所以,我很困惑,我的应用程序中的一切都运行良好,我做了一些更改,然后在我的应用程序中恢复。但是,这导致我的登录页面出现错误:
SyntaxError at /login/
invalid syntax (views.py, line 41)
Request Method: GET
Request URL: http://X.X.X.X:8000/login/
Django Version: 1.8
Exception Type: SyntaxError
Exception Value:
invalid syntax (views.py, line 41)
Exception Location: /path/to-app/project-name/app (same name as project)/urls.py in <module>, line 22
这是我的 url.py 文件:
from django.conf.urls import *
from django.contrib import admin
from django.conf import settings
from django.conf.urls.static import static
from import views
import overtime.views
import schedule.views
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
url(r'^login/', views.login_user),
url(r'^index/', views.index),
url(r'^$', views.index, name='index'),
url(r'^login_success/$', views.login_success, name='login_success'),
url(r'^overtime/',include('overtime.urls')),
url(r'^schedule/',include('schedule.urls')),
url(r'^logout/$', 'django.contrib.auth.views.logout', {'next_page': '/login'}),
]
我尝试了许多 View 导入的变体,所有这些都导致相同的语法错误:
来自 .导入 View 从 MonolithEmployee 导入 * 导入 View 导入 MonolithEmployee.views
其中每一项都会导致语法错误。
这是否可能是其他地方的问题?
这似乎是引用的观点:
# user login
@login_required
def login_user(request):
if request.user.is_authenticated():
pass
if request.method == 'POST':
username = request.POST['username']
password = request.POST['password']
user = authenticate(username=username, password=password)
if user is not None:
if user.is_active:
login(request, user)
return redirect("login_success")
else:
form = LoginForm()
context = {'form': form}
return render_to_response('login.html', context, context_instance=RequestContext(request))
else:
''' user is not submitting the form, show the login form '''
form = LoginForm()
context = {'form': form}
return render_to_response('login.html', context, context_instance=RequestContext(request))
def login_success(request):
"""
Redirects users based on the group they are in
"""
if request.user.groups.filter(name="BAML").exists():
return redirect("")
else:
return redirect("index")
def logout_user(request):
logout(request)
return HttpResponseRedirect('/login/')
def index(request):
context = {}
return render_to_response('index.html', context, RequestContext(request))
def index_BAML(request):
context = {}
return render_to_response('.html', context, RequestContext(request
最佳答案
这是您的代码:
def index_BAML(request):
context = {}
return render_to_response('index_BAML.html', context, RequestContext(request
你刚刚留下了最后一个括号:
def index_BAML(request):
context = {}
return render_to_response('index_BAML.html', context, RequestContext(request))
关于python - 语法错误,DEBUG 在 urls.py 中显示错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34478195/