As part of resolving XECS-1852 - "remote invocation of yum update through sipXsupervisor", I need to launch a Python script that resides in the SipX bin directory. I, obviously cannot hard code the path to this location and SipXecsService does not provide a way of obtaining it either. Based on the paths that SipXecsService provides today, it doesn't look like this type of path should be added here either.
Is there a recommended programmatic method of obtaining this path from C++ or should I add it to the sipxsupervisor-config file and then obtain it from the Configuration Database? Thanks in advance Raymond _______________________________________________ sipx-dev mailing list [email protected] List Archive: http://list.sipfoundry.org/archive/sipx-dev Unsubscribe: http://list.sipfoundry.org/mailman/listinfo/sipx-dev
