下面的脚本获取 URL 列表中的元数据。
URL是在我的前端输入的,我设法将数据获取到另一个页面(此脚本),但现在不是将表格回显到脚本所在的同一页面上,而是我想将该数据反馈给我的前端并将其放在一个漂亮的桌子上供用户查看。
如何让 php 脚本在另一个页面上回显数据?
谢谢
瑞奇
<?php
ini_set('display_errors', 0);
ini_set( 'default_charset', 'UTF-8' );
error_reporting(E_ALL);
//ini_set( "display_errors", 0);
function parseUrl($url){
//Trim whitespace of the url to ensure proper checking.
$url = trim($url);
//Check if a protocol is specified at the beginning of the url. If it's not, prepend 'http://'.
if (!preg_match("~^(?:f|ht)tps?://~i", $url)) {
$url = "http://" . $url;
}
//Check if '/' is present at the end of the url. If not, append '/'.
if (substr($url, -1)!=="/"){
$url .= "/";
}
//Return the processed url.
return $url;
}
//If the form was submitted
if(isset($_POST['siteurl'])){
//Put every new line as a new entry in the array
$urls = explode("\n",trim($_POST["siteurl"]));
//Iterate through urls
foreach ($urls as $url) {
//Parse the url to add 'http://' at the beginning or '/' at the end if not already there, to avoid errors with the get_meta_tags function
$url = parseUrl($url);
//Get the meta data for each url
$tags = get_meta_tags($url);
//Check to see if the description tag was present and adjust output accordingly
$tags = NULL;
$tags = get_meta_tags($url);
if($tags)
echo "<tr><td>$url</td><td>" .$tags['description']. "</td></tr>";
else
echo "<tr><td>$url</td><td>No Meta Description</td></tr>";
}
}
?>
我认为最好为此使用 Ajax 吗?所以不会刷新