如何使用Python从启动文件中获取ROS xml参数

Question

I have a launch file which extension is xml, and I would like to get the value of a parameter. This launch file is called ardrone.launch

<!-- This is a sample lanuch file, please change it based on your needs -->
<launch>
    <node name="ardrone_driver" pkg="ardrone_autonomy" type="ardrone_driver" output="screen" clear_params="true">
        <param name="outdoor" value="1" />
        <param name="flight_without_shell" value="1" />
        <param name="max_bitrate" value="4000" />
        <param name="bitrate" value="4000" />
        <param name="navdata_demo" value="0" />
        <param name="altitude_max" value="10000" />
        <param name="altitude_min" value="50" />
        <param name="euler_angle_max" value="0.35" />
        <param name="control_vz_max" value="2000" />
        <param name="control_yaw" value="1.75" />
    </node>
</launch>

For example, I would like to get the value from altitude_max, altitude_min, and the others using python. I have to add that this file is inside a directory called launch and the file where I am calling it is at a directory called scripts, and scripts and launch are both in the same directory.

Answer 1

Parameters set in a launch file are stored at the ROS parameter server .

Accessing these parameters from a python node is quite easy as is shown on this wiki page . In your case the parameters are defined as private parameters of the node (because they are defined inside the <node> tag), so you have to prefix them with ~ when accessing them:

altitude_max = rospy.get_param('~altitude_max')

Answer 2

对我来说，rospy.get_param（“/ ardrone_driver / altitude_max”）有效