Question

I was trying to make a popup php form and login popup pages and I fortunately I got some notes and resources from the net on how to design and create what I was up to. The only problem now is that whenever I submit the form or the login popup page, the error messages throw the page back from the popup to original page, which is not what I want. I believe there is a way how to do the validation in the same login/ form page that can take place in the same popup window. any idea how to do that??

Looking forward to hearing from you.

if a code or anything related is required, I would post it here and update the original question.

update#1: here is the code. P.S. this code goes between the body tages

<?php

if ($username && $userid){
    echo "You are already logged in as <b>$username</b>. <a href='./member.php'>Click here</a> to go to the member page...or <a href='./logout.php'>Logout</a>";
}
else{
    $form = "<form action='./login.php' method='post' id='loginform'>
    <table>
    <tr>
        <td>Username:</td>
        <td><input type='text' name='user' /></td>
    </tr>
    <tr>
        <td>Password:</td>
        <td><input type='password' name='password' /></td>
    </tr>
    <tr>
        <td></td>
        <td><input type='submit' name='loginbtn' value='Login' /></td>
    </tr>
    <tr>
        <td><a href='./register.php'>Register</a></td>
        <td><a href='./forgotpass.php'>Forgot your password?</a></td>
    </tr>
    </table>
    </form>";

    if ($_POST['loginbtn']){
        $user = $_POST['user'];
        $password = $_POST['password'];

        if ($user){
            if ($password){
                require("connect.php");

                $password = md5(md5("kj87fiJAR46ufj".$password."Fj754456fj"));
                // make sure login info correct
                echo"$password";
                $query = mysql_query("SELECT * FROM users WHERE username='$user'");
                $numrows = mysql_num_rows($query);
                if ($numrows == 1){
                    $row = mysql_fetch_assoc($query);
                    $dbid = $row['id'];
                    $dbuser = $row['username'];
                    $dbpass = $row['password'];
                    $dbactive = $row['active'];

                    if ($password == $dbpass){
                        if ($dbactive == 1){
                            // set session info
                            $_SESSION['id'] = $dbid;
                            $_SESSION['username'] = $dbuser;

                            echo "You have been logged in as <b>$dbuser</b>. <a href='./member.php'>Click here</a> to go to the member page.";

                        }
                        else
                            echo "You must activate your account to login. $form";
                    }
                    else
                        echo "You did not enter the correct password. $form";
                }
                else
                    echo "The username you entered was not found. $form";

                //mysql_close();
            }
            else
                echo "You must enter your password. $form";
        }
        else
            echo "You must enter your username. $form";
    }
    else
        echo $form;
}
?>

and there is a code for the popup.

update#2: If found a similar tut that show how to build a complete login page with popup so I;m going to start from there: enter link description here

Thanks,

Answer 1

Use ajax to do so.

For the login example:

send and ajax request with the credentials

do an action via javascript, depending on the result received by the server. For example, if request result is false show an error, if result is true close the popup and redirect the parent window if needed.

Just an example (neither tested nor working) to guide you do this

1) Include the ajax requests on the js code included in your popup

//create the request object
if (window.XMLHttpRequest) {
  request = new XMLHttpRequest();
}
else if (window.ActiveXObject) {
  request = new ActiveXObject("Microsoft.XMLHTTP");
}

//send the request (you can do it via post or get)
if (request) {
  request.onreadystatechange = checkAjaxResponse;
  request.open("POST", "your_server_file_that_generates_ajax_response.php", true);
  request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
  request.send("post params you want to include like param1=value");
}

/**
* Ajax callback function, this should handle your ajax response
*/
function checkAjaxResponse() {

  //check what was de ajax response
  if (request.readyState == 4) {
    if (request.status == 200) {
      xml_response = request.responseXML;   //you can do this with xml, json, ...
    }
  }

  //manipulate your response as needed, and take actions depending on if result is correct or not
  if (manipulated_data_from_the_response !== 'the answer you specified as right') {
    close_popup();
  }
  else {
    show_wrong_credentials_message();
  }
}

2) Implement your php file (or preferred server side languaje) to login the user and generate the http response
//your_server_file_that_generates_ajax_response.php

<?php

  function authenticateUser($user, $password) 
  {
    $query = "SELECT user_id FROM users_table WHERE user = $user AND password = $password LIMIT 1";

    //connect to your database, then make the query 
    $result = mysql_query($query);
    $row = mysql_fetch_row($result);

    //return the response you want    
    if ($row['user_id']) {
      $res = true; 
    }
    else {
      $res = false;
    }

    return $res;
  }


  //main actions
  echo authenticateUser($_POST['user'], $_POST['password']);

?>