Vimeo API Integration

我正在尝试创建一个使用 viemo api 搜索视频，然后以 xml 格式输出它们的网站。我有一个良好的开端，但现在我被困住了。用户输入正在发布，但火虫控制台上没有显示任何结果...

以下是我对vimeo api集成.php：

    <?
include('connect.php');
$video_id= $_POST['text'];
$url = 'http://vimeo.com/api/rest/v2';
$url .= '?';
$url .= 'method=vimeo.videos.search&';
$url .= 'oauth_consumer_key='.$api_key2.'&';
$url .= 'per_page=10&';
$url .= 'query='.$video_id.'&';
$url .= 'sort=relevant&';
$url .= 'full_response=1';
$ch = curl_init();
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_URL, $url);
$curl_response = curl_exec($ch);
curl_close($ch);

$xmlObject = simplexml_load_string($curl_response);
$title= 'title';
$id = 'id';
$video_url = 'videosurl';
$thumbnails = 'thumbnail[1]';
$outputXML .= "";
$outputXML .= "<rsp>'n";
foreach($xmlObject->videos as $video) {
    $videoTagBegin = "'t<video>'n";
    $urlXML = "'t't<videosurl>".$video->attributes()->$video_url."</videosurl>'n";
    $titleXML = "'t't<title>".$video->attributes()->$title."</title>'n";
    $thumbXML = "'t't<thumbnail>".$video->attributes()->$thumbnails."</thumbnail>'n";
    $videoTagEnd = "'t</video>'n";
    $outputXML .= $videoTagBegin.$titleXML.$thumbXML.$urlXML.$videoTagEnd;
}
$outputXML .= "</rsp>";
print $outputXML;
?>

任何帮助将不胜感激。我完全被困住了。

这是我的JS，如果它有帮助的话....

function closeDivs(e) {
    e.preventDefault();
    $('').empty();
};
$(document).ready(function() {
    $('#searchbtn').bind('click' || 'enter',function(e) {
        if ($.trim($('#searchBox').val()) !== '') {
            $('#videos').empty();
            closeDivs(e);
            $('#videos').append('<img src="img/loading.gif" alt="loading" class="loading" />');
            getEvents(e);
            getVideos(e);
        }
    });
function getEvents(e) {
    e.preventDefault();
    var text = 'event_id='+$('#searchBox').val();
    $.ajax({
        url: 'getEvents.php',
        dataType: 'xml',
        type: 'POST',
        data: text,
        success: function(data) {                   
            },
        error: function(data) {
                console.log('Error: ' + data);
        }
    })
};
function getVideos(e) {
    e.preventDefault();
    var text = 'video_id='+$('#searchBox').val();
    $.ajax({
        url: 'getVideos.php',
        dataType: 'xml',
        type: 'POST',
        data: text,
        success: function(data) {                               
        },
        error: function(data) {
                console.log('Error: ' + data);
        }
    })
};
});