Question

I’m working on validating an image submission form with jQuery. I need it to check if there is an instance of the given submission url in the database to reduce reposts. I used the following code:

$("#submit_form").validate({
  rules: {
   right_form_title: {
   required: true,
   maxlength: 105
  },
 right_form_url: {
  required: true,
  image: true,
  gif: true,
  remote: "/~lyons/imagecheck.php"

 }
 },
 messages: {
   right_form_url: {
   remote: "That image has already been submitted under that URL."
   }
  }       
 });
});

Imagecheck.php:

<?php

//Database Information
$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database
mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

$title = mysql_real_escape_string($right_form_url);

$query = "SELECT filename FROM images WHERE filename='$title';";
$res = mysql_query($query);
if (mysql_num_rows($res) > 0) {
return false;
} else {
return true;
}
echo json_encode($output);


?>

The problem is that now, no matter what, everything comes back as having been submitted before. What is wrong with this? I have literally no clue after looking for awhile.

Answer 1

When you do return true and return false, these values are only in the php environment (server side). Thus, the value that you get in jquery is the

echo json_encode($output)

This will always evaluate to false.