Why would it be HTML encoded? I send large XML documents across SOAP as GZIP compressed strings. But even without compression I do not see the need for HMTL encoding unless you mean tags would have to be escaped??
-----Original Message----- From: David B. Bitton [mailto:[EMAIL PROTECTED]] Sent: 08 April 2002 20:40 To: [EMAIL PROTECTED] Subject: Re: Send XML file via SOAP if you did that than it would be HTML encoded, and that would increase the size of the data -- David B. Bitton [EMAIL PROTECTED] www.codenoevil.com Code Made Fresh DailyT ----- Original Message ----- From: "Roumeliotis, Pete" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Monday, April 08, 2002 3:19 PM Subject: RE: Send XML file via SOAP > > Could you just return the entire xml document as a string? > > Pete R. > > > -----Original Message----- > From: Maris Orbidans [mailto:[EMAIL PROTECTED]] > Sent: Monday, April 08, 2002 2:33 PM > To: [EMAIL PROTECTED] > Subject: RE: Send XML file via SOAP > > > > It is more specific :-) > > My question is - how to send my own XML files. > > I know how to do RPCs with simple datatypes like this: > > Parameter ret = resp.getReturnValue(); > Object val = ret.getValue(); > > int res = ((Integer) val).intValue(); > > But what to do if I want to get XML files: > > Parameter ret = resp.getReturnValue(); > Object val = ret.getValue(); > > XMLFILE res = (XMLFILE) val; > > What should I write instead of XMLFILE ? Or should I use > different API ? > > > Maris Orbidans > > > // ---------- remote procedures --------------------------------------- > > public int getNumberOfUsers() throws MalformedURLException,SOAPException > { > if (!isURLValid) throw new MalformedURLException(); > Call call = new Call(); > call.setTargetObjectURI(SOAP_URN); > call.setMethodName("getNumberOfUsers"); > call.setParams(new Vector()); > > Response resp = call.invoke(this.rpcrouter,""); > > if (!resp.generatedFault()) > { > Parameter ret = resp.getReturnValue(); > Object val = ret.getValue(); > > int res = ((Integer) val).intValue(); > return res; > } else > { > Fault fault = resp.getFault(); > throw new > SOAPException(fault.getFaultCode(),fault.getFaultString()); > } > } > > > -----Original Message----- > > From: Richard L Williams [mailto:[EMAIL PROTECTED]] > > Sent: Monday, April 08, 2002 7:40 PM > > To: [EMAIL PROTECTED] > > Subject: Re: Send XML file via SOAP > > > > > > > > You can begin by reading the documentation provided at > > http://xml.apache.org/soap/index.html. > > > > The download and install the soap distribution. > > > > Then you'll be able to ask more specific questions. > > > > > > > > > > > > > > > > "Maris Orbidans" <[EMAIL PROTECTED]> on 04/08/2002 10:31:48 AM > > > > Please respond to [EMAIL PROTECTED] > > > > > > To: <[EMAIL PROTECTED]> > > cc: > > bcc: > > Subject: Send XML file via SOAP > > > > > > > > hello > > > > I need to send and receive XML files via SOAP. > > > > How can I do it ? > > > > Maris > > > > > > > > > > > > > > > > >
