Retain Field Values After Submit

I need the action of this form to call my script, then if the email address is already in the database it should show the alert. This works but obviously I'm directed to my blank

Solution 1:

Yes you could implement (some kind of) a flash session in this case:

So upon submission:

<?php

session_start(); // don't forget
require_once('../scripts/connect.php');

$con = mysqli_connect(DBHOST, DBUSER, DBPASS, DBNAME) or die('Could not connect to database server.');

if(isset($_POST['submit'])) {

    $var_Email1 = mysqli_real_escape_string($con, $_POST['field_email1']);
    $var_Email2 = mysqli_real_escape_string($con, $_POST['field_email2']);

    if ($var_Email1 == $var_Email2){

        $sql = mysqli_query($con, "SELECT * FROM membership WHERE Email = '$var_Email1' "); 
        if(mysqli_num_rows($sql) > 0){
            // set session
            $_SESSION['email_exists'] = $var_Email1;
            header('Location: the_starting_php.php');
            exit;
        }
    }

    echo "not in database";

}
?>

And then in the form page add this as well:

<?php
session_start();

// check if there is
$email = '';
if(isset($_SESSION['email_exists'])) {
    $email = $_SESSION['email_exists'];
    unset($_SESSION['email_exists']); // unset it
    echo "
      <script type='text/javascript'>
      alert('The email address $email already exists');
      </script>
    ";
}

?>


<form action="register_script2.php" method="POST" name="form_register" id="form_registerID" accept-charset="UTF-8">

    <aside class="field_reg_form">
          <input name="field_email1" type="text" required id="field_email1ID" value="<?php echo $email; ?>" />
          <br /><br />
          <input name="field_email2" type="text" required id="field_email2ID" value="<?php echo $email; ?>" />
          <br /><br />
          <input type="submit" value="submit" id="submit" name="submit" />
    </aside>

</form>

Sidenote: I suggest use prepared statements.

<?php

session_start(); // don't forget
require_once('../scripts/connect.php');

$con = mysqli_connect(DBHOST, DBUSER, DBPASS, DBNAME) or die('Could not connect to database server.');

if(isset($_POST['submit'])) {

    $var_Email1 = $_POST['field_email1'];
    $var_Email2 = $_POST['field_email2'];

    if ($var_Email1 == $var_Email2){

        $sql = 'SELECT * FROM membership WHERE Email = ?'; 
        $select = $con->prepare($sql);
        $select->bind_param('s', $var_Email1);
        $select->execute();
        if($select->num_rows > 0){
            // set session
            $_SESSION['email_exists'] = $var_Email1;
            header('Location: the_starting_php.php');
            exit;
        }
    }

    echo "not in database";

}
?>

Solution 2:

I know you posted code, but I'm still; not sure on workflow. However, I'll give you a "for instance" and see if you can at least run with the idea.

Given your form, also provide the value attributes and pull them in from the $_POST values:

register_form.php

<!-- keeping it brief... -->
<form method="POST">
  <input type="email" name="email1" value="<?= $_POST['email1']; ?>" />
  <input type="email" name="email2" value="<?= $_POST['email2']; ?>" />
  <input type="submit" value />
</form>

Then on the original request, output it:

<!-- ... -->
<?php include('register_form.php'); ?>
<!-- ... -->

Then, within the page you're submitting to also call it (this time it'll be pre-populated by the incoming values):

<!-- ... -->
<?php if (isset($_POST['submit'])) { ... } ?>
<!-- ... -->
<?php include('register_form.php'); ?>
<!-- ... -->

Solution 3:

You can use ajax to send data and then showing the return of your php script, like this :

$.ajax({
    type: 'POST',
    url: 'register_script2.php',
    data: $('form#form_register').serialize()
})
.done(function(msg) {
      if(msg != 'error'){
          alert('The email address '+msg+' is already in our database');
      } else {
          // init your form
      }
});

In your PHP code do this ;

...

if(mysqli_num_rows($sql) > 0 ){
    echo $var_Email1;
} else {
    echo 'error';
}

...