python - 无法识别 Django 错误

Question

看到以下语法错误，但 urls.py 文件似乎是正确的。我错过了什么？

SyntaxError at /admin/
invalid syntax (urls.py, line 6)
Request Method: GET
Request URL:    http://127.0.0.1:8000/admin/
Django Version: 1.4.1
Exception Type: SyntaxError
Exception Value:    
invalid syntax (urls.py, line 6)
Exception Location: /Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/utils/importlib.py in import_module, line 35
Python Executable:  /Library/Frameworks/Python.framework/Versions/2.7/Resources/Python.app/Contents/MacOS/Python
Python Version: 2.7.2

这是 urls.py:

from django.conf.urls.defaults import *
from events import views

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    (r'^events/', include('events.urls')),
    (r'^admin/', include(admin.site.root)),
)

...

from django.conf.urls.defaults import *
from events import views

urlpatterns = patterns('',
    url(r'^create/$', views.create, name='ev_create')
    url(r'^tonight/$', views.tonight, name='ev_tonight'),   
)

Answer 1

您缺少一个逗号:

urlpatterns = patterns('',
    url(r'^create/$', views.create, name='ev_create')     # <-- comma missing
    url(r'^tonight/$', views.tonight, name='ev_tonight'),   
)

请注意，这是 urls.py 模块的第 5 行，但是当 Python 遇到不适合 url(. ..) 调用。