Jquery input form will not pass variable when form submitted

Question

I'm having trouble getting my variable to post in my form. I use jquery to load a box if the user wants to enter a new sub category.

Here is the select in the form

<select id="sub_cat1" name="sub_cat1">
   <option value=""></option>
   <option value="new">New Sub Cat</option>
<?php
$query = "SELECT sub_category FROM blog_posts GROUP BY sub_category";
$data = mysqli_query($dbc, $query);
    if (mysqli_num_rows($data) > 0) {
      while ($row = mysqli_fetch_array($data)) {
    $sub_category = $row['sub_category'];
    echo '<option value="' . $sub_category . '">' . $sub_category . '</option>';
    }
     }
?>
        </select>
<div id="load_sub_cat_box"></div>

This is the jquery

$(document).ready(function(){ 
$("#sub_cat1").on("keyup change", function() {  
  var prod = $("#sub_cat1").val();
  if(prod == 'new'){
  $("#load_sub_cat_box").append('<div style="padding-top:15px;">Enter New Sub Category</div><input type="text" name="sub_cat2" />');
  }
    }); 
});

And here is the form validate code

  if($_POST['sub_cat1'] == "new"){
$sub_cat = mysqli_real_escape_string($dbc, trim($_POST['sub_cat2']));
  }
  else{
$sub_cat = mysqli_real_escape_string($dbc, trim($_POST['sub_cat1']));  
  }

If I select new, the box does load on the page but for some reason when form is submitted it will not load a variable for $sub_cat1. Is there something I don't know about jquery and forms?

kufudo · Accepted Answer · 2013-02-14 02:34:17Z

2

You have two select elements with id="sub_cat1".

ids must be unique per page.

It is also invalid HTML, with a select inside a select.

answered Feb 14, 2013 at 2:34

kufudo

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