Use Transformer.transform(DOMSource, StreamResult):
http://www.theserverside.com/discussions/thread.tss?thread_id=26060
If Android included org.apache.xml.serialize you could use the simpler
XMLSerializer.
On Jul 30, 5:45 pm, beacon indiantalkiedhi...@gmail.com wrote:
Thanks for your reply.
Hi Filip,
It doesn't seem to be the case, as the web service does not give me
any option it just sends XML response whatever it is.
On Jul 31, 9:29 am, Filip Havlicek havlicek.fi...@gmail.com wrote:
Hi,
just a quick thought about that - some web services have a request property
to define
Hello Ed,
Thanks for your response. I added the code snippet to me requesting/
receiving class . But I still get the response in the same format as
mentioned above. I am not sure why.
beacon
On Aug 1, 10:56 pm, Ed edscha...@gmail.com wrote:
Add this for all the classes you're using. This
Hello CMF,
Thanks for your reply. But I dont think I can use this
kind of approach as I need to send a soap object to the web service
and nothing else and it also implements OAuth.
beacon
On Aug 1, 11:26 pm, CMF manf...@gmail.com wrote:
I dont know if it is suitable for your case,
Hi Beacon,
it would probably be the best if you could give as a link to the service you
are trying to use. We could at least look into the website to see if you are
not missing something, because this just doesn't look like XML at all afaik.
Best regards,
Filip Havlicek
2010/8/2 beacon
Hi Filip,
Thanks for your reply. Sure I could give that. This is the what I am
trying to access.
private static final String NAMESPACE = http://tempuri.org/;;
private static final String METHOD_NAME = GetProfileDataXML;
private static final String URL =
What Ed is telling you (and the other people trying to help) is that
what you are seeing is NOT the format that the server is ending.
In fact, it would be somewhat difficult for you to actually see that
data in your program and also parse it.
What you are seeing is the result of (implicitly)
Thanks for your reply Bob. I got hold of whats happening now. I will
look in to it. Thank you,
Beacon
On Aug 2, 1:23 pm, Bob Kerns r...@acm.org wrote:
What Ed is telling you (and the other people trying to help) is that
what you are seeing is NOT the format that the server is ending.
In
I think Bob Kearns may have it right. I was also wondering why you'd
be seeing any XML if you're using SOAP, which is basically an RPC
framework.
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Add this for all the classes you're using. This will let KSoap2 know
what java classes handle the each response element.
envelope.addMapping(request.getNamespace(),
MyDotNetClassName,
MyAndroidClass.class);
For others trying to help this is the
I dont know if it is suitable for your case, but I can retrieve the
xml document for the web with the following method
URL mURL= new URL(http://X/a.xml;);
http = (HttpURLConnection) mURL.openConnection();
int nRC = http.getResponseCode();
if (nRC == HttpURLConnection.HTTP_OK)
{
Hi,
just a quick thought about that - some web services have a request property
to define in what format you want the result. Are you sure this isn't your
case?
Best regards,
Filip Havlicek
2010/7/31 beacon indiantalkiedhi...@gmail.com
Hello Frank,
This is what I am doing on the client side
Thanks for your reply. But it is not similar to JSON. My sample data
looks something like this
anyType{ProductName=anyType{Text=Nadolol; Code=anyType{Value=11755;
CodingSystem=FDB_routed; }; Code=anyType{Value=20.55472;
CodingSystem=Google; }; Code=anyType{Value=7870; CodingSystem=FDB; };
I don't recognize what data format that is, but I would guess that
your code has an error or is not sending the right HTTP headers. What
client side code are you using?
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Hello Frank,
This is what I am doing on the client side
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
request.addProperty(token,WebPage.token);
SoapSerializationEnvelope envelope = new
SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
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