Question

好的，所以我必须为JAVA中的给定WSDL创建一个SOAP客户端，如标题所示。现在我使用NetBeans构建它，问题是，当我运行它并输入我想要的IP时，我得到以下响应 “net.webservicex.GeoIP@564809be”

我在他们的站点测试了WSDL，对于相同的IP，我得到以下

<GeoIP xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"       xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="http://www.webservicex.net/">
<ReturnCode>1</ReturnCode>
<IP>178.128.33.188</IP>
<ReturnCodeDetails>Success</ReturnCodeDetails>
<CountryName>Greece</CountryName>
<CountryCode>GRC</CountryCode>
</GeoIP>

任何想法？虽然我必须“解码”消息才能正常打印出来？提前谢谢

这是客户端的代码

public static void main(String[] args) {
    try {
        System.out.println("Enter the IP Adress");
        InputStreamReader converter = new InputStreamReader(System.in);
        BufferedReader in = new BufferedReader(converter);
        String ipad = in.readLine();
        System.out.println(getGeoIP(ipad));

    } catch (IOException ex) {
        Logger.getLogger(Geoipad.class.getName()).log(Level.SEVERE, null, ex);
    }
}

private static GeoIP getGeoIP(java.lang.String ipAddress) {
    net.webservicex.GeoIPService service = new net.webservicex.GeoIPService();
    net.webservicex.GeoIPServiceSoap port = service.getGeoIPServiceSoap();
    return port.getGeoIP(ipAddress);

Answer 1

net.webservicex.GeoIP@564809be
您似乎正在打印对象的引用（net.webservicex.GeoIP未覆盖toString）。难道他们没有String getIP()来获取IP吗？

Answer 2

以下作品

GeoIPService ipService = new GeoIPService();
GeoIPServiceSoap geoIPServiceSoap = ipService.getGeoIPServiceSoap();
GeoIP geoIp = geoIPServiceSoap.getGeoIP("10.34.55.1");
System.out.println(geoIp.getCountryName());

SOAP客户端对JAVA中给定WSDL的响应奇怪

2 个答案: