ialsoagree · October 18, 2009

I don't know if this is your problem but your PHP isn't written well

<?php
   $con=mysql_connect('127.0.0.1','root','root');

// DO THE FOLLOWING ONLY ON SUCCESSFUL DATABASE CONNECTION
    if($con){
        echo "Connection Successfuly \n";
    }

// DO THE FOLLOWING EVEN IF THERE IS NO DATABASE CONNECTION
    $db=mysql_select_db('Ajax',$con);
    $sql="SELECT *  FROM `users` WHERE `user` ='".$_GET['user']."'";
    $r=mysql_query($sql);
    $result=mysql_num_rows($r);
    echo( $result);
?>

See the problem? You check if $con successfully connected, but even if it didn't, you still try to run a query.

Also, I hope you realize that your $db variable is a boolean. mysql_select_db returns a boolean that represents whether or not it worked (TRUE if it worked, FALSE otherwise).

Also, you're concatenating your variable to a string that's being parsed for variables:

$sql="SELECT *  FROM `users` WHERE `user` ='".$_GET['user']."'";

Because you use double quotes, PHP is searching the string for variables (parsing the string). But you remove your variable from the string (".$_GET['user'].") so there's nothing for PHP to find!

You should do this instead:

$sql="SELECT *  FROM `users` WHERE `user` ='$_GET[user]'";

You also aren't escaping the user input! This makes it easy for a malicious user to inject SQL into your form that your database will then process!

msafvati · October 19, 2009

thank u,

i do your tells but dont give any answer from Server,responseText from Server is true(Example if my Data Exist in Db Return 1)but if block have problem,

plz guid me!!!

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