Print Python Exception Type (Raised in Fabric)
exception, python, python-2.7
Solution
The issue is that fabric uses subprocess for doing these sorts of things. If you look at the source code for `local` you can see it doesn't actually raise an exception. It calls suprocess.Popen and uses `communicate()` to read stdout and stderr. If there is a non-zero return code then it returns a call to either `warn` or `abort`. The default is abort. So, to do what you want, try this:
def init():
with settings(warn_only=True):
local('mkdir ./www')
If you look at the source for `abort`, it looks like this:
10 def abort(msg):
21 from fabric.state import output
22 if output.aborts:
23 sys.stderr.write("\nFatal error: %s\n" % str(msg))
24 sys.stderr.write("\nAborting.\n")
25 sys.exit(1)
So, the exception would be a SystemExit exception. While you could catch this, the proper way to do it is outlined above using `settings`.
Problem
I'm using Fabric to automate, including the task of creating a directory. Here is my fabfile.py: ``` #!/usr/bin/env python from fabric.api import * def init(): try: local('mkdir ./www') except ##what exception?##: #print exception name to put in above ``` Run fab `fabfile.py` and f I already have `./www` created an error is raised, but I don't know what kind, so I don't know how to handle the error yet. Fabric only prints out the following: ``` mkdir: cannot create directory ‘./www’: File exists Fatal error: local() encountered an error (return code 1) while executing 'mkdir ./www' Aborting. ``` What I want to do is be able to find out the error type so that I can except my errors properly without blanket statements. It would be really helpful if an answer does not just tell me how to handle a `mkdir` exception, but print (or otherwise find the name to) any exception I may run into down the line (mkdir is just an example). Thank you!