将输出保存在xml文件中

Question

如何将以下链接的网页内容保存到xml文件中？下面的代码不适合我。

<?php 
$url = "https://services.boatwizard.com/bridge/events/ae0324ff-e1a5-4a77-9783-f41248bfa975/boats?status=on"; 
copy($url, "file.xml"); 
?> 

Answer 1

这是可以做到如下：

function savefile($filename,$data){ 
$fh = fopen($filename, 'w') or die("can't open file"); 
fwrite($fh, $data); 
fclose($fh); 
} 

$data = file_get_contents('your link'); 
savefile('file.xml',$data); 

Answer 2

使用2个功能file_get_contents()和file_put_contents();

file_get_contents - 获取文件的内容，如果fopen包装已启用，则可以使用该URL作为文件名。

file_put_contents - 把第二个参数的内容，在第一个参数指定的文件

<?php 
    $url = "https://services.boatwizard.com/bridge/events/ae0324ff-e1a5-4a77-9783-f41248bfa975/boats?status=on"; 
    file_put_contents('file.xml',file_get_contents($url)); 
    @chmod('file.xml', 0755); 
?> 

Answer 3

http://de2.php.net/manual/en/book.dom.php

$dom = new DOMDocument(); 
$dom->load('http://www.example.com'); 
$dom->save('filename.xml'); 

Answer 4

您也可以下载给定链路通过到file.xml的内容卷曲：

<?php 
$url = 'https://services.boatwizard.com/bridge/events/ae0324ff-e1a5-4a77-9783-f41248bfa975/boats?status=on'; 
$fp = fopen (dirname(__FILE__) . '/file.xml', 'w+'); 
$ch = curl_init($url); 
curl_setopt($ch, CURLOPT_TIMEOUT, 50); 
curl_setopt($ch, CURLOPT_FILE, $fp); 
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
curl_exec($ch); 
curl_close($ch); 
fclose($fp); 
?>