It's pretty easy actually
.
I'm running a bit late today, so ill just give you pseudo-code.
Danshadow wrote:
$array = LOAD_DB_ROW_DATA;
echo "<form action='PAGE.PHP' method='post'>";
echo "<select name='FORM SUBMISSION POST NAME'>";
for ($i = 0; $i < count($array); $i++) {
echo "<option value='" . $array[$i] . "'>";
echo $array[$i];
echo "</option>";
}
echo "</select>";
echo "<input type='Submit' value='Submit Selection!'>";
echo "</form>";
echo "<br><br>";
if (!empty($_POST["FORM SUBMISSION POST NAME"])) {
echo "<input type='text' value='" . $_POST["FORM SUBMISSION POST NAME"] . "'>";
}
You load the DB row values into an array, populate a dropdown menu with the items (doesnt have to be a dropdown menu), and add a submission form.
The bottom checks if a post value has been submitted, and if it has it will set the value of the text box with the selected "option" in the previous form.
Hope that helps.