[英]How to send xml file content to webservice using php curl
我有一個需要發送到webservice的datafile.xml。 我正在使用php curl發送文件。 問題是我不知道如何訪問和發送datafile.xml的內容。
這是datafile.xml的內容 -
<?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>
這是我發送的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>
這是使用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;
}
?>
任何幫助將非常感激!
如果您不需要解析XML文件,那么我建議使用fopen()或file_get_contents() 。
為簡單起見,假設datafile.xml
位於同一目錄中,請嘗試:
$xml = file_get_contents('datafile.xml');
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