Question

I have a variable called $position which has a value selected from a database.
I am having an issue displaying the $position value in my dropdown menu as the selected value on page load.
The $position variable does echo out the correct value outside of this code.

I read about the Ternary Operator method so I have used that here instead of normal IF statment but it still does not work. Is my code correct? Is there another method I should use?

Thanks

My Code

$number = $_GET['id'];
$number = mysqli_real_escape_string($connect,$number);

$stmt = $connect->prepare("SELECT s_code, s_type, s_position, s_idno, s_firstnames, s_email, s_cellno FROM smco WHERE s_ainumber = ?") or die(mysqli_error());

$stmt->bind_param('s', $number);

$stmt->execute();

$stmt->bind_result($code, $type, $position, $idno, $name, $email, $cell);

$stmt->fetch();

$stmt->close();

echo "<td width=\"62%\" align=\"left\">
<p><b>Position:</b></p>
<select name=\"position\">
<option value=\"AQ\" (($position == 'AQ') ? \"selected='selected'\")>Account Queries</option>
<option value=\"KM\" (($position == 'KM') ? \"selected='selected'\")>Key Account Manager</option>
<option value=\"MD\" (($position == 'MD') ? \"selected='selected'\")>Managing Director</option>
<option value=\"RB\" (($position == 'RB') ? \"selected='selected'\")>Rebates</option>
<option value=\"BY\" (($position == 'BY') ? \"selected='selected'\")>Store Buyer</option>
<option value=\"OW\" (($position == 'OW') ? \"selected='selected'\")>Store Owner</option>
</select>
</td>";

Answer 1

echo '<option value="">' . (true ? 'true' : 'false') . '</option>';

In you case:

echo "<option value=\"AQ\"" . (($position == 'AQ') ? ' selected="selected"' : '') . ">Account Queries</option>";