Launching xml file from python launch file
I'm trying to launch an xml launch file from a python launch file, but it's getting stuck on the <launch>
section of the xml.
Is there a different way of doing this?
Code:
def generate_launch_description():
return LaunchDescription([
IncludeLaunchDescription(PythonLaunchDescriptionSource(
os.path.join(get_package_share_directory('rosbridge_server'),
'launch/rosbridge_websocket_launch.xml')))
])
Error:
File "/opt/ros/foxy/share/rosbridge_server/launch/rosbridge_websocket_launch.xml", line 1
<launch>
^
SyntaxError: invalid syntax
Asked by SonicBoom on 2022-07-26 05:14:05 UTC
Answers
To include an XML file, you need to use FrontendLaunchDescriptionSource
instead of PythonLaunchDescriptionSource
, which can be imported with from launch.launch_description_sources import FrontendLaunchDescriptionSource
.
So the correct code would be:
def generate_launch_description():
return LaunchDescription([
IncludeLaunchDescription(FrontendLaunchDescriptionSource(
os.path.join(get_package_share_directory('rosbridge_server'),
'launch/rosbridge_websocket_launch.xml')))
])
Asked by JeremieBourque on 2022-07-28 10:15:31 UTC
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