java - Android + WS-* 服务？

Question

如何创建Android具有 WS-* 服务连接的应用程序？

从哪里获取适合初学者的信息 - 它应该包含视频教程和说明不支持哪些类型的绑定(bind)。

Answer 1

简单且基本的想法是建立与 Web 服务 URL 的连接并将 XML 数据发布到该 URL。下面的代码假设 Web 服务 URL 为 http://www.example.com/example.asmx您要调用的方法是 WebServiceMethod1，它接受一个字符串参数。返回字符串是来自Web Service的响应，您必须解析XML才能获取数据。

public static String callWebServiceMethod(String email){
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://www.example.com/example.asmx");
String  webServiceXml = "";
String  response = "";

try
{
    webServiceXml += "<?xml version=\"1.0\" encoding=\"utf-8\"?>";
    webServiceXml += "<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/\">";
    webServiceXml += "<soap:Header>";
    webServiceXml += "<AuthHeader xmlns=\"http://www.example.com/\">";
    webServiceXml += "</AuthHeader>";
    webServiceXml += "</soap:Header>";
    webServiceXml += "<soap:Body>";
    webServiceXml += "<WebServiceMethod1 xmlns=\"http://www.example.com/\">";
    webServiceXml += "<emailId>" + email + "</emailId>";
    webServiceXml += "</WebServiceMethod1>";
    webServiceXml += "</soap:Body>";
    webServiceXml += "</soap:Envelope>";

    httpPost.setHeader("content-type","text/xml; charset=utf-8");
    httpPost.setHeader("SOAPAction","http://www.example.com/WebServiceMethod1");

    httpPost.setEntity(new StringEntity(webServiceXml));

    HttpResponse httpResponse = httpClient.execute(httpPost);

    response = EntityUtils.toString(httpResponse.getEntity());
}
catch(Exception ex)
{
    Log.i("error", ex.getMessage());
}

return response;

}