2017-06-24 38 views
0

我尝试了一个soap weservice示例。 该代码使用GeoIp webservice从ip地址确定国家。 当我执行代码时,出现以下异常。运行代码时未将对象引用设置为对象execption的实例

例外:

Exception in thread "main" com.sun.xml.internal.ws.fault.ServerSOAPFaultException: Client received SOAP Fault from server: System.Web.Services.Protocols.SoapException: Server was unable to process request. ---> System.NullReferenceException: Object reference not set to an instance of an object. 
at WebserviceX.Service.Adapter.IPAdapter.CheckIP(String IP) 
at WebserviceX.Service.GeoIPService.GetGeoIP(String IPAddress) 
--- End of inner exception stack trace --- Please see the server log to find more detail regarding exact cause of the failure. 
    at com.sun.xml.internal.ws.fault.SOAP11Fault.getProtocolException(SOAP11Fault.java:178) 
    at com.sun.xml.internal.ws.fault.SOAPFaultBuilder.createException(SOAPFaultBuilder.java:116) 
    at com.sun.xml.internal.ws.client.sei.StubHandler.readResponse(StubHandler.java:238) 
    at com.sun.xml.internal.ws.db.DatabindingImpl.deserializeResponse(DatabindingImpl.java:189) 
    at com.sun.xml.internal.ws.db.DatabindingImpl.deserializeResponse(DatabindingImpl.java:276) 
    at com.sun.xml.internal.ws.client.sei.SyncMethodHandler.invoke(SyncMethodHandler.java:104) 
    at com.sun.xml.internal.ws.client.sei.SyncMethodHandler.invoke(SyncMethodHandler.java:77) 
    at com.sun.xml.internal.ws.client.sei.SEIStub.invoke(SEIStub.java:147) 
    at com.sun.proxy.$Proxy31.getGeoIP(Unknown Source) 
    at org.manjosh.demo.IPlocationFinder.main(IPlocationFinder.java:19) 

我的代码:

import net.webservicex.GeoIP; 
import net.webservicex.GeoIPService; 
import net.webservicex.GeoIPServiceSoap; 

public class IPlocationFinder { 

    public static void main(String[] args) { 

     if (args.length != 1){ 
      System.out.println("you need to pass atleast 1 IP address"); 
     } 
     else 
     { 
      String ipAddress = args[0]; 
      GeoIPService ipService = new GeoIPService(); 
      GeoIPServiceSoap geoIpserviceSoap = ipService.getGeoIPServiceSoap(); 
      GeoIP geoIp = geoIpserviceSoap.getGeoIP(ipAddress); 
      System.out.println((String)geoIp.getCountryName()); 
     } 
    } 

} 
+0

试试IP 212.58.246.79看看你是否得到英国。如果这样做意味着您输入的IP无法映射。这是一个很糟糕的服务,因为许多人失踪。 有数以千计的IP免费数据库。我发现其中一个,我在自己的MySQL中执行查找,而不是依赖Web服务。 – Arminius

+0

异常说明发生了什么。你到达了主机,你发送了你的请求,但是这对服务并不好,或者它不对。要进一步挖掘,您需要捕获您发送的内容以及实际的Soap故障。尝试使用SOAPUI和tcpMonitor。当您使SOAPUI工作时,您可以比较您的代码请求和正在运行的SOAPUI请求。 – Vadim

+0

谢谢,它适用于英国。 – manjosh

回答

1

使用的soapUI创建XML和发送请求时,如果你在sopui响应第一。然后检查您发送的xml请求是否包含所有字段,并在soapui请求中匹配一个。

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