Question

I’ve created a php form to insert values into a database.
One of my form options is a dynamic list populated with fields from another table.
I first created the form without the dynamic option, and all data inserted just fine (and still does).
Now I’m attempting to include the code below, and while it displays the option values properly, the value fails to insert. Any advice?

        <?php
/*
 * LIST ALL CATEGORIES
 ****************************************/
 include('../dbconnection.php');

$query = 'SELECT category_id, category_name FROM ingredient_categories';
$result = mysql_query($query);

echo '<select>';
while ($ingredientCategoryOption = mysql_fetch_array($result)) {
  echo '<option value="'.$ingredientCategoryOption[category_id].'">'.$ingredientCategoryOption[category_name].'</option>';
}
echo '</select>';
?>

Answer 1

I had created something similar yesterday. The $polls array is passed to the view in CodeIgniter in the $this->load->view('poll.php', $data['polls']), while you do it in the page itself. However, you can have the general idea.

<FORM id="formPoll" class="question" name="createpoll" action="<?php echo base_url()?>index.php/poll/selectOption/" method="POST">
     <select name="poll_list">
        <?php
        foreach($polls as $poll){
            echo "<option name='poll_table'>$poll->name</option>";
        }
        ?>
    </select>
     <div id="input">
        Poll name: <input type="text" name="name"></input>
        Title: <input type="text" name="title"></input>
     </div>
     <div id="options">
        Option: <input type="text" name="option"></input>
     </div>

     <input type="submit"></input>
</FORM>

Some ideas:

1. Check if $results is not empty before using it
2. Give your <select> form a name, as I showed above
3. Check if $ingredientCategoryOption is not null or is something returned.
4. Check your database connection