Php Jquery自动完成不显示结果

我正在使用JQuery自动完成，如下所示：

<ul id="searchFilter">
    <li><input name="searchtext" id="searchtext" placeholder="Search..." /><input type="submit" class="prontoSubmit"  id="submit" name="submit" value="Go" />
    </li>
</ul>

它调用search.php，返回基于$_请求['term']的结果，我可以在我的下一个工作调用中看到它，它正在工作并给出预期的结果：

[{"label":"Parkside Semi II, Hamilton, Greenhill Glen","value":"Parkside Semi II"},{"label":"Parkside Semi, Hamilton, Greenhill Glen","value":"Parkside Semi"},{"label":"Parkside, Hamilton, Greenhill Glen","value":"Parkside"},{"label":"Parkview, Beamsville, Discovery","value":"Parkview"}]

但是我的自动完成没有显示结果…我做错了什么

下面是search.php中的代码

<?php

include("classes/FunctionsClass.php");
include("classes/Class.php");

$functionClass = new FunctionsClass;
$class = new Class;

$connection = $functionClass->getConnection();

echo json_encode($class->getSearch($connection, $_REQUEST['term']));

?>

---

了解search.php中的代码。用php代码回答我的问题，这是你的类真正命名的类，还是仅仅是一个例子？仅仅是一个例子我的朋友如果你检查DOM是为它生成的HTML？以防万一一些CSS可能会阻止它显示。

<?php

include("classes/FunctionsClass.php");
include("classes/Class.php");

$functionClass = new FunctionsClass;
$class = new Class;

$connection = $functionClass->getConnection();

echo json_encode($class->getSearch($connection, $_REQUEST['term']));

?>

function getSearch($connection, $term){
                $query = mysqli_query($connection, "SELECT home_title, rb_locations.locationLabel, rb_communities.rb_communityLabel FROM readyBuilt INNER JOIN rb_communities ON rb_communities.rb_communityId = readyBuilt.home_community INNER JOIN rb_locations ON rb_locations.locationId = readyBuilt.home_location WHERE home_title LIKE '" . $term . "%'");
                $results = array();
                while($row = mysqli_fetch_array($query)){
                        $results[] = array(
                                'label' => $row['home_title'] . ', ' . $row['locationLabel'] . ', ' . $row['rb_communityLabel'],
                                'value' => $row['home_title']
                        );
                }
                return $results;
        }