处理 PHP 中的 Web 服务错误

Question

我有一些简单的 SOAP 客户端来从 WSDL 获取数据并显示它。

<?php 
    //Data, connection, auth
    $dataFromTheForm = $_POST['fieldName']; // request data from the form
    $soapUrl = "https://connecting.website.com/soap.asmx?op=DoSomething"; // asmx URL of WSDL
    $soapUser = "username";  //  username
    $soapPassword = "password; // password

    // xml post structure

    $xml_post_string = '<?xml version="1.0" encoding="utf-8"?>
                        <soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
                          <soap:Body>
                            <GetItemPrice xmlns="http://connecting.website.com/WSDL_Service"> // xmlns value to be set to your's WSDL URL
                              <PRICE>'.$dataFromTheForm.'</PRICE> // data from the form, e.g. some ID number
                            </GetItemPrice >
                          </soap:Body>
                        </soap:Envelope>';

        $headers = array(
                    "Content-type: text/xml;charset=\"utf-8\"",
                    "Accept: text/xml",
                    "Cache-Control: no-cache",
                    "Pragma: no-cache",
                    "SOAPAction: http://connecting.website.com/WSDL_Service/GetPrice", // your op URL
                    "Content-length: ".strlen($xml_post_string),
                );

        $url = $soapUrl;

        // PHP cURL  for https connection with auth
        $ch = curl_init();
        curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
        curl_setopt($ch, CURLOPT_USERPWD, $soapUser.":".$soapPassword); // username and password - declared at the top of the doc
        curl_setopt($ch, CURLOPT_HTTPAUTH, CURLAUTH_ANY);
        curl_setopt($ch, CURLOPT_TIMEOUT, 10);
        curl_setopt($ch, CURLOPT_POST, true);
        curl_setopt($ch, CURLOPT_POSTFIELDS, $soapUrl); // asmx URL of WSDL - declared at the top of the doc
        curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);

        // converting
        $response = curl_exec($ch); 
        curl_close($ch);

        // converting
        $response1 = str_replace("<soap:Body>","",$response);
        $response2 = str_replace("</soap:Body>","",$response1);

        // convertingc to XML
        $parser = simplexml_load_string($response2);        



?>

当 WSDL 工作正常时，一切都工作正常。

但是，当 WSDL 失败（服务器上的运行时错误）时，我的网站在第 152 行抛出数百个警告，即：

$parser = simplexml_load_string($response2);

如何检测该 WSDL已返回无效的 XML（html 错误消息）并在此基础上显示简单的错误消息？

Answer 1

查看文档页面上的提示。

提示
使用 libxml_use_internal_errors() 抑制所有 XML 错误，然后使用 libxml_get_errors() 迭代它们。

这里有一些示例。

---

UPDATE:

对于本例，示例如下所示那：

libxml_use_internal_errors(true);   //enable error handling
$parser = simplexml_load_string($output2);

if(!$parser){  // if $parser is not valid XML response      
     echo '<p>Sorry. This service is currently unavailable. Please try again later.</p>';
} else {
     //if $parser is valid XML response, do something
}

Answer 2

只需检查标头 (ContentType) 或更简单地检查 $response2（如果它以 开头）。