Python3. Async hook doesn't seem to work

[peppelinux]

According to the documentation I tryed first of all this configuration, considering that I placed that hook function in the same file and, as you can read later, in the file called hooks.py in its working directory:

tid = async('subprocess.run', 
                ['bash', '/tmp/test.sh', 'test'], 
                stdout=PIPE, 
                task_name='example-{}'.format(uuid4().hex),
                hook='hooks.hook_example' # optional, run action when result will be available
                )

This way it returns, as I can read from stdout:

[Q] ERROR malformed return hook 'hooks.hook_example' for [example-a4b92b55f392441d8fee85f22c5311d3]

So I placed that hooks, in the async init, as an callable object (not a string):

    import application_name.hooks
    tid = async('subprocess.run', 
                ['bash', '/tmp/test.sh', 'test'], 
                stdout=PIPE, 
                task_name='example-{}'.format(uuid4().hex),
                hook=application_name.hooks.hook_example # optional, run action when result will be available
                )

It doesn't notify any errors or warning, this way it should be executed succesfully but... But nothing was called, I put a file creation in a temporary folder to be sure that it was called or not.

So, I think it could be usefull having these:

django.VERSION = (1, 10, 7, 'final', 0)
sys.version           = '3.5.2 (default, Nov 17 2016, 17:05:23) \n[GCC 5.4.0 20160609]'
django-q (0.8.0)

[peppelinux]

Fixed, my mistake:

According to signals.py I read that the hook declaration inside async(**kwargs) must be as a string BUT the namespace of this must be very accurate.

I fixed this in this way, specifing the application_name.file_name.funcion_name as a string.