I have a autocomplete function that works fine but where I want to extend the drop down details a bit. It looks like this:
Datafile (form-lookup.php):
if ($db)
{
$fetch = mysqli_query($db,"SELECT * FROM uni_labels where label_name like '%" . $_GET['term'] . "%'");
/* Retrieve and store in array the results of the query.*/
while ($row = mysqli_fetch_array($fetch, MYSQL_ASSOC)) {
$row_array['label'] = htmlspecialchars_decode($row['label_name']);
$row_array['lookupid'] = $row['id'];
$row_array['address'] = $row['label_address'];
$row_array['number'] = $row['label_number'];
$row_array['postalcode'] = $row['label_postalCode'];
$row_array['country'] = $row['label_country'];
array_push($return_arr,$row_array);
}
}
/* Free connection resources. */
mysqli_close($db);
/* Toss back results as json encoded array. */
echo json_encode($return_arr);
My page that retreives the data looks like this:
$(function() {
$("#step1Name").autocomplete({
source: "/pages/form-lookup.php?country=dk",
minLength: 2,
select: function(event, ui)
{
$('#step1Name').val(ui.item.address);
$('#lookupid').val(ui.item.lookupid);
$('#vej').val(ui.item.address);
}
});
});
<input maxlength="100" type="text" required="required" class="form-control input-lg" name="step1Name" id="step1Name" />
Everyting works just great, but I would like for my drop down to show both $row_array['label'] and $row_array['address'], and when I select a value in the drop down, the input box will only output the $row_array['label'] value.
In the datafile I have tried to add the address to the label, like this:
$row_array['label'] = htmlspecialchars_decode($row['label_name'])." - ".$row['label_address'];
This displays the value fine in the drop down, but when selecting the choice, the input box of course contains too much data, where I only want it to display the label_name.
Can someone point me in the right direction?
Thanks in advance.