Pop up message after successful registration

Question

I want to have a pop-up message when the user has successfully submitted a registration form. But being a newbie in PHP, I'm having a hard time with some validations. Need help :( Here's my code.

my buttons here <div class="row">
<div class="col-lg-12" style="text-align: center;">
<input class="btn btn-success" type="submit" value="<?php echo __('Register'); ?>"/>
<input class="btn btn-warning" type="reset" value="<?php echo __('Reset');?>"/>
<input class="btn btn-danger" type="button" value="<?php echo __('Cancel'); ?>" onclick="javascript:
    window.location.href='index.php';"/>
</div>

enter code <div class="modal fade success-popup" id="success" tabindex="-1" role="dialog" aria-labelledby="modal-label" style="transition: .3s ease all;"><div class="modal-dialog modal-sm" role="document">
<div class="modal-content">
  <div class="modal-header">
    <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">x</span></button>
    <h4 class="modal-title" id="modal-label">Thank You!</h4>
  </div>
  <div class="modal-body text-center">
    <i class="fa fa-check" aria-hidden="true" style="border-radius: 50%; font-size: 35px;"></i>
    <p class="lead">Registration form successfully submitted.</p>
    <a href="index.php" class="btn btn-default" style="text-decoration: none;">Go To Home</a>
  </div>
</div>

The modal is what I want to show after the code is validated and the user submitted the form successfully.

Welcome to Stack Overflow! Can you elaborate on how your code "doesn't work"? What were you expecting, and what actually happened? If you got an exception/error, post the line it occurred on and the exception/error details. Please [edit] these details in or we may not be able to help. — Blue, Jun 20 '17 at 02:29

score 0 · Answer 1 · answered Jun 20 '17 at 04:29

There are various methods to show the popup after registration. You can set it on the session or you can return '1' for the successful registration in your php file.
//example 1
if(insert_query){
return '1';
}

//example 2
if(insert_query){
 session_start();
 $_SESSION['registration_success']='1';
}

score 0 · Answer 2 · answered Jun 21 '17 at 09:30

I tell you a standard approach which works everywhere not just your current code so you can master such issues.

In you server side function you should return a flag i.e a number or text for success and error

// success
if($success)
    return 1
else
    return 0

then in you ajax request check what the server response.

You asked about validation but i didn't really get that. Do you mean form validation or something ? lets do some basics:

<?php

if( $_POST['your_input'] == 'your value' )
    return 1
else
    return 0

Where $_POST['your_input'] is the data sent from client and 'your value' is the value to test against.

Your question is somehow vague. I hope it helps and feel free to comment here if you steel need help :)

Pop up message after successful registration

2 Answers2