Question

I am working from this example: http://jqueryui.com/demos/autocomplete/#remote and I am encoding the output like this:

    $rows = array();
    while($r = mysql_fetch_assoc($category_result))
    {

        $rows[] = $r;
        error_log ("rows: ".$rows[0]);
    }

    echo json_encode($rows);

But the dropdown on the other side shows nothing. Here is my test page: http://problemio.com/test.php – if you enter “ho” it matches 2 results in the database, but they are not getting displayed for some reason. Any idea why?

Thanks!!

Answer 1

The properties should be named label and value. From the JQuery UI demo page you linked to:

The local data can be a simple Array of Strings, or it contains
Objects for each item in the array, with either a label or value
property or both. The label property is displayed in the suggestion
menu.

So you would need to rename category_name to label either in PHP or later on in your JavaScript source handler function. The latter would require you to replace the PHP URL with a callback function like in the remote example. That way you could get the data any way you want (e.g. by jQuery.getJSON()) and work with it before it gets handed over to the suggestion box.

Hope this helps.

Regarding your comment, this should do it:

$rows = array();
while ($r = mysql_fetch_array($category_result)) {
   $rows[] = array("label" => $r["category_name"]);
}
echo json_encode($rows);