Fehler beim Ausführen von Server Django Syntaxfehler

Question

Ich versuche, durch ein Django-Tutorial zu bekommen, und ich renne in einen Syntaxfehler, wenn ich versuche, manage.py runserver ausführen. Der Fehler ist unten.

(C:\Users\Scott\Anaconda3) C:\Users\Scott\Desktop\django 
tutorials\mysite>manage.py runserver 
Performing system checks... 

Unhandled exception in thread started by <function   
check_errors.<locals>.wrapper at 0x000001705F6D48C8> 
Traceback (most recent call last): 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\utils\autoreload.py", line 227, in wrapper 
fn(*args, **kwargs) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\management\commands\runserver.py", line 125, in inner_run 
self.check(display_num_errors=True) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\management\base.py", line 359, in check 
include_deployment_checks=include_deployment_checks, 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\management\base.py", line 346, in _run_checks 
return checks.run_checks(**kwargs) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\checks\registry.py", line 81, in run_checks 
new_errors = check(app_configs=app_configs) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\checks\urls.py", line 16, in check_url_config 
return check_resolver(resolver) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\core\checks\urls.py", line 26, in check_resolver 
return check_method() 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\urls\resolvers.py", line 254, in check 
for pattern in self.url_patterns: 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\utils\functional.py", line 35, in __get__ 
res = instance.__dict__[self.name] = self.func(instance) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\urls\resolvers.py", line 405, in url_patterns 
patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\utils\functional.py", line 35, in __get__ 
res = instance.__dict__[self.name] = self.func(instance) 
File "C:\Users\Scott\Anaconda3\lib\site-packages\django\urls\resolvers.py", line 398, in urlconf_module 
return import_module(self.urlconf_name) 
File "C:\Users\Scott\Anaconda3\lib\importlib\__init__.py", line 126, in import_module 
return _bootstrap._gcd_import(name[level:], package, level) 
File "<frozen importlib._bootstrap>", line 978, in _gcd_import 
File "<frozen importlib._bootstrap>", line 961, in _find_and_load 
File "<frozen importlib._bootstrap>", line 950, in _find_and_load_unlocked 
File "<frozen importlib._bootstrap>", line 655, in _load_unlocked 
File "<frozen importlib._bootstrap_external>", line 674, in exec_module 
File "<frozen importlib._bootstrap_external>", line 781, in get_code 
File "<frozen importlib._bootstrap_external>", line 741, in source_to_code 
File "<frozen importlib._bootstrap>", line 205, in _call_with_frames_removed 
File "C:\Users\Scott\Desktop\django tutorials\mysite\mysite\urls.py", line 22 
] 
^ 
SyntaxError: invalid syntax 

mysite \ urls.py ist unten. Ich bin nicht sicher, was mit der Syntax falsch ist, oder warum es auf die Klammer zeigt.

from django.conf.urls import url 
from . import views 

urlpatterns = [ 
url(r'^$', views.index, name='index')] 

Answer 1

Bitte verwenden Sie einen Listentyp für URL-Muster. Lassen Sie ein Komma am Ende der Liste:

urlpatterns = [ 
    url(r'^$', views.index, name='index'), 
]