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" /> >> > 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 > >