PHP将数据回显到HTML下拉菜单中

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>';

        }
    }
}

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>';

From the top in your php:

// not seeing uses of the $form I removed it from my answer

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

$workshop = $_POST['workshop'];
$lowerWorkshop = strtolower($workshop);
// neither of $_POST['Forex'] nor $_POST['Binary'] are defined in your POST. you could as well remove those lines?    

//Retrieve Binary Workshops HERE we can define the sql in a single line:
$sql = "SELECT id, course, location FROM courses WHERE course LIKE '%$workshop%' OR  course LIKE '&$lowerWorkhop%'";       
$query = mysqli_query($link, $sql); // here no need to have two results

// Here lets build our select first, we'll echo it later.
$select = '<select id="$workshop" name="$workshop" style="display: none">';
while($result = mysqli_fetch_assoc($query)){
    $select.= '<option value="' . $result['id'] . '">' . $result['course'] . '</option>';
// here I replaced the outer containing quotes around the html by single quotes.
// since you use _fetch_assoc the resulting array will have stroing keys. 
// to retrieve them, you have to use quotes around the key, hence the change
    }
$select.= '</select>';
}
echo $vSelect;

this will output a select containing one option for each row returned by either of the queries. by the way this particular exemple won't echo anything on screen (since your select display's set to none). but see the source code to retrieve the exemple.