來自Android soap服務電話的奇怪回應

Question

我們有一個Android應用程序，我們正在嘗試用於Web服務通信。

我們可以使用以下代碼段成功發送請求並獲得響應。

public void onCreate(Bundle savedInstanceState)
{
    super.onCreate(savedInstanceState);

    TextView textView = new TextView(this);

    setContentView(textView);
    //textView.setText();

    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);

    request.addProperty("foo", "bar");
    request.addProperty("foo", "bar");
    request.addProperty("foo", "bar");
    request.addProperty("foo", "bar");
    request.addProperty("foo", "bar");
    request.addProperty("foo", "bar");


    Log.w(TAG, request.toString());
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.dotNet = true;

    envelope.setOutputSoapObject(request);
    envelope.encodingStyle = "utf-8";
    envelope.implicitTypes = false;
    AndroidHttpTransport httpTransport = new AndroidHttpTransport(URL);

    try

    {
    httpTransport.call(SOAP_ACTION, envelope);

    KvmSerializable ks = (KvmSerializable)envelope.bodyIn;
    for(int i=0;i < ks.getPropertyCount();i++)
    {
        ks.getProperty(i);
        SoapObject soap = (SoapObject)ks.getProperty(i);
        String tmp = soap.getProperty(0).toString();
        textView.setText(tmp);
        Log.w(TAG2, envelope.toString());
        Log.w(TAG3, ks.getProperty(i).toString());

    }

    }
    catch (Exception exception)
    {
        textView.setText(exception.toString());
        Log.w(TAG4, exception);
    }


 }

現在，來自textView.setText(tmp);呈現的請求的響應textView.setText(tmp); 提出如下答復：

anyType{foo=bar;foo=bar;foo=bar;}

而我們希望它以foo = bar, foo = bar

任何點擊和指針將受到高度贊賞。 提前致謝。

Answer 1

你有這些：

    SoapObject soap = (SoapObject)ks.getProperty(i);
    String tmp = soap.getProperty(0).toString();

你需要使用這些：

    SoapObject soap = (SoapObject)ks.getProperty(i);
    SoapObject tmp = soap.getProperty(0);
    String foo = tmp.getProperty(0).toString();

我看到你使用的是復雜類型，你應該使用HttpTransportSE而不是AndroidHttpTransport 。

希望能幫助到你