Question

我在MAMP做了一个测试网站。一切正常。

然而，当我上传它时，我在一个关键页面上收到了很多错误消息。

查看错误消息的链接是：

http://ryanmurphy.org.uk/musicsite/artist.php

我认为查看该页面的代码可能很有用：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Ryan Murphy's Music Website</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<img src="http://upload.wikimedia.org/wikipedia/commons/thumb/0/0a/BBC_Radio_1.svg/175px-BBC_Radio_1.svg.png">
<link rel="stylesheet" href="stylesheet.css" type="text/css" media="screen" />
<img src="http://blogs.howtogeek.com/mysticgeek/files/2008/08/618pxlast.fm-logo.svg.preview.png" style="padding-left:10px;">

</head>
<body>
<h1>Most popular artist this week</h1>
<div id="navbar"><ul>
<li><a href="index.php">Home</a></li>
<li><a href="music.php">Track Playlist from the last week</a></li>
<li><a href="artist.php">Most played Artist from the last week</a></li>
<li><a href="sources.html">Sources Used</a></li>

</ul>
</div>

<table>
<th>Artist</th>
<th>Times played this week</th>
<th>Previous plays</th>

<?php
foreach ($xml->artists->children() as $child)
{
  echo "<tr>";
  $artist = (string)$child->name;
  echo "<td>$artist</td>";
  $playedweek = (string)$child->plays;
  echo "<td>$playedweek</td>";
  $previousplays = (string)$child->previous_plays;
  echo "<td>$previousplays</td>"; 
  echo "</tr>";
}?>
</table>
</body>
</html>

Answer 1

你做这样的方法调用：

$obj->children()

然而，$ obj不是一个对象。请尝试var_dump($obj)查看其中包含的内容。

上传我的网站时混淆错误消息

1 个答案: