Django URLs TypeError：对于include（），视图必须是可调用的或列表/元组

升级到Django 1.10后，出现错误：

TypeError: view must be a callable or a list/tuple in the case of include().

我的urls.py如下：

from django.conf.urls import include, url

urlpatterns = [
    url(r'^$', 'myapp.views.home'),
    url(r'^contact/$', 'myapp.views.contact'),
    url(r'^login/$', 'django.contrib.auth.views.login'),
]

完整的回溯是：

Traceback (most recent call last):
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/autoreload.py", line 226, in wrapper
    fn(*args, **kwargs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/commands/runserver.py", line 121, in inner_run
    self.check(display_num_errors=True)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py", line 385, in check
    include_deployment_checks=include_deployment_checks,
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py", line 372, in _run_checks
    return checks.run_checks(**kwargs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/registry.py", line 81, in run_checks
    new_errors = check(app_configs=app_configs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py", line 14, in check_url_config
    return check_resolver(resolver)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py", line 24, in check_resolver
    for pattern in resolver.url_patterns:
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py", line 35, in __get__
    res = instance.__dict__[self.name] = self.func(instance)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py", line 310, in url_patterns
    patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py", line 35, in __get__
    res = instance.__dict__[self.name] = self.func(instance)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py", line 303, in urlconf_module
    return import_module(self.urlconf_name)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py", line 37, in import_module
    __import__(name)
  File "/Users/alasdair/dev/urlproject/urlproject/urls.py", line 28, in <module>
    url(r'^$', 'myapp.views.home'),
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/conf/urls/__init__.py", line 85, in url
    raise TypeError('view must be a callable or a list/tuple in the case of include().')
TypeError: view must be a callable or a list/tuple in the case of include().

Django 1.10不再允许您'myapp.views.home'在URL模式中将视图指定为字符串（例如）。

解决方案是更新您urls.py的视图以包含可调用的视图。这意味着您必须在中导入视图urls.py。如果您的URL模式没有名称，那么现在是添加名称的好时机，因为用虚线python路径进行反向操作不再起作用。

from django.conf.urls import include, url

from django.contrib.auth.views import login
from myapp.views import home, contact

urlpatterns = [
    url(r'^$', home, name='home'),
    url(r'^contact/$', contact, name='contact'),
    url(r'^login/$', login, name='login'),
]

如果有很多视图，则不方便分别导入它们。一种替代方法是从您的应用程序导入视图模块。

from django.conf.urls import include, url

from django.contrib.auth import views as auth_views
from myapp import views as myapp_views

urlpatterns = [
    url(r'^$', myapp_views.home, name='home'),
    url(r'^contact/$', myapp_views.contact, name='contact'),
    url(r'^login/$', auth_views.login, name='login'),
]

请注意，我们已经使用as myapp_views和as auth_views，这允许我们views.py从多个应用程序导入，而不会发生冲突。

有关的更多信息，请参见Django URL调度程序文档urlpatterns。

该错误仅意味着myapp.views.home不能调用它，就像函数一样。它实际上是一个字符串。当您的解决方案在django 1.9中运行时，它仍然发出警告，说它将从版本1.10开始弃用，这就是发生的一切。通过@Alasdair先前的溶液中导入必要的视图函数到脚本通过任一 from myapp import views as myapp_views 或 from myapp.views import home, contact

如果视图和模块的名称冲突，也可能会出现此错误。我将我的视图文件分发到views文件夹下，/views/view1.py, /views/view2.py并在view2.py中导入了一个名为table.py的模型时遇到了错误，该模型恰巧是view1.py中的视图名称。因此，命名视图功能 v_table(request,id) 很有帮助。

您的代码是

urlpatterns = [
    url(r'^$', 'myapp.views.home'),
    url(r'^contact/$', 'myapp.views.contact'),
    url(r'^login/$', 'django.contrib.auth.views.login'),
]

在导入include()功能时将其更改为以下内容：

urlpatterns = [
    url(r'^$', views.home),
    url(r'^contact/$', views.contact),
    url(r'^login/$', views.login),
]