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Hello,
I would want to call a service using python. Currently this command line is working :
rosservice call /camera/start_capture
but I can't figure out how to do the same thing with python. I already tried this without succes :
rospy.ServiceProxy('/camera/start_capture',True)
Any idea ?
Matt
As @yzheng0310 said, you must use the class that defines your service.
Let's suppose that the service used by /camera/start_capture is a StartCatpure that is defined in your package, and let's suppose that your package is named my_package.
In this case you would do the following:
import rospy
from my_package.srv import StartCatpure
start_capture = rospy.ServiceProxy('start_capture_node', StartCapture)
Now, let's call the service. In this case, supposing that the service doesn't accept parameters:
start_capture()
You can check the ROS Tutorial Writing a SImple Publisher and Subscriber in Python that explains step by step.
Asked: 2017-06-02 05:27:21 -0600
Seen: 6,958 times
Last updated: Aug 04 '17
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Hi, the second argument in SeriveProxy should be your service class associated with the name '/camera/start_capture'
from command line, use rosservice info /camera/start_capture to find the service type from the service type, find out the input/output of this service when call this using python: thisService = rospy.ServiceProxy('/camera/start_capture', type) result = thisService(input) #input is to trigger the server, result is the output/acknowledgement from server.