Using variable with php from javascript onsubmit event

Question

I'm trying to use a value from onsubmit() in php.

More specific: An HTML form is being submitted by the client. The onSubmit passes it to a couple of javascript functions. One of these functions is supposed to fetch a list of registered emails from the database to check if it is already used.

<?php
$email='?> frm.add_fd2.value <?php ';
// execute query
$num_rows = mysql_num_rows($result); ?>
var num_rows = <?php echo (is_int($num_rows)) ? $num_rows : 0; ?>;
if (num_rows > 0) {
    // JAVASCRIPT ERROR
}

What exactly is it that you want to know?

Pointy
– Pointy

2011-02-08 14:37:30 +00:00
Commented Feb 8, 2011 at 14:37 — Pointy
– Pointy, Commented Feb 8, 2011 at 14:37

Brad Christie · Accepted Answer · 2011-02-08 14:39:22Z

1

JavaScript is a client-side language. It looks like you're trying to retrieve a value from it through PHP, which will never happen. PHP will be long-executed before JavaScript even comes in to play.

You need to either submit the value to PHP through a form, or use something like an AJAX call and send PHP the value so it can process it after the page has been loaded.

In your case, AJAX appears to be a better option. In which case something like the following should work:

$.getJSON('verify_email.php',{email:frm.add_fd2.value},function(data){
  if (data.matches>0){
    // matches were found
  }else{
    // no matches found
  }
});

then in verify_email.php:

<?php
  // ... establish connection ...
  $email = mysql_real_escape_string($_GET['email']); // note, this should be checked for existance
  $result = mysql_query("SELECT * FROM table WHERE email='{$email}'");
  header('Content-Type: application/json');
  echo '{matches:'.mysql_num_rows($result).'}';
?>

Or something of that sort. (Note I'm using jQuery to make this syntactically easier).

answered Feb 8, 2011 at 14:39

Brad Christie

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2 Comments

Marius Over a year ago

Hi Brad, this seems like it makes a lot of sense. Thanks for pointing me in the right direction!

Brad Christie Over a year ago

@Marius: You're welcome, not a problem. If you need any further help, just leave a comment (or post a new question). ;-)

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