将xml加载到php中

我需要从rss.xml提要在我的站点中显示顶级单曲音乐图表（标题名），并且图像显示有问题

<?php
$i=1;
$rss = new DOMDocument();
$rss->load('http://ax.phobos.apple.com.edgesuite.net/WebObjects/MZStore.woa/wpa/MRSS/topsongs/sf=143449/limit=889991/rss.xml');
$feed = array();
foreach ($rss->getElementsByTagName('item') as $node) {
    $item = array ( 
        'title' => $node->getElementsByTagName('title')->item(0)->nodeValue,
        'desc' => $node->getElementsByTagName('description')->item(0)->nodeValue,
        );
    array_push($feed, $item);
}
$limit = 41;
for($x=1;$x<$limit;$x++) {
    $title = str_replace(' & ', ' &amp; ', $feed[$x]['title']);
    $SongTitle = substr($title, 3);
    $SongTerm =  stristr($SongTitle, '-');
    $SongTerm = substr($SongTerm, 2);
    $description = $feed[$x]['desc'];
    $AlbumArt = stristr($description, 'img src="');
    $AlbumArt = substr($AlbumArt, 9);
    $AlbumArt = explode(" ", $AlbumArt);
    $AlbumArt  = $AlbumArt[0];

    echo '<div class="Music-Tumbs"><a href="music.php?search='.$SongTerm.'" title="#'.$i++.'. '.$SongTitle.'"><img src="'.$AlbumArt.' width="92" height="92"></a></div>';

}?>

---

您忘记关闭IMG标记中的SRC

替换此项：

echo '<div class="Music-Tumbs"><a href="music.php?search='.$SongTerm.'" title="#'.$i++.'. '.$SongTitle.'"><img src="'.$AlbumArt.' width="92" height="92"></a></div>';

echo'；

为此：

echo '<div class="Music-Tumbs"><a href="music.php?search='.$SongTerm.'" title="#'.$i++.'. '.$SongTitle.'"><img src="'.$AlbumArt.'" width="92" height="92"></a></div>';

echo'；

它显示了什么错误？？？正如@PavelDenisevich所说，您忘记关闭

src

属性