ajax popup

piet bieruik · March 1, 2014

Hi,

Im making an inlog form.
With ajax i want to check if the inlog is correct and display the message.
Im using this script.

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Naamloos document</title>
<style type="text/css">
body{background-color:#CCC;}
#element_to_pop_up {
background-color:#fff;
border-radius:15px;
color:#000;
display:none;
padding:20px;
min-width:400px;
min-height: 180px;
}
.b-close{
cursor:pointer;
position:absolute;
right:10px;
top:5px;
}
</style>
</head>

<body>
<!-- Button that triggers the popup -->
<button id="my-button">POP IT UP</button>
<!-- Element to pop up -->
<div id="element_to_pop_up">
<a class="b-close">x<a/>
<div id="content"></div>
</div>



<script type="text/javascript" src="javascript/Jquery.js"></script>
<script type="text/javascript" src="javascript/bpopup.js"></script>
<script>

(function($) {

// DOM Ready
$(function() {

// Binding a click event
// From jQuery v.1.7.0 use .on() instead of .bind()
$('#my-button').bind('click', function(e) {

// Prevents the default action to be triggered.
e.preventDefault();

// Triggering bPopup when click event is fired
// $('#element_to_pop_up').bPopup({
// speed: 650,
//transition: 'slideIn'

$('#element_to_pop_up').bPopup({
contentContainer:'#content',
speed: 650,
transition: 'slideIn',
loadUrl: 'action.php' //Uses jQuery.load()
});

});

});

})(jQuery);

</script>
</body>
</html>

I need to send the username and pasword to "action.php"

How can i do this ?

Thanks,

Skewled · March 5, 2014

You can do it by posting the data, I don't see an attempt on your part to post the actual data, Are you wanting it written for you? Here is an example of posting data from a form called "login_form" which has 2 input fields of "username" and "password", the var postData = $(this).serialize() builds the array which I would use PHP to grab the $_POST['field'] from.

Typically:

<script type="text/javascript">
   $(function() {

    $("#login_form").submit(function(o) {
      o.preventDefault();
      var url = $(this).attr('action');
      var postData = $(this).serialize();

      $.post(url, postData, function(e) {
        if (o.result == 1) { // We get the result from PHP/SERVER SIDE
           window.location.href = ''; // Where to go on success
        } else {
            // What to do if the login was not found
        }

        }, 'json');

      });
   });
</script>

piet bieruik · March 6, 2014

im using this html form.

<h1>Login</h1>

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" target="_self" title="Registratie">

<label>Email adres: </label><input name="email" type="email" id="email">

<br>

<label>Wachtwoord: </label><input name="wachtwoord" type="password" id="wachtwoord">

<br>

<p><a href="#">wachtwoord vergeten ?</a></p>

<br>

</form>

</div>

the outcome from the message i want to place in the popup.

in action.php i check the username and password

Skewled · March 7, 2014

Ok in action php you'd need to add a return value for either success or failure and json_encode it. So here is an example:

<?php

public function login()
{
     // Handle Your POST input and assign your variables

     // Do your query
     $result = "SELECT * FROM ....";

     if($result)
     {
          // You would compile your json_encoded statement here for success
          json_encode(['result'] => 1]); // This will not work it's just to guide you
          return_false; // exit the function
     } else {
          // You would compile the same as above without returning false
          json_encode(['result'] => 0]); // This will not work it's just to guide you
}

Now you'd edit your form as follows to add a new div to populate the html output:

<div id="login">
                <h1>Login</h1>

    <div id="error"><!--Dynamic jQuery Return --></div> 
                
    <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" target="_self"  title="Registratie">
     <label>Email adres: </label><input name="email" type="email" id="email">
     <br>
     <label>Wachtwoord: </label><input name="wachtwoord" type="password" id="wachtwoord">
     <br>
     <p><a href="#">wachtwoord vergeten ?</a></p>
     <br>
     <input name="login_submit" type="submit" id="login_submit" value="Login">
    
    </form>
</div>

Now all you'd need to do is finish off the JavaScript I posted above in my previous post.

<script type="text/javascript">
   $(function() {
 
    $("#login_form").submit(function(o) {
      o.preventDefault();
      var url = $(this).attr('action');
      var postData = $(this).serialize();
      $("error").hide();
 
      $.post(url, postData, function(e) {
        if (o.result == 1) { // We get the result from PHP/SERVER SIDE
           window.location.href = ''; // Where to go on success
        } else {
            // What to do if the login was not found
            // Populate the div for login error
            $("error").append('Username / Password Invalid.');
        }
 
        }, 'json');
 
      });
   });
</script>

Most of this is pseudo code to help you get on track and most of it is accurate except for how you deal with your database. I hope this helps you and puts you in the right direction.

piet bieruik · March 19, 2014

thx skewled for your help.

But the problems i have you skip.

How to call out the username.

normally its like

$result = select * from user where username = $_POST['user_name'];

but there is nog post, so $_POST['user_name'] would be empty ?

And

in your script there is no popup

or am i missing the point ?

thank you

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