PHP RSS 提要帮助获取要显示的 XML 标签

如何解决PHP RSS 提要帮助获取要显示的 XML 标签

我正在使用以下代码为我的网站生成 RSS 提要。运行代码时，出现错误：Error on line 1 at column 14: XML declaration allowed only at the start of the document。如果我取出 header()，页面会运行，但数据格式不正确，包括整个数据数组。当我从我的 WordPress 站点请求 RSS 提要时，我得到了格式正确的 XML 响应。对我缺少的东西有什么想法吗？

谢谢，

达雷尔

代码：

if($result){
       //header( "Content-type: text/xml");
        echo "<?xml version='1.0' encoding='UTF-8'?>
        <rss version='2.0'>
       <channel>
       <title>Mom Minder | RSS</title>
       <link>www.darrellh89.com</link>
       <description>Mom Minder Notes</description>
       <language>en-us</language>";

   
    while($row = mysqli_fetch_array($result)){
        //print_r($row);
        $rssTitle=$row["Title"];
        $rssNote=$row["Note"];
        $rssDate=$row["postDate"];
        $rssTime=$row["postTime"]; 
      
       echo "<item>
        <title>$rssTitle</title>
        <note>$rssNote</note>
       <postdate>$rssDate</postdate>
       <posttime>$rssTime</posttime>
       </item>";
   }
   echo "</channel></rss>";

    } 

错误响应：源代码看起来正确，带有...

www.darrellh89.com Mom Minder Notes en-us Did you brush your teeth? 2020-12-21 08:30:00 What restaurant do you want to eat at? 2020-12-21 04:56:05 I'll pick you up at 5 2020-12-21 04:56:05 I'll pick you up at 4PM on Wednesday for the foot doctor 2020-12-22 04:56:05 Did you eat all your eggs? 2021-01-03 12:06:00

我想要的

<channel>
    <title>Adventure Scuba</title>
    <atom:link href="https://ohioadventurescuba.com/feed/" rel="self" type="application/rss+xml" />
    <link>https://ohioadventurescuba.com</link>
    <description>Individual Instruction on Your Schedule</description>
    <lastBuildDate>Sun,20 Dec 2020 15:37:02 +0000</lastBuildDate>