如何從webService響應中獲得響應XML？

Question

我必須創建一個獲取webService響應的方法，如xml。我知道如何使用Java類創建，但問題是從webService獲取響應爲xml。

這些web服務是基於soap的。

在此先感謝。

Answer 1

我剛剛解決了我的問題。 HttpURLConnection幫助我。

下面的代碼塊顯示我如何在java（如Mozilla海報）中獲得xml響應的海報。

public static void main(String[] args) { 
    try { 
     String uri = "http://test.com/IntegratedServices/IntegratedServices.asmx?op=GetUserInfo"; 

     String postData = new XmlTest().xmlRequest("QWERTY10"); 

     URL url = new URL(uri); 

     HttpURLConnection connection = (HttpURLConnection) url.openConnection(); 
     connection.setDoOutput(true); // This is important. If you not set doOutput it is default value is false and throws java.net.ProtocolException: cannot write to a URLConnection exception 
     connection.setRequestMethod("POST"); // This is method type. If you are using GET method you can pass by url. If method post you must write 
     connection.setRequestProperty("Content-Type", "text/xml;charset=UTF-8"); // it is important if you post utf-8 characters 

     DataOutputStream wr = new DataOutputStream(connection.getOutputStream()); // This three lines is importy for POST method. I wrote preceding comment. 
     wr.write(postData.getBytes()); 
     wr.close(); 

     InputStream xml = connection.getInputStream(); 

     BufferedReader reader = new BufferedReader(new InputStreamReader(xml)); 
     String line = ""; 
     String xmlResponse = ""; 
     while ((line = reader.readLine()) != null) { 
      xmlResponse += line; 
     } 

     File file = new File("D://test.xml"); // If you want to write as file to local. 
     FileWriter fileWriter = new FileWriter(file); 
     fileWriter.write(xmlResponse); 
     fileWriter.close(); 

    } catch (Exception e) { 
     e.printStackTrace(); 
    } 
} 

public String xmlRequest(String pin) { 
    return "<?xml version=\"1.0\" encoding=\"utf-8\"?>\n" 
      + "<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/\">\n" 
      + " <soap:Body>\n" 
      + " <GetUserInfo xmlns=\"http://tempuri.org/\">\n" 
      + "  <pin>" + pin + "</pin>\n" 
      + " </GetUserInfo>\n" 
      + " </soap:Body>\n" 
      + "</soap:Envelope>"; 
} 

我希望這可以幫助誰想要得到xml作爲迴應。我還寫了對我的代碼的詳細評論。

Answer 2

用於制肥皂類型的web服務：

Parsing SOAP Response in Java

休息一下這個鏈接：

http://duckranger.com/2011/06/jaxb-without-a-schema/