如何使用 PHP 显示来自 RSS 提要的图像

Question

我正在尝试显示来自 rss 提要项目附件的图像。我当前的代码仅使用提要源的 url 作为图像源，因此不显示任何内容。我试图编辑正在寻找图像的行，但我没有成功。有人在 php 中看到错误吗？

这是我要查找图片的地方...

$has_image = preg_match('/<img.+src=[\'"](?P<src>.+?)[\'"].*>/i', $content, $image);

这是显示的地方...

if ($has_image == 1) {
            $description = '<div id="image"><img class="feed-item-image" src="' . $image['src'] . '" /></div>' . $description;
        }

完整 PHP...

function get_rss_feed_as_html($feed_url, $max_item_cnt = 1, $show_date = true, $show_description = true, $max_words = 0, $cache_timeout = 7200, $cache_prefix = "/tmp/rss2html-")
{
    $result = "";
    // get feeds and parse items
    $rss = new DOMDocument();
    $cache_file = $cache_prefix . md5($feed_url);
    // load from file or load content
    if ($cache_timeout > 0 &&
        is_file($cache_file) &&
        (filemtime($cache_file) + $cache_timeout > time())) {
            $rss->load($cache_file);
    } else {
        $rss->load($feed_url);
        if ($cache_timeout > 0) {
            $rss->save($cache_file);
        }
    }
    $feed = array();
    foreach ($rss->getElementsByTagName('item') as $node) {
        $item = array (
            'title' => $node->getElementsByTagName('title')->item(0)->nodeValue,
            'desc' => $node->getElementsByTagName('description')->item(0)->nodeValue,
            'content' => $node->getElementsByTagName('description')->item(0)->nodeValue,
            'link' => $node->getElementsByTagName('link')->item(0)->nodeValue,
            'date' => $node->getElementsByTagName('pubDate')->item(0)->nodeValue,
        );
        $content = $node->getElementsByTagName('encoded'); // <content:encoded>
        if ($content->length > 0) {
            $item['content'] = $content->item(0)->nodeValue;
        }
        array_push($feed, $item);
    }
    // real good count
    if ($max_item_cnt > count($feed)) {
        $max_item_cnt = count($feed);
    }
    $result .= '<ul class="feed-lists">';
    for ($x=0;$x<$max_item_cnt;$x++) {
        $title = str_replace(' & ', ' &amp; ', $feed[$x]['title']);
        $link = $feed[$x]['link'];
        $result .= '<li class="feed-item">';
        $result .= '<div class="feed-title"><strong><a href="'.$link.'" title="'.$title.'">'.$title.'</a></strong></div>';
        if ($show_date) {

            $date = date('l F d, Y', strtotime($feed[$x]['date']));
            $result .= '<small class="feed-date"><em>'.$date.'</em></small>';
        }
        if ($show_description) {
            $description = $feed[$x]['desc'];
            $content = $feed[$x]['content'];
            // find the img
            $found = preg_match('/<img.+src=[\'"](?P<src>.+?)[\'"].*>/i', $content, $match);
            // no html tags
            $description = strip_tags(preg_replace('/(<(script|style)\b[^>]*>).*?(<\/>)/s', "", $description), '');
            // whether cut by number of words
            if ($max_words > 0) {
                $arr = explode(' ', $description);
                if ($max_words < count($arr)) {
                    $description = '';
                    $w_cnt = 0;
                    foreach($arr as $w) {
                        $description .= $w . ' ';
                        $w_cnt = $w_cnt + 1;
                        if ($w_cnt == $max_words) {
                            break;
                        }
                    }
                    $description .= " ...";
                }
            }
            // add img if it exists


                if ($found)
                {
                echo '<div id="image"><img class="feed-item-image" src="' . (string)$match[1] . '" /></div>';
                }
            $result .= '<div class="feed-description">' . $description;

        }
        $result .= '</li>';
    }
    $result .= '</ul>';
    return $result;
}

function output_rss_feed($feed_url, $max_item_cnt = 1, $show_date = true, $show_description = true, $max_words = 100)
{
    echo get_rss_feed_as_html($feed_url, $max_item_cnt, $show_date, $show_description, $max_words);
}

?>

谢谢

Answer 1

试试这个

$found = preg_match("/img src=\"([^\"]+)\"/", $content, $match);

if ($found)
{
    echo '<div id="image"><img class="feed-item-image" src="' . (string)$match[1] . '" /></div>';
}

您的 css 文件和 feed-item-image class 可能有问题。你能试试吗

"<img src=\"" . (string)$match[1] . "\">";

而不是

'<div id="image"><img class="feed-item-image" src="' . (string)$match[1] . '" /></div>'

Answer 2

我无法使用内容元素找到图像文件。我把它一块一块地分解，并在项目描述中找到了它。这不是解决问题的最佳方法，但现在可以使用。

$found = preg_match('/<img.+src=[\'"](?P<src>.+?)[\'"].*>/i', $description, $match);

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