我有一个小的 PHP 脚本，从原子提要中提取信息。将回显推送到文件或数据库的最佳方法是什么？我希望能够每小时参考该信息。
举个例子。
PHP grabs HTML page.
Webserver stores the resulting HTML locally.
Webserver references that locally stored HTML.
Webserver displays a template version of the local information.
这是拉取 html 的 php：
<?php include('simple_html_dom.php'); echo file_get_html('http://www.website.com')->plaintext[1] ; ?>
您可以使用
file_put_contents()file_put_contents("myFile", file_get_html('http://www.website.com')->plaintext[1]);
希望这会有所帮助呃...
<html>
<head>
<title>Online PHP Script Execution</title>
</head>
<body>
<?php require'simple_html_dom.php'; ?>
<?php
$url_path='http://www.google.com/';
$create=fopen("folder/"."index.html",'w+') or die("can't open file");
$src=file_get_html($url_path)->plaintext ;
$write=fwrite($create,$src);
fclose($create);
foreach($html->find('script') as $script) //get script
{
$scriptPath=$script->src;
$js = explode("/", $scriptPath);
$jsName = end($js);
$path="";
for ($i=0;$i<(count($image)-1);$i++)
{
$path .= $image[$i] . "/";
if(!file_exists('folder/'.$path))
{
mkdir('folder/'.$path, 0777, true); //jquery folder created
}
}
file_put_contents('folder/'.$path.$jsName,file_get_contents($url_path.$scriptPath)); //all jquery downloaded
}
foreach($html->find('img') as $img) //image
{
$imgpath=$img->src;
$image = explode("/", $imgpath);
$path="";
for ($i=0;$i<(count($image)-1);$i++)
{
$path .= $image[$i] . "/";
if(!file_exists('folder/'.$path))
{
mkdir('folder/'.$path, 0777, true); //img folder created
}
}
$imgName = end($image);
file_put_contents('folder/'.$path.$imgName,file_get_contents($url_path.$imgpath)); //img downloaded
}
foreach($html->find('link') as $link) //get link
{
if(strtolower($link->getAttribute('rel')) == "stylesheet" )
{
$linkpath=$link->getAttribute('href');
}
$links = explode("/", $linkpath);
$linkName = end($links);
$path="";
for ($i=0;$i<(count($image)-1);$i++)
{
$path .= $image[$i] . "/";
if(!file_exists('folder/'.$path))
{
mkdir('folder/'.$path, 0777, true); //css folder created
}
}
file_put_contents('folder/'.$path.$linkName,file_get_contents($url_path.$linkpath)); //download css
}
?>
</body>
</html>