PHP SoapClient-异常，我做错了什么？

Question

我最近开始寻找使用 PHP 的 Web 服务调用。 我无法让以下相当简单的代码片段工作，而是 Web 服务返回异常。

<?php
$client = new SoapClient("http://www.webservicex.net/geoipservice.asmx?wsdl");
print($client->__soapCall("GetGeoIP", array('IP' => '83.251.30.62')));
?>

我还尝试了更简单的 $client->GetGeoIP('83.251.30.62'); 但它们都产生相同的结果。

我的代码有问题吗？ 异常如下：

Fatal error: Uncaught SoapFault exception: [soap:Server] 
System.Web.Services.Protocols.SoapException: Server was unable to process request. --->     System.ArgumentNullException: Value cannot be null. 
Parameter name: input at System.Text.RegularExpressions.Regex.IsMatch(String input) at 
WebserviceX.Service.Adapter.IPAdapter.CheckIP(String IP) at 
WebserviceX.Service.GeoIPService.GetGeoIP(String IPAddress) --- End of inner exception 
stack trace --- in C:\Program Files (x86)\Apache Software 
Foundation\Apache2.2\htdocs\isolda\test.php:16 Stack trace: #0 C:\Program Files 
(x86)\Apache Software Foundation\Apache2.2\htdocs\isolda\test.php(16): SoapClient-
>__soapCall('GetGeoIP', Array) #1 {main} thrown in C:\Program Files (x86)\Apache 
Software Foundation\Apache2.2\htdocs\isolda\test.php on line 16

Answer 1

您应该编写 "IPAddress" 而不是 'IP' （在数组内），它将起作用。

$client = new SoapClient("http://www.webservicex.net/geoipservice.asmx?wsdl"); 

$result = $client->GetGeoIP(array("IPAddress" => "83.251.30.62")); //change was made here

echo $result->GetGeoIPResult->CountryName;