Android, отправка XML через HTTP POST (SOAP)

ЗДЕСЬ Альтернатива для отправки soap msg.

public String setSoapMsg(String targetURL, String urlParameters){

        URL url;
        HttpURLConnection connection = null;  
        try {
          //Create connection
          url = new URL(targetURL);

         // for not trusted site (https)
         // _FakeX509TrustManager.allowAllSSL();
         // System.setProperty("javax.net.debug","all");

          connection = (HttpURLConnection)url.openConnection();
          connection.setRequestMethod("POST");


          connection.setRequestProperty("SOAPAction", "**** SOAP ACTION VALUE HERE ****");

          connection.setUseCaches (false);
          connection.setDoInput(true);
          connection.setDoOutput(true);


          //Send request
          DataOutputStream wr = new DataOutputStream (
                       connection.getOutputStream ());
          wr.writeBytes (urlParameters);
          wr.flush ();
          wr.close ();

          //Get Response    
          InputStream is ;
          Log.i("response", "code="+connection.getResponseCode());
          if(connection.getResponseCode()<=400){
              is=connection.getInputStream();
          }else{
              /* error from server */
              is = connection.getErrorStream();
        } 
         // is= connection.getInputStream();
          BufferedReader rd = new BufferedReader(new InputStreamReader(is));
          String line;
          StringBuffer response = new StringBuffer(); 
          while((line = rd.readLine()) != null) {
            response.append(line);
            response.append('\r');
          }
          rd.close();
          Log.i("response", ""+response.toString());
          return response.toString();

        } catch (Exception e) {

         Log.e("error https", "", e);
          return null;

        } finally {

          if(connection != null) {
            connection.disconnect(); 
          }
        }
      }

надеюсь, что это помогает. если кому интересно allowAllSSL() метод, google it :).

Итак, если вы используете:

httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

это все еще отдых, но если вы используете:

StringEntity se = new StringEntity(SOAPRequestXML,HTTP.UTF_8);
httppost.setEntity(se);  

Это мыло???

пример отправки XML в WS через HTTP POST.

DefaultHttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://foo/service1.asmx/GetUID");     

        //XML example to send via Web Service.
        StringBuilder sb = new StringBuilder();
        sb.append("<myXML><Parametro><name>IdApp</name><value>1234567890</value></Parameter>");
        sb.append("<Parameter><name>UID1</name><value>abc12421</value></Parameter>");
                sb.append("</myXML>");

        httppost.addHeader("Accept", "text/xml");
        httppost.addHeader("Content-Type", "application/x-www-form-urlencoded");

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1); 
        nameValuePairs.add(new BasicNameValuePair("myxml", sb.toString());//WS Parameter and    Value           
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);

вот мой код для отправки HTML.... Вы можете увидеть данные nameValuePairs.добавлять.(..)

        HttpClient httpclient = new DefaultHttpClient();
        // Your URL
        HttpPost httppost = new HttpPost("http://192.71.100.21:8000");

        try {
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            // Your DATA
            nameValuePairs.add(new BasicNameValuePair("id", "12345"));
            nameValuePairs.add(new BasicNameValuePair("stringdata","AndDev is Cool!"));

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            HttpResponse response;
            response = httpclient.execute(httppost);
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

мне пришлось отправить XML через HTTP Post на Android тоже.

String xml = "xml-block";
StringEntity se = new StringEntity(xml,"UTF-8");
se.setContentType("application/atom+xml");
HttpPost postRequest = new HttpPost("http://some.url");
postRequest.setEntity(se);

надеюсь, что это работает!

здесь фрагменты кода кода, которые я использую для публикации xml в SOAP services и взамен получения Inputstream из интернета.

 private InputStream call(String soapAction, String xml) throws IOException {

    byte[] requestData = xml.getBytes("UTF-8");
    URL url = new URL(URL);

    connection = (HttpURLConnection) url.openConnection();
    connection.setRequestProperty("Accept-Charset", "UTF-8");
    // connection.setRequestProperty("Accept-Encoding","gzip,deflate");
    connection.setRequestProperty("Content-Type", "text/xml; UTF-8");
    connection.setRequestProperty("SOAPAction", soapAction);
    connection.setRequestProperty("User-Agent", "android");
    connection.setRequestProperty("Host",
            "base_urlforwebservices like - xyz.net");
    // connection
    // .setRequestProperty("Content-Length", "" + requestData.length);
    connection.setRequestMethod("POST");
    connection.setDoOutput(true);
    connection.setDoInput(true);

    os = connection.getOutputStream();
    os.write(requestData, 0, requestData.length);
    os.flush();
    os.close();
    is = connection.getInputStream();
    return is; // inputStream
}

Here xml: встроенный xml-запрос, используемый для вызова служб.

получать удовольствие;

другой способ сделать это с помощью Вызов Apache. URL Api, URI действия и тело API должны быть предоставлены

InputStream input = new ByteArrayInputStream(apiBody.getBytes());
Service service = new Service();
Call call = (Call) service.createCall();
SOAPEnvelope soapEnvelope = new SOAPEnvelope(input);

call.setTargetEndpointAddress(new URL(apiUrl));
call.setUseSOAPAction(true);
if(StringUtils.isNotEmpty(actionURI)){
 call.setSOAPActionURI(actionURI);
}

soapEnvelope = call.invoke(soapEnvelope);
return soapEnvelope.toString();

С помощью OkHTTP это просто

String soap_string = "soap string";

public static final MediaType SOAP_MEDIA_TYPE = MediaType.parse("application/xml");
final OkHttpClient client = new OkHttpClient();
RequestBody body = RequestBody.create(SOAP_MEDIA_TYPE, soap_string);

final Request request = new Request.Builder()
       .url("Your URL")
       .post(body)
       .addHeader("Content-Type", "application/xml")
       .addHeader("Authorization", "Your Token")
       .addHeader("cache-control", "no-cache")
       .build();

client.newCall(request).enqueue(new Callback() {
    @Override
    public void onFailure(Call call, IOException e) {
          String mMessage = e.getMessage().toString();
          Log.w("failure Response", mMessage);
    }

    @Override
    public void onResponse(Call call, Response response) throws IOException {

         String mMessage = response.body().string();

          //code = response.code();
          getResponse(mMessage, response);
    }
});

Читать полный учебник здесь - https://www.studytutorial.in/android-sending-xml-or-soap-rest-api-request-using-okhttp-tutorial