我正在尝试从 media:content 标签 RSS FEED 中检索图像 URL 并将其存储在 mysql 中，但我已经厌倦了尝试

Question

这是 RSS 提要的一部分

<item>
    <title>
        <![CDATA[
        Japan backs coronavirus drug; Moscow keeps lockdown: Live updates
        ]]>
    </title>
    <description>
        <![CDATA[
        <div><img src="https://www.aljazeera.com/mritems/Images/2020/5/9/b738bc043efb400296eff230bcfd444f_18.jpg" style="width: 100%;"><div>Over 275,000 have died worldwide while more than 1.3 million have recovered as countries weigh easing of lockdown rules.</div></div>
        ]]>
    </description>
    <link>
        https://www.aljazeera.com/news/2020/05/japan-backs-coronavirus-drug-moscow-lockdown-live-updates-200508232257831.html
    </link>
    <guid isPermaLink="false">592c6747e057269acbb31f383c9c110d</guid>
    <dc:creator>
        <![CDATA[ AJENews ]]>
    </dc:creator>
    <pubDate>Sat, 09 May 2020 01:18:50 GMT</pubDate>
    <media:content medium="image" url="https://www.aljazeera.com/mritems/Images/2020/5/9/b738bc043efb400296eff230bcfd444f_18.jpg"/>
</item>

此处通过 DOMDocument() 加载 rss 提要内容并定义数组以保存内容并使用 getElementsByTagName

    <?php
       $rss = new DOMDocument();
        $rss->load('https://rss.app/feeds/7e40Pu6EiBz9ourE.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,
                'link' => $node->getElementsByTagName('link')->item(0)->nodeValue,
                'date' => $node->getElementsByTagName('pubDate')->item(0)->nodeValue,
                //'dcUri' => $rss->lookupNamespaceUri('dc'),
                //'authors' => $node->getElementsByTagNameNS($item[dcUri], 'creator'),
                //'cre' => $item[authors]->item(0)->nodeValue,
                );
            array_push($feed, $item);
        }

在 mysql 数据库中存储数组内容的代码，我已成功保存（标题、des、日期）

    $mysqli = new mysqli("localhost","root","","hmsdb");

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $stmt = $mysqli->prepare("INSERT INTO `rssitems` (`title`, `description`, `link`, `pubDate`) VALUES (?, ?, ?, ?)");

    $stmt->bind_param('ssss', $title, $description, $link, $pubDate);

    foreach( $feed as $RssItem){
        $title = $RssItem["title"];
        $description = $RssItem["desc"];
        $link = $RssItem["link"];
        $pubDate = $RssItem["date"];
        //$creator= $RssItem["cre"];

        $stmt->execute();
    }

Answer 1

这是一个最小的可重现示例，说明如何提取图像 URL。

遍历所有节点
为所有子节点嵌套循环
获取特定类型所需的属性

$rss = new DOMDocument();
$rss->load('https://rss.app/feeds/7e40Pu6EiBz9ourE.xml');
$feed = array();
foreach ($rss->getElementsByTagName('item') as $node) {
    foreach($node->childNodes as $childNode) {
        if($childNode->tagName == 'media:content') {
            var_dump($childNode->getAttribute('url'));
        }
    }
}

string(90) "https://www.aljazeera.com/mritems/Images/2020/5/10/1024c76c23214805817eff50f9eb45c0_18.jpg"
string(90) "https://www.aljazeera.com/mritems/Images/2020/4/24/d8cba9945a604b88a1dbb413197d0db0_18.jpg"
string(90) "https://www.aljazeera.com/mritems/Images/2020/5/10/bb5f767a4baf4d4098cf766f71a7b347_18.jpg"
string(90) "https://www.aljazeera.com/mritems/Images/2020/5/10/6adcf99eaeda40bea68a519d95ad9fa5_18.jpg"
string(90) "https://www.aljazeera.com/mritems/Images/2020/3/23/4cb97ca92a4543799b3b0c83e8d6df3f_18.jpg"

我正在尝试从 media:content 标签 RSS FEED 中检索图像 URL 并将其存储在 mysql 中，但我已经厌倦了尝试

I am trying to retrieve the image URL from media:content tag RSS FEED and storing it in the mysql, but I am tired of trying

php

xml

sql

database

rss