如何在PHP中将RSS feed转换为JSON

Question

我正在尝试将rss供稿转换为JSON对象，在转换为JSON对象方面取得了一些成功，但是结构很少。

以下是我得到的输出。

{
  "title": "Woot",
  "description": "One Day, One Deal",
  "link": "http:\/\/www.woot.com\/?utm_source=version1&utm_medium=rss&utm_campaign=api.woot.com",
  "item": [
    {
      "title": "Bounty Hunter Snooper II Metal Detector"
    },
    {
      "price": "$64.99"
    },
    {
      "type": "New"
    },
    {
      "title": "GIV Mobile Phones with One Month Unlimited Service"
    },
    {
      "price": "$249.99"
    },
    {
      "type": "Refurbished"
    }
  ]
}

输出我想要的

{
  "title": "Woot",
  "description": "One Day, One Deal",
  "link": "http:\/\/www.woot.com\/?utm_source=version1&utm_medium=rss&utm_campaign=api.woot.com",
  "item": [
    {
      "title": "Bounty Hunter Snooper II Metal Detector",
      "price": "$64.99",
      "type": "New"
    },
    {
      "title": "GIV Mobile Phones with One Month Unlimited Service",
      "price": "$249.99",
      "type": "Refurbished"
    }
  ]
}

我正在使用的代码

<?php
header('Content-Type: application/json');
$feed = new DOMDocument();
$feed->load('RSS Feed Url');
$json = array();

$json['title'] = $feed->getElementsByTagName('channel')->item(0)->getElementsByTagName('title')->item(0)->firstChild->nodeValue;
$json['description'] = $feed->getElementsByTagName('channel')->item(0)->getElementsByTagName('description')->item(0)->firstChild->nodeValue;
$json['link'] = $feed->getElementsByTagName('channel')->item(0)->getElementsByTagName('link')->item(0)->firstChild->nodeValue;

$items = $feed->getElementsByTagName('channel')->item(0)->getElementsByTagName('item');

$json['item'] = array();
$i = 0;


foreach($items as $item) {

   $title = $item->getElementsByTagName('title')->item(0)->firstChild->nodeValue;
   $description = $item->getElementsByTagName('description')->item(0)->firstChild->nodeValue;
   $purchaseurl = $item->getElementsByTagName('purchaseurl')->item(0)->firstChild->nodeValue;
   $standardimage = $item->getElementsByTagName('standardimage')->item(0)->firstChild->nodeValue;
   $shipping =      $item->getElementsByTagName('shipping')->item(0)->firstChild->nodeValue;
   $price =         $item->getElementsByTagName('price')->item(0)->firstChild->nodeValue;
   $condition  =    $item->getElementsByTagName('condition')->item(0)->firstChild->nodeValue;
   $guid = $item->getElementsByTagName('guid')->item(0)->firstChild->nodeValue;


   $json['item'][$i++]['title'] = $title;
   $json['item'][$i++]['description'] = $description;
   $json['item'][$i++]['purchaseurl'] = $purchaseurl;
   $json['item'][$i++]['image'] = $standardimage;
   $json['item'][$i++]['shipping'] = $shipping;
   $json['item'][$i++]['price'] = $price;
   $json['item'][$i++]['type'] = $condition;
   $json['item'][$i++]['guid'] = $guid;  

}


echo json_encode($json);

?>

让我知道您的想法，预先感谢！

Answer 1

用这个测试

尝试这种方式进行项目迭代

foreach($items as $item) {

  $json['item'][] = array("title"=>$title,"price"=>$price,"description"=>$description)
}
echo json_encode($json)

之所以会这样，是因为如果您在最后给出$json['item'][$i]并且$i=$i+1话，您就为每个数组索引都给出了$i++ 。

所以这也会起作用

$i=0;
 foreach($items as $item) {

  $json['item'][$i] = array("title"=>$title,"price"=>$price,"description"=>$description)
  $i=$i+1; 
 }
echo json_encode($json)

第三种方法

$json['item'][$i++]['guid'] = $guid;

仅将$i++应用于最后一个元素，将$i应用于其余元素，因为$i++是发布增量