send_email（）接受1个位置参数，但给出了3个位置参数-Django Python

我正试图从我的Django python项目发送一封测试电子邮件，并遵循Django文档。当执行下面的函数时，我得到一个错误（附图片）。我做错了什么

错误：

回溯： 为了澄清我的问题，我还添加了一个回溯。希望能有帮助

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/core/handlers/exception.py" in inner
  35.             response = get_response(request)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/core/handlers/base.py" in _get_response
  128.                 response = self.process_exception_by_middleware(e, request)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/core/handlers/base.py" in _get_response
  126.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/contrib/admin/options.py" in wrapper
  574.                 return self.admin_site.admin_view(view)(*args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/utils/decorators.py" in _wrapped_view
  142.                     response = view_func(request, *args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/views/decorators/cache.py" in _wrapped_view_func
  44.         response = view_func(request, *args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/contrib/admin/sites.py" in inner
  223.             return view(request, *args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/utils/decorators.py" in _wrapper
  62.             return bound_func(*args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/utils/decorators.py" in _wrapped_view
  142.                     response = view_func(request, *args, **kwargs)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/utils/decorators.py" in bound_func
  58.                 return func.__get__(self, type(self))(*args2, **kwargs2)

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/contrib/admin/options.py" in changelist_view
  1596.                 response = self.response_action(request, queryset=cl.get_queryset(request))

File "/home/jimtiaz/PycharmProjects/TestAoo/venv/lib/python3.6/site-packages/django/contrib/admin/options.py" in response_action
  1330.             response = func(self, request, queryset)

Exception Type: TypeError at /admin/home/something
Exception Value: send_email() takes 1 positional argument but 3 were given

---

正如您在评论中指出的，此代码用于管理操作。然而，在文档的一节中，指出这样的函数应该有三个参数。举个例子：

def make_published(modeladmin, request, queryset):
    queryset.update(status='p')

您的函数只接受一个参数，但使用三个参数调用，因此会出现错误

将函数定义更改为：

def send_email(modeladmin, request, queryset):

---

什么是呼叫

发送电子邮件

？很明显，函数签名需要一个参数，不清楚的是用三个参数调用它。@JaredGoguen，让我详细检查跟踪。会回到you@JaredGoguen，我在问题中添加了追溯。这对您有帮助吗？显示您的

URL.py

，您在其中将此函数连接到URL。@ThijsvanDien我没有为此函数分配任何URL。我在管理页面中添加了这样的操作：“操作=[编辑所选员工，发送邮件]”。当我选择send_mail并按下“Go”按钮时，它会被调用，然后它会随着一个

递归错误
@NickT崩溃，这是怎么回事？他的函数名非常相似，但不等于导入的函数。啊，错过了e。关于删除答案的评论根本没有提及这一点all@ThijsvanDien工作得很好。现在我得到这个错误，这是有关发送电子邮件。ConnectionRefusedError。。create中的位置是/usr/lib/python3.6/socket.py_connection@ThijsvanDien我想我错过了settings.py文件中的SMTP设置，这就是连接被拒绝的原因
def send_email(modeladmin, request, queryset):