Validation of password using javascript

Question

Following is code for username.php

    <html>
  <head>
       <script type="text/javascript" src="js/jquery-1.7.1.min.js"></script>
      <!-- <script type="text/javascript" src="js/jquery-ui-1.8.17.custom.min.js"></script> -->

       <script type="text/javascript">

         $(document).ready(function(){
            $("#username").change(function(){
                 $("#message").html("<img src='ajax-loader.gif' /> checking...");


            var username=$("#username").val();

              $.ajax({
                    type:"post",
                    url:"check.php",
                    data:"username="+username,
                        success:function(data){
                        if(data==0){
                            $("#message").html("Username available");
                        }
                        else{
                            $("#message").html("Username already taken");
                        }
                    }
                 });

            });

         });

       </script>
  </head>

  <body>

       <table>
        <tr>
              <td>Username</td>
              <td>:</td>
              <td><input type="text" name="id" id="username""/><td>
                <td id="message"><td>
        </tr>

        <tr>
              <td>Password</td>
              <td>:</td>
              <td><input type="text" name="password" id="password" /><td>
        </tr>
       </table>
  </body>
</html>

And the code for check.php

    <?php

  mysql_connect("localhost","healhmate","healthmate");
  mysql_select_db("hmdb");
  if(isset($_POST['id']))
  $username=$_POST['id'];
  $query=mysql_query("SELECT * from user where id='$username' ");

  $find=mysql_num_rows($query);

  echo $find;

?>

this code gives output as username and password boxes. I have included all the 3 files ajax-loader.gif, username.php and check.php in one single folder.On entering username no validation is performed. Can anyone help me to figure out why is this happening?

dont you wish to to validate password as well? your code only does username check. — Mohit
– Mohit, Commented Oct 17, 2013 at 4:04

Rituraj ratan · Accepted Answer · 2013-10-17 03:58:15Z

4

 <?php

  mysql_connect("localhost","healhmate","healthmate");
  mysql_select_db("hmdb");
  if(isset($_POST['username']))// because in ajax you send username not id
  $username=$_POST['username'];
  $query=mysql_query("SELECT * from user where id='$username' ");

  $find=mysql_num_rows($query);

  echo $find;

?>

answered Oct 17, 2013 at 3:58

Rituraj ratan

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1 Comment

Flight Odyssey Over a year ago

This still has a sql injection vulnerability.

Gopi Krishna · Accepted Answer · 2013-10-17 04:07:37Z

2

Use data:"id="+username in Ajax request because that is the POST variable your checking in PHP.

Also on a side note: Make sure you handle a case where $_POST['username'] is not set.

<?php
mysql_connect("localhost","healhmate","healthmate");
mysql_select_db("hmdb");
if(isset($_POST['username'])) {// because in ajax you send username not id
    $username=$_POST['username'];
    $query=mysql_query("SELECT * from user where id='$username' ");
    $find=mysql_num_rows($query);
   echo $find;
} else {
    echo "-1";
}
?>

And do not use mysql_* functions. They are deprecated. Use mysqli_* functions or PDO.

edited Oct 17, 2013 at 4:07

answered Oct 17, 2013 at 3:59

Gopi Krishna

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