Django Server stecken auf Beanstalkd

Question

Ich versuche, einen Django-Server mit vorhandenem Code einzurichten. Aber ich schien auf diese geklebt werden, während Server zu starten versucht python manage.py runserver

dies die Zurückverfolgungs mit:

(camdyvirtualenv) Mac-mini-3:big-picture-api Sqooge_Ahmed$ python manage.py runserver 
Traceback (most recent call last): 
    File "manage.py", line 10, in <module> 
    execute_from_command_line(sys.argv) 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/django/core/management/__init__.py", line 338, in execute_from_command_line 
    utility.execute() 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/django/core/management/__init__.py", line 303, in execute 
    settings.INSTALLED_APPS 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/django/conf/__init__.py", line 48, in __getattr__ 
    self._setup(name) 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/django/conf/__init__.py", line 44, in _setup 
    self._wrapped = Settings(settings_module) 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/django/conf/__init__.py", line 92, in __init__ 
    mod = importlib.import_module(self.SETTINGS_MODULE) 
    File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py", line 37, in import_module 
    __import__(name) 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/Camdy-APIs/big-picture-api/bigpicture/settings.py", line 5, in <module> 
    from pystalkd.Beanstalkd import Connection as beanstalkdConnect 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/pystalkd/__init__.py", line 20, in <module> 
    from . import Beanstalkd, Job 
    File "/Users/Sqooge_Ahmed/Documents/Django/camdyvirtualenv/lib/python2.7/site-packages/pystalkd/Beanstalkd.py", line 156 
    def send_command(self, command, *args, ok_status=None, error_status=None): 
               ^
SyntaxError: invalid syntax 

Answer 1

In Python 2 sollten Sie vor * args Standardargumente setzen:

def send_command(self, command, ok_status=None, error_status=None, *args):