java - 线程 "main"com.sun.xml.internal.ws.fault.ServerSOAPFaultException : Client received SOAP Fault from server 中的异常

Question

我正在尝试学习 SOAP 协议(protocol)。因此，在使用以下命令借助 wsimport 生成 net.webservicex 代码后

C:\Program Files\Java\jdk1.8.0_74\bin>wsimport -keep -s src http://www.webservicex.net/geoipservice.asmx?WSDL

我将生成的代码放在我的项目中(请看下面的屏幕截图)。我在 run configurations --> java application --> arguments --> programm arguments--> 216.58.213.238

添加了一些 google 的 ip

但是在 IPLocationFinder 类中执行 main 方法时，出现以下错误:

IPLocationFinder 类

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 in one IP address");
        } else {
            String ipAddress = args[0];
            GeoIPService ipService = new GeoIPService();
            GeoIPServiceSoap geoIPServiceSoap = ipService.getGeoIPServiceSoap();
            // Here is line 14.
            GeoIP geoIP  = geoIPServiceSoap.getGeoIP(ipAddress);
            System.out.println(geoIP.getCountryName());
        }
    }
}

错误

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(Unknown Source)
    at com.sun.xml.internal.ws.fault.SOAPFaultBuilder.createException(Unknown Source)
    at com.sun.xml.internal.ws.client.sei.StubHandler.readResponse(Unknown Source)
    at com.sun.xml.internal.ws.db.DatabindingImpl.deserializeResponse(Unknown Source)
    at com.sun.xml.internal.ws.db.DatabindingImpl.deserializeResponse(Unknown Source)
    at com.sun.xml.internal.ws.client.sei.SyncMethodHandler.invoke(Unknown Source)
    at com.sun.xml.internal.ws.client.sei.SyncMethodHandler.invoke(Unknown Source)
    at com.sun.xml.internal.ws.client.sei.SEIStub.invoke(Unknown Source)
    at com.sun.proxy.$Proxy31.getGeoIP(Unknown Source)
    at IPLocationFinder.main(IPLocationFinder.java:14)

Answer 1

Some addresses will not appear in the database and therefore cannot be mapped. This is one of the limitation of IP Address location (Geolocation).

你应该试试这个IP地址212.58.246.79

它应该显示“英国”

您可以在此处阅读有关一般地理定位的更多信息:https://www.lifewire.com/does-ip-address-geolocation-really-work-818154

希望对你有帮助。