Hey,
I have been playing around with a python SimpleXMLRPCServer service
that I created and have a flex gui that uses functions from it.
Everything works except I get the security error accessing url when I
run it across a domain. Basically I am not sure where this file is
supposed to go. Right now I have it in the root directory of the web
server where the swf file is located. This seems to only work when I
turn my python web service into a cgi webservice.
# example cgi web service
handler = SimpleXMLRPCServer.CGIXMLRPCRequestHandler()
However, performance really takes a hit when I run it as a cgi web
service. So instead I want to keep it as a normal web service
# Example simple web service
server = SimpleXMLRPCServer.SimpleXMLRPCServer((serverip,8888))
So again the security problem comes back to life because the
webservice is no longer run from apache but instead when I invoke the
python script which can be anywhere on the filesystem. I am confused
of where the crossdomain.xml needs to be?
Any body have experience with this?
This is my crossdomain.xml file
<?xml version="1.0"?>
<!DOCTYPE cross-domain-policy SYSTEM
"http://www.macromedia.com/xml/dtds/cross-domain-policy.dtd";>
<cross-domain-policy>
<allow-access-from domain="*" />
</cross-domain-policy>
thanks,
Corey Osman
-------------------------------------------------------------
To unsubscribe from this list, simply email the list with unsubscribe in the
subject line
For more info, see http://www.affug.com
Archive @ http://www.mail-archive.com/discussion%40affug.com/
List hosted by http://www.fusionlink.com
-------------------------------------------------------------
