How do I test the ROS version in Python code?

Question

This has been asked for C++ (http://answers.ros.org/question/9562/how-do-i-test-the-ros-version-in-c-code/) but it would be good to know what the correct way to do this is in Python.

There is a python script called rosversion which does this in a bit of an awkward way IMO. It does it by checking whether the "ROS_DISTRO" environment variable exists and, if not, it parses an XML file (roscore.xml). This seems quite complicated and it only gives you the release name (cturtle, electric, fuerte, etc.).

Is there a way to get access to those nice looking DEFINES like in the C++ code? Even having just the functionality of the rosversion script somewhere in an official ROS Python API function would be better than nothing (just getting the release name).

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Originally posted by StFS on ROS Answers with karma: 182 on 2012-06-13

Post score: 3

Answer 1

If you look further down in rosversion, it only prints the distro name when invoked with the -d or --distro option. To get the ROS version string, invoke:

$ rosversion ros

On Fuerte, that returns 1.8.7 (without a newline), on Electric 1.6.9 (with a newline).

From a Python script, you can do that yourself:

import rospkg
version = rospkg.RosStack().get_stack_version('ros')

That uses the new distro-independent rospkg module. Get it like this:

sudo pip install -U rospkg

It works for other stacks, as well:

messages_version = rospkg.RosStack().get_stack_version('common_msgs')

In most cases, it is better to test for the existence of some new API instead:

TimeReference = None
try:
    from sensor_msgs.msg import TimeReference
except ImportError:
    pass

Then, make your logic dependent on the existence of that message:

if TimeReference:
    tr_pub = rospy.Publisher('gps_time', TimeReference)
    

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Originally posted by joq with karma: 25443 on 2012-06-14

This answer was ACCEPTED on the original site

Post score: 4