我有PHP代码,它正在创建xml feed并为bootstap创建一个html
<?php
$x = 0;
// get how many feeds are displayed
if (!$_GET) {
$numfeeds = 13;
} else {
$numfeeds = $_GET["numfeeds"];
}
// feed information
$html = "";
$url = " // a url to xml feed \\ ";
$xml = simplexml_load_file($url);
for ($i = $x; $i < $numfeeds; $i++) {
$portitem = "<div class='col-md-4 portfolio-item'>";
if ($x <= 3) echo $portitem;
$image = $xml->channel->item[$i]->children('media', True)->content->attributes();
if (empty($image)) echo 'images/banana.jpg';
$title = $xml->channel->item[$i]->title;
$link = $xml->channel->item[$i]->link;
$pubDate = $xml->channel->item[$i]->pubDate;
$author = $xml->channel->item[$i]->author;
$description = $xml->channel->item[$i]->description;
if (!$x++) echo $portitem;
if ($x % 4 == 0) echo $portitem;
$html .= html_entity_decode("
<br>
$portitem
<div>
<p>
<p>$pubDate, $author</p>
<a href='$link'><H2>$title</H2></a></div> <br>
<a href='$link'>
<img class='img-responsive' src='$image' alt=''>
</a>
<p>$description</p>
</p>
</div>
</div>
<br>");
$x++;
if ($x % 4 == 0) echo '</div>';
}
echo print_r($html);
?>
一切似乎都在起作用,我希望以格式提供te,但它也给了我两个错误:
我该如何处理它们?有快速解决方法吗?