我想在我的Django项目中使用Python type hints。在Django中注释简单class-based view的get
/ post
方法的正确方法是什么?
我已经搜索过Django代码本身,但它似乎没有包含任何类型的提示。
答案 0 :(得分:3)
存在您可能感兴趣的此存储库:https://github.com/machinalis/mypy-django
这将允许您使用这样的注释:
def get(self, request: HttpRequest, question_id: str) -> HttpResponse:
答案 1 :(得分:0)
如果使用基于功能的视图,并且您不想要或不需要mypy-django,您可以这样做:
from django.http import HttpRequest, HttpResponse, JsonResponse
def some_fbv(request: HttpRequest) -> HttpResponse:
....
return foo
def some_json_producing_fbv(request: HttpRequest) -> JsonResponse:
...
return foo
答案 2 :(得分:0)
Django stubs 是维护良好的包,https://github.com/typeddjango/django-stubs。
import typing as t
from django.http import HttpResponseRedirect
from django.shortcuts import render
from django.views import View
from django.http import HttpRequest, HttpResponse, JsonResponse,
HttpResponseRedirect
from .forms import MyForm
# type alias when response is one of these types
RedirectOrResponse = t.Union[HttpResponseRedirect, HttpResponse]
class MyFormView(View):
form_class = MyForm
initial = {'key': 'value'}
template_name = 'form_template.html'
def get(self, request: HttpRequest, *args: Any, **kwargs: Any) -> HttpResponse:
form = self.form_class(initial=self.initial)
return render(request, self.template_name, {'form': form})
def post(self, request: HttpRequest, *args: tuple[Any],
**kwargs: dict[str, t.Union[int, str]]) -> RedirectOrResponse:
form: MyForm = self.form_class(request.POST)
if form.is_valid():
# <process form cleaned data>
return HttpResponseRedirect('/success/')
return render(request, self.template_name, {'form': form})
HttpRequest
映射到函数或方法中的请求变量。HttpResponse, JsonResponse, StreamingResponse, Redirect
将是视图函数/方法返回的值。*args, **kwargs
既简单又棘手,因为它可以是 any 值的元组或值的字典。 *args: Any
或 *args: tuple[Any]
(如果您知道,也可以使用特定类型)。这同样适用于 **kwargs
。type[cls]
。更多示例:https://github.com/typeddjango/django-stubs/tree/master/tests