Python 3 support for fabric
Python 3.xFabricPython 3.x Problem Overview
Does fabric (http://docs.fabfile.org/en/1.7/) support Python 3 yet. As per Python 3 Wall of Superpowers it does not yet. If not what is the best alternative if using Django 1.5 with Python 3.
Python 3.x Solutions
Solution 1 - Python 3.x
Current advice
It seems there is already official support for Python 3.4+ in Fabric v2+ and I guess it should be preferred although there may be some incompatible changes. So in an ideal world, nobody should have this problem anymore :)
pip3 install -U "fabric>2.0"
Maintained old API in Python 3
Because some people were happy with the old API of the Fabric, there is a maintained fork supporting Python 3 called fab-classic. See also discussion in the Github issue Recreating the Fabric 1 API in Fabric 2?
The previous advice
There is Python3 port of Fabric named Fabric3 which supports Python 2.7 and 3.4+ (dropped support for Python <2.7). I use it some time as a replacement and it works for me without any issues.
pip3 install fabric3
Solution 2 - Python 3.x
Current Answer which has already been done by @miso.belica
pip install -U "fabric>2.0.0"
OLD Answer (as of 2013)
From the docs:
> Please note that all documentation is currently written with Python > 2.5 users in mind, but with an eye for eventual Python 3.x compatibility
Or to be more specific:
> eventual Python 3.x compatibility
It looks like Python3 support for fabric is a little sticky and depends upon another open source library, paramiko being compatible with python3. As of now, the master version supports python 3.
It might be worth messing around with and then building your fabric instance on top of that instead of getting it from pip.
Solution 3 - Python 3.x
You could also create a separate python2.7 virtualenv just for fabric:
sudo virtualenv --no-site-packages -p /usr/bin/python fabric-env
source /var/www/APPNAME/fabric-env/bin/activate
sudo /var/www/APPNAME/fabric-env/bin/pip install fabric
deactivate
Then:
/var/www/APPNAME/fabric-env/bin/fab -f /var/www/APPNAME/fabfile.py FAB_COMMAND
Solution 4 - Python 3.x
It seems that Fabric is going through a rewrite and that Fabric 2.0 will be python 3.x compatible. At least this is what I read between the line of the roadmap below. However, I cannot find the Fabric 2.0 repository or branch, so, hard to fully confirm.
Solution 5 - Python 3.x
For those looking for a supported version of Fabric 1.x, there's a fork which is maintained: https://github.com/ploxiln/fab-classic
I recommend using it in case you don't want to rewrite all the existing code. And it works with Python 3.7.
We've switched from fabric3==1.14.post1 to fab-classic==1.15.2 without any problems.
Solution 6 - Python 3.x
It is actually possible to get Fabric to work with Python 3. You can see how I did it in this gist.
In the comments above, @rnevius suggests that invoke is meant as a successor of Fabric 1.x, however that is not completely correct. From the invoke docs:
> Thus, Invoke was created to focus exclusively on local and abstract > concerns, leaving Fabric 2.x concerned only with servers and > network commands.
It thus seems that Fabric 2.x will be built on top of invoke. Hopefully, Fabric 2.x will be available soon, so Python2 can die as quickly as possible.
Solution 7 - Python 3.x
By this day "fabric 2" is awailable to work with python 3+.
"fabric 3" is unofficial fork.
To install "fabric 2" from https://pypi.org/project/Fabric/ :
pip install Fabric