PHP将数据回显到HTML下拉菜单中

Question

I have a HTML etc.. tags now what I want to achieve is upon a selection of ie. i want to load the related info from database to in a new tag with as many tags.

I am using PHP to do achieve this now at this point if for example i choose option1 then the query behind it retrieves relevant information and stores it in a array, and if I select option2 exactly the same is done.

The next step I made is to create a loop to display the results from array() but I am struggling to come up with the right solution to echo retrieved data into etc. As its not my strongest side.

Hope you understand what I am trying to achieve the below code will clear thing out.

HTML:

<select id="workshop" name="workshop" onchange="return test();">
                <option value="">Please select a Workshop</option>
                <option value="Forex">Forex</option>
                <option value="BinaryOptions">Binary Options</option>
            </select>

PHP:

   $form = Array();

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

$form['workshop'] = $_POST['workshop'];
$form['forex'] = $_POST['Forex'];
$form['binary'] = $_POST['Binary'];


//Retrieve Binary Workshops 
if($form['workshop'] == 'Forex'){
    $sql2 = "SELECT id, course, location FROM courses WHERE course LIKE '%Forex%' OR  course LIKE '&forex%'";
        $query2 = mysqli_query($link, $sql2);
            while($result2 = mysqli_fetch_assoc($query2)){


               //The problem I am having is here :/ 

            echo "<select id='Forex' name='Forex' style='display: none'>";
            echo "<option value='oko'>.$result[1].</option>";
            echo "</select>";
            print_r($result2);echo '</br>';
        }
    }else{
        $sql = "SELECT id, course, location FROM courses WHERE course LIKE '%Binary%' OR course LIKE '%binary%'";
        $query = mysqli_query($link, $sql);
            while($result = mysqli_fetch_assoc($query)){
                print_r($result);echo '</br>';

        }
    }
}

Answer 1

Try this code:

$query2 = mysqli_query($link, $sql2);

echo "<select id='Forex' name='Forex' style='display: none'>";
while($result2 = mysqli_fetch_assoc($query2)){
  echo "<option value='oko'>{$result['course']}</option>";

}
echo "</select>";
echo '</br>';