display SQL data in drop down menu

[fasmy98]

I have a select option in form and stored the value in SQL. I would like to generate the drop down menu with the selected value in a form for update.

<select name="ud_kategori" id="kategori">
<option value="" selected>- - Pilihan kategori - -</option>
<option value="Meyuarat">Mesyuarat</option>
<option value="Seminar">Seminar</option>
<option value="Bengkel">Bengkel</option>
<option value="Kursus">Kursus</option>
<option value="Lain-lain">Lain-lain</option>
</select>

[bogu]

U first need an array with all the list box values, something like:

$list_box = array('Meyuarat', 'Seminar',...);

Then u need to know your selected value:

$selected_value = 'Seminar';

When u have this; u go on generating listbox:

<select name="ud_kategori" id="kategori">
    <option value="">- - Pilihan kategori - -</option>
    <?php
    foreach ($list_box as $value) {
        $selected = '';
        if ($value == $selected_value) $selected = 'selected="selected"';
        echo "<option value=\"{$value}\" {$selected}>{$value}</option>";
    }
    ?>
</select>

and voilà, u should have your list box generated with the option u need, selected ...

[fasmy98]

how if all the values were stored in the database?

how do I get the selected value?

[Sangre]

Then you need to do it like this:

$command = "select * from whatever";
$result = mysql_query($command, $connection) or die(mysql_error());

$html = "<select name=\"boxname\">\n";

while ($nextrecord = mysql_fetch_array($result)) {
	$id = $nextrecord['id'];
	$name = $nextrecord['name'];
	$html .= "<option value=\"$id\">$name</option>\n";
}

$html .= "</select>\n";

the .= makes the $html longer

then all you have to do is post somewhere in your Body:

<? print $html ?> and the dropbox will show