python - Django 1.11.17 TypeError : 'context must be a dict rather than Context' , 除了它是一个字典

Question

我最近从 Django 1.9 切换到 1.11.17，有一件事让我很困扰。有这样的错误:

TypeError at /somepath
context must be a dict rather than Context

抛出它的行是:

return render(request=request, template_name="mytemplate.html", context={"form": form, "update": updateType})

有很多关于 SO 的答案，人们使用 RequestContext 或 Context 而不是 dict 作为 context 并切换到 dict 解决了他们的问题。但不适合我。在这里，我非常确定我的 context 实际上是一个字典。如果我将其更改为:

return render(request=request, template_name="mytemplate.html", context={})

错误消失了，但显然稍后会导致另一个错误。你们知道我在这里做错了什么吗？

编辑:我的进口:

from django.shortcuts import render, render_to_response
from django.template.context import RequestContext, Context

我尝试过 bot render 和 render_to_response 具有类似的效果。还使用 Context 或 RequestContext 给出了类似的错误。

EDIT2:更多引用代码

from django.http import (
    HttpResponseRedirect,
    HttpResponseBadRequest,
)
from django.shortcuts import render, render_to_response
from django.template import RequestContext, Context
from django.utils.html import escape

# some more imports, but from local files, not django

def update_my_template(request):
    user = request.user
    # preform some checks for user
    ...

    if request.method == "GET":
        updateType = request.GET.get("id")
        if updateType:
            form = None
            if updateType == "something":
                form = SomeForm(user)
            if updateType == "something else":
                form = DifferentForm()
            if form is None:
                return HttpResponseRedirect("/somepage")

            # This was the code that worked in 1.9
            rctx = RequestContext(
                request, {"form": form, "update": updateType}
            )
            return render_to_response("mytemplate.html", rctx)



    # some different cases, but the error is thrown already
...

这些都不起作用:

dictctx = {"form": form, "update": updateType}
return render(request=request, template_name="mytemplate.html", dictctx)

.

ctx = Context({"form": form, "update": updateType})
return render(request=request, template_name="mytemplate.html", ctx)

.

ctx = Context({"form": form, "update": updateType})
return render(request=request, template_name="mytemplate.html", ctx.flatten())

.

rctx = RequestContext(request, {"form": form, "update": updateType})
return render_to_response("mytemplate.html", rctx.flatten())

Answer 1

render 逻辑有所不同，具体取决于您传递给 render 的内容:

def render(self, context):
        "Display stage -- can be called many times"
        with context.render_context.push_state(self):
            if context.template is None:
                with context.bind_template(self):
                    context.template_name = self.name
                    return self._render(context)
            else:
                return self._render(context)

看起来您可以将参数template_name更改为name，但您的对象没有context.render_context 值，这就是为什么最好创建和使用 Context

的实例

https://docs.djangoproject.com/en/1.11/_modules/django/template/base/#Template.render

文档显示传递 Context 的实际实例，因此我建议您在代码中执行此操作，而不仅仅是传递字典:

>>> from django.template import Context, Template
>>> template = Template("My name is {{ my_name }}.")

>>> context = Context({"my_name": "Adrian"})
>>> template.render(context)
"My name is Adrian."

>>> context = Context({"my_name": "Dolores"})
>>> template.render(context)

因此修复代码的最简单方法如下:

from django.template import Context
...
return render(request=request, template_name="mytemplate.html", context=Context({"form": form, "update": updateType}))