I think you'll have to post your launch file, and probably a "rosparam get MY_NAMESPACE". It could be your configuration isn't set up properly. On Wed, Jul 14, 2010 at 1:18 PM, Mike Purvis wrote: > Hi Kevin, > > I had gotten that far, but by what name do I then access the parameter from > my python source? Neither rospy.get_param('port') nor rospy.get_param('~port') > seem to work. > > M. > > > > On 14 July 2010 15:53, Kevin Watts wrote: > >> That's not the correct syntax for private parameters. Your example puts >> them into the same namespace as the node, but not the private namespace. >> >> You can find full details here: >> >> http://www.ros.org/wiki/roslaunch/XML/param >> >> For your code, it might look like: >> >> >> >> >> >> On Wed, Jul 14, 2010 at 12:41 PM, Mike Purvis < >> mpurvis@clearpathrobotics.com> wrote: >> >>> Hi everyone, >>> >>> My current launch file structure is like so: >>> >>> >>> >>> >>> >>> >>> >>> ... >>> >>> >>> >> output="screen" /> >>> .... >>> >>> >>> >>> And then inside of horizon.py, I am able to access the parameter >>> with rospy.get_param('port'). However, that parameter is really not relevant >>> to any node except horizon.py. From my reading of the documentation, I >>> should be able to make it private by putting the tag inside the >>> relevant node tag, and then calling it with rospy.get_param('~port'). >>> However, this doesn't seem to work---I consistently just get the default >>> parameter. >>> >>> What am I doing wrong? >>> >>> Thanks. >>> >>> _______________________________________________ >>> ros-users mailing list >>> ros-users@code.ros.org >>> https://code.ros.org/mailman/listinfo/ros-users >>> >>> >> >> _______________________________________________ >> ros-users mailing list >> ros-users@code.ros.org >> https://code.ros.org/mailman/listinfo/ros-users >> >> > > _______________________________________________ > ros-users mailing list > ros-users@code.ros.org > https://code.ros.org/mailman/listinfo/ros-users > >