So senden Sie SOAP XML -Anfrage und lesen Sie die Antwort ⇐ Java

[Anonymous]

Ich habe Code mit Java erstellt, die SOAP XML -Anforderung senden und die Antwort lesen sollte, aber diese Fehler empfangen. https://test.api.system.simplyclub.co.i ... countLogin.
Was könnte das Problem sein?

Code: Select all

 package com.test.com;

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;

public class Send_Xml {

public static void main(String[] args) {

try {
String url = "https://test.api.system.simplyclub.co.il/TerminalService.asmx?op=UserAccountLogin";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("Content-Type","text/xml; charset=utf-8");
String xml =
""+
""+
""+
""+
""+
"string"+
"string"+
"string"+
""+
""+
""+
"string"+
"string"+
""+
""+
"string"+
"string"+
""+
""+
""+
""+
"";
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(xml);
wr.flush();
wr.close();
String responseStatus = con.getResponseMessage();
System.out.println(responseStatus);
BufferedReader in = new BufferedReader(new InputStreamReader(
con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
System.out.println("response:" + response.toString());
} catch (Exception e) {
System.out.println(e);
}
}
}