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	Hi Pete


local a,b =  xmlrpc.http.call ("http://username:password@localhost:8080/RPC";,

"show","data")

print(a)

nil

print(b)

<HTML><HEAD><TITLE>Error 405</TITLE></HEAD><BODY><H1>Error 405</H1><P>POST
is the only HTTP method this server understands</P><p><HR><b><i><a href="
http://xmlrpc-c.sourceforge.net";>ABYSS Web Server for XML-RPC For
C/C++</a></i></b> version Xmlrpc-c 1.14.99<br></p></BODY></HTML>





I thought call() is already doing the POST or is it not?

	Yes it is.  I think it is not possible to make a XMLRPC
using a GET.  You can check the code of xmlrpc/http.lua.


  Can anyone
suggest how this problem can be resolved?

	I am not sure, but your URL seems very strange.  Maybe
it is confusing the server...

	Regards,
		Tomás