如何在php中将JSON源转换为rss

lf3rwulv  于 2022-12-28  发布在  PHP
关注(0)|答案(1)|浏览(174)

我想在php中将thisJSON Feed转换为RSS,然后获得以下信息
titledateimages
我得到了这样的数据,它给了我一个数组

<?php
function convert_jsonfeed_to_rss($content = NULL, $max = NULL){

 $jsonFeed = json_decode($content, TRUE);
 echo "<pre>";
 print_r($jsonFeed);
 echo "</pre>";

}
$content = @file_get_contents("shorturl.at/gDHY9");
convert_jsonfeed_to_rss($content);

请教育我在正确的方向to get the rss feed from the given json,我不知道如何做到这一点谢谢

iszxjhcz

iszxjhcz1#

下面是一个PHP函数的示例,您可以使用它将JSON提要转换为RSS提要,此函数将JSON提要的URL作为输入,并将其转换为RSS提要:

function convert_json_to_rss( $json_feed_url ) {
    // Parse the JSON feed
    $json_feed = file_get_contents( $json_feed_url );
    $json      = json_decode( $json_feed );

    // Create the RSS feed object
    $rss = new SimpleXMLElement( '<rss version="2.0"></rss>' );

    // Add the channel element to the RSS feed
    $channel = $rss->addChild( 'channel' );

    // Add the channel data to the RSS feed
    $channel->addChild( 'title', $json->title );
    $channel->addChild( 'description', $json->description );
    $channel->addChild( 'link', $json->link );

    // Add the items to the RSS feed
    foreach ( $json->items as $item ) {
        $rss_item = $channel->addChild( 'item' );
        $rss_item->addChild( 'title', $item->title );
        $rss_item->addChild( 'description', $item->description );
        $rss_item->addChild( 'link', $item->link );
    }

    // Output the RSS feed as a string
    header( 'Content-Type: application/rss+xml' );
    echo $rss->asXML();
}

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