我有一个数据文件.xml需要发送到网络服务。我正在使用 php curl 来发送文件。 问题是我不知道如何访问和发送数据文件的内容.xml。
这是数据文件的内容.xml -
<?xml version="1.0" encoding="ISO-8859-1"?>
<inventoryUpdateRequest version="1.0">
<action name="bookupdate">
<username>user</username>
<password>issecret</password>
</action>
<WebList>
<Website>
<transactionType>delete</transactionType>
<vendorBookID>FaNuPh1</vendorBookID>
</Website>
</WebList>
</inventoryUpdateRequest>
这是我发送的 php 文件 -
<?php require_once('post_xml.php');?>
<?php
$xml = HERE IS MY LACK OF UNDERSTANDING HOW TO EXTRACT datafile.xml CONTENT;
$url ='https://inventoryupdate.website.com';
$port = 80;
$response = xml_post($xml, $url, $port);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN""http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Untitled Document</title>
</head>
<body>
<P><?=nl2br(htmlentities($response));?></P>
</body>
</html>
这是使用 cURL 的post_xml.php文件 -
<?php
// open a http channel, transmit data and return received buffer
function xml_post($post_xml, $url, $port)
{
$user_agent = $_SERVER['HTTP_USER_AGENT'];
$ch = curl_init(); // initialize curl handle
curl_setopt($ch, CURLOPT_URL, $url); // set url to post to
curl_setopt($ch, CURLOPT_FAILONERROR, 1); // Fail on errors
if (ini_get('open_basedir') == '' && ini_get('safe_mode' == 'Off'))
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1); // allow redirects
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1); // return into a variable
curl_setopt($ch, CURLOPT_PORT, $port); //Set the port number
curl_setopt($ch, CURLOPT_TIMEOUT, 15); // times out after 15s
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_xml); // add POST fields
curl_setopt($ch, CURLOPT_USERAGENT, $user_agent);
if($port==443)
{
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 2);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE);
}
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
?>
任何帮助将不胜感激！
如果你不需要解析XML文件，那么我建议使用fopen（）或file_get_contents（）。
为简单起见，假设datafile.xml位于同一目录中，请尝试：
$xml = file_get_contents('datafile.xml');