我在 php 中创建 RSS FEED，并且提要也显示在页面上，但他们现在被评论爱板所困扰，我该如何修复它

Question

在此代码中，当我$description=$row['description'];本节发表评论但不通过评论获取时会显示提要 爱板请帮助http://ravisah.in/feed这是我的代码请帮助

<?php
        $host = 'localhost';
        $user = 'root';
        $password = 'Qwaszx12';
        $dbname = 'ravisah';
        $con=mysqli_connect($host,$user,$password,$dbname);
    if (mysqli_connect_error()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $sql = "SELECT * FROM posts ORDER BY id DESC";
    $query = mysqli_query($con, $sql) or die('error');
    @header("Content-type: text/rss+xml");
    echo "<?xml version='1.0' encoding='UTF-8'?>
            <rss version='2.0'>
            <channel>
            <title>Freelance Web Designer in Delhi, India - Ravi Sah</title>
            <link>http://www.ravisah.in</link>
            <description>I am Delhi based responsive website designer and website developer. I am passionate about designing and worked with various clients</description>";
    while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
        $title=$row['title'];
        $link=$row['slug_url'];
        $description=$row['description'];
        echo "<item>
                <title>$title</title>
                <link>http://ravisah.in/blog/$link</link>
                <description>$description</description>
             </item>";
    }
    echo "</channel>";
    echo "</rss>";
    ?>

Answer 1

RSS提要，用于简单地以XML格式发布任何数据。在上面的代码中没有提到"爱情板"。您可以使用 like 运算符从数据库中获取查询。

您有更改数据库查询，例如

$sql = "SELECT * FROM posts where description like '%love boards%' ORDER BY id DESC";
$query = mysqli_query($con, $sql) or die('error');