Php Javascript Youtube API error

Question

我有这个代码，它在 Html 代码的div 上添加 YouTube 频道的视频（和视频信息）（_POST 美元）：

    var args= "url="+urlchaine;
    xhr_object.open("POST", "traitement.php", true);
xhr_object.onreadystatechange = function() { 
    if(xhr_object.readyState == 4) {
            eval(xhr_object.responseText);
        }
        return xhr_object.readyState;
}
xhr_object.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhr_object.send(args);

叛徒.php ：

<?php
 if ( isset($_POST["url"]) && !empty($_POST["url"]) )
$urlchaine = $_POST["url"];
 else
$urlchaine = null;
 $stringresult = str_replace("http://www.youtube.com/user/", "",$urlchaine);
 require_once "Zend/Loader.php";
 Zend_Loader::loadClass('Zend_Gdata_YouTube');
 $yt = new Zend_Gdata_YouTube();
 //Get Video info with channel name $stringresult
 $videoFeed = $yt->getVideoFeed('http://gdata.youtube.com/feeds/users/...');
 if ( $stringresult != null){
  echo "var mydiv = document.getElementById('vids');";
  echo "var newcontent = document.createElement('div');";
  foreach ($videoFeed as $v): $thumbs = $v->getVideoThumbnails();
    $videoId = $v->getVideoId();
    $thumb = $thumbs[0]['url'];
    $videoViewCount = $v->getVideoViewCount();
    $videoTitle = $v->getVideoTitle();
echo "newcontent.innerHTML = 
 '<div class="videos">' +
 ' <div class="img_videos">' +
 '  <img class="img_video" width="250" idvideo="".$videoId."" ' +
 '  src="".$thumb.""/>' +
 ' </div>' +
 ' <h3>$videoTitle</h3>' +
 ' <p>$videoViewCount views</p>' +
 '</div>' ;";
echo "mydiv.appendChild(newcontent.firstChild);";
endforeach;
?>

问题是，当我想这样做时，它与某些通道完美配合，而其他通道则存在错误（未捕获的语法错误：意外标识符）。经过几次测试，我发现通过删除$ videoTitle的显示，每个频道测试都有效。我的代码有什么问题？O.o

Answer 1

在某些情况下$videoTitle似乎包含一个简单的引号，这会破坏您的JS代码。

尝试用这样的东西来转义这个引号：

str_replace("'", "‚", $videoTitle)