Question

如果有人将链接输入到文本输入字段并按“提交”以获取实时内容，我如何将RSS提要保存到数据库？对不起我的英语很差

<?php
$url = "";
if(isset($_POST['submit'])){
    if($_POST['feedurl'] != ''){
        $url = $_POST['feedurl'];
    }
}

$invalidurl = false;
if (@simplexml_load_file($url)) {
    $feeds = simplexml_load_file($url);
} else {
    $invalidurl = true;
    echo "<h2>Invalid RSS feed URL.</h2>";
}


$i=0;
if (!empty($feeds)) {
    $site = $feeds->channel->title;
    $sitelink = $feeds->channel->link;

    echo "<h1>".$site."</h1>";
    foreach ($feeds->channel->item as $item) {
        $title = $item->title;
        $link = $item->link;
        $description = $item->description;
        $postDate = $item->pubDate;
        $pubDate = date('D, d M Y',strtotime($postDate));

        if($i>=5) break;
?>
<div class="post">
    <div class="post-head">
        <h2><a class="feed_title" href="<?php echo $link; ?>"><?php echo $title; ?></a></h2>
        <span><?php echo $pubDate; ?></span>
    </div>
    <div class="post-content">
        <?php echo implode(' ', array_slice(explode(' ', $description), 0, 20)) . "..."; ?> <a href="<?php echo $link; ?>">Read more</a>
    </div>
</div>
<?php
$i++;
   }
} else {
    if (!$invalidurl) {
        echo "<h2>No item found</h2>";
    }
}
?>

我应该如何修改它才能完成它？上面的代码完成了从已发布的RSS URL获取内容并将其显示给用户的事情。

Answer 1

这段代码只是为了让你开始，考虑这是执行与你拥有的项目一样多的插入，一个更好的替代方法是做insert with multiple values

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$stmt = $conn->prepare('INSERT INTO RSS_FEED (title, item, link, descripton, pub_date) VALUES (?, ?, ?, ?, ?)');

$values = array();
foreach ($feeds->channel->item as $item) {
  $stmt->bind_param(
    'sssss',
    $item->title,
    $item->link,
    $item->description,
    date('D, d M Y',strtotime($item->pubDate))
  );
  $stmt->execute();
}
$stmt->close();
$conn->close();

了解更多details click here

Answer 2

这取决于您想要在数据库中采用的方法，如果您不需要在Feed中搜索特定元素，您甚至可以将整个内容保存为单个blob

如何使用Php从粘贴的URL保存rss feed到数据库？

2 个答案: