[ros-users] private roslaunch parameters

[Mike Purvis]

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 <watts at willowgarage.com> 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:
>
> <node pkg="something" ... >
>   <param name="port" value="/dev/ttyUSB0" />
> </node>
>
> On Wed, Jul 14, 2010 at 12:41 PM, Mike Purvis <
> mpurvis at clearpathrobotics.com> wrote:
>
>> Hi everyone,
>>
>> My current launch file structure is like so:
>>
>> <launch>
>>   <env name="PYTHONPATH" value="$(env PYTHONPATH)" />
>>   <group ns="clearpath/robots/$(optenv ID default)" clear_params="true">
>>     <param name="port" value="$(optenv CPR_SERIAL_PORT /dev/ttyUSB0)" />
>>     <param name="turn_radius" value="0.4" />
>>
>>     ...
>>
>>     <node pkg="clearpath_horizon" type="horizon.py" name="$(anon horizon)"
>> output="screen" />
>>     <node pkg="clearpath_2dnav" type="simple.py" name="$(anon nav)"
>> output="screen" />
>>     ....
>>   </group>
>> </launch>
>>
>> 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 <param> 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.
>>
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>>
>>
>
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