如何在 django 1.3 中获取 POST 数据

发布于 2024-11-07 17:35:59 字数 2318 浏览 0 评论 0 原文

嘿,我正在按照本教程学习使用 Django 制作 wiki 页面。然而,它是用 django 0.96 制作的,而我使用 Django 1.3,所以有些东西是不同的。有些我已经自己修复了,但是这个我似乎无法让它发挥作用。

我制作了一个将数据提交到视图的表单。 这是形式:

<form method="post" action"/wikicamp/{{page_name}}/save/">{% csrf_token %}
    <textarea name="content" rows="20" cols="60">{{content}}</textarea><br>
    <input type="submit" value="Save Page"/>
</form>

并且 /wikicamp/{{page_name}}/save/ url 重定向到 save_page 视图:

from django.http import HttpResponseRedirect
from django.core.context_processors import csrf

def save_page(request, page_name):
    c = {}
    c.update(csrf(request))
    content = c.POST["content"]
    try:
        page = Page.objects.get(pk=page_name)
        page.content = content
    except Page.DoesNotExist:
        page = Page(name=page_name, content=content)
    page.save()
    return HttpResponseRedirect("wikicamp/" + page_name + "/")

但是问题是我收到此错误:

Help

Reason given for failure:

    CSRF token missing or incorrect.


In general, this can occur when there is a genuine Cross Site Request Forgery, or when Django's CSRF mechanism has not been used correctly. For POST forms, you need to ensure:

    The view function uses RequestContext for the template, instead of Context.
    In the template, there is a {% csrf_token %} template tag inside each POST form that targets an internal URL.
    If you are not using CsrfViewMiddleware, then you must use csrf_protect on any views that use the csrf_token template tag, as well as those that accept the POST data.

You're seeing the help section of this page because you have DEBUG = True in your Django settings file. Change that to False, and only the initial error message will be displayed.

You can customize this page using the CSRF_FAILURE_VIEW setting.

所以我阅读了一些文档,例如 http://docs.djangoproject.com/en/dev/ref/contrib/csrf/#how-to-use-it。但我尝试这样做,但仍然出现相同的错误。

那么:有人知道如何使用 Django 1.3 很好地处理表单发布数据吗?

我认为这与以下内容有关:视图函数使用 RequestContext 作为模板,而不是 Context。但我现在不知道它是什么。

顺便说一句,在我的终端中显示本地主机的 http 请求,它说:模板中使用了 {% csrf_token %},但上下文未提供该值。这通常是由于没有使用 RequestContext 造成的。

Hey, I am following this tutorial to learn to make a wiki page with Django. However, it is made in django 0.96 and I use Django 1.3 so there are some things that are different. Some I already fixed myself, however this one I can't seem to make it work.

I made a form that submits data to a view.
This is the form:

<form method="post" action"/wikicamp/{{page_name}}/save/">{% csrf_token %}
    <textarea name="content" rows="20" cols="60">{{content}}</textarea><br>
    <input type="submit" value="Save Page"/>
</form>

and the /wikicamp/{{page_name}}/save/ url redirects to the save_page view:

from django.http import HttpResponseRedirect
from django.core.context_processors import csrf

def save_page(request, page_name):
    c = {}
    c.update(csrf(request))
    content = c.POST["content"]
    try:
        page = Page.objects.get(pk=page_name)
        page.content = content
    except Page.DoesNotExist:
        page = Page(name=page_name, content=content)
    page.save()
    return HttpResponseRedirect("wikicamp/" + page_name + "/")

However the problem is that I get this error:

Help

Reason given for failure:

    CSRF token missing or incorrect.


In general, this can occur when there is a genuine Cross Site Request Forgery, or when Django's CSRF mechanism has not been used correctly. For POST forms, you need to ensure:

    The view function uses RequestContext for the template, instead of Context.
    In the template, there is a {% csrf_token %} template tag inside each POST form that targets an internal URL.
    If you are not using CsrfViewMiddleware, then you must use csrf_protect on any views that use the csrf_token template tag, as well as those that accept the POST data.

You're seeing the help section of this page because you have DEBUG = True in your Django settings file. Change that to False, and only the initial error message will be displayed.

You can customize this page using the CSRF_FAILURE_VIEW setting.

So I read through some of the documentation, like http://docs.djangoproject.com/en/dev/ref/contrib/csrf/#how-to-use-it. I tried to do that however and it still gave the same error.

So: Anyone an idea how to handle form post data well with Django 1.3?

I think it has something to do with: The view function uses RequestContext for the template, instead of Context. but i don't now what it is.

btw, in my terminal which shows the http request of the localhost it says this: A {% csrf_token %} was used in a template, but the context did not provide the value. This is usually caused by not using RequestContext.

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评论(4

孤檠 2024-11-14 17:35:59

您将需要在标签之间添加 {% csrf_token %} 模板标签,并包含

   django.middleware.csrf.CsrfViewMiddleware
   django.middleware.csrf.CsrfResponseMiddleware

在应用程序 settings.py 中的 MIDDLEWARE_CLASSES 中

添加一些示例发布数据处理:

这是我在其中使用 POST 数据的一次示例一个视图。我通常会依靠表单类通过 clean_data 数组进行验证。

if request.method == 'POST':
        form = ForgotPassword(data=request.POST)
        if form.is_valid():
            try:
                new_user = backend.forgot_password(request, **form.cleaned_data)
            except IntegrityError:
                context = {'form':form}
                form._errors[''] = ErrorList(['It appears you have already requested a password reset, please \
                check ' + request.POST['email2'] + ' for the reset link.'])
                return render_template(request,'passwordReset/forgot_password.html',context)
            if success_url is None:
                to, args, kwargs = backend.post_forgot_password(request, new_user)
                return redirect(to, *args, **kwargs)
            else:
                return redirect(success_url)

You will need the {% csrf_token %} template tag in between your tags as well as including

   django.middleware.csrf.CsrfViewMiddleware
   django.middleware.csrf.CsrfResponseMiddleware

in your MIDDLEWARE_CLASSES in the applications settings.py

Adding some example post data handling:

This is an example of one of the times I am using POST data in a view. I will generally rely on the form class to do verification via the cleaned_data array.

if request.method == 'POST':
        form = ForgotPassword(data=request.POST)
        if form.is_valid():
            try:
                new_user = backend.forgot_password(request, **form.cleaned_data)
            except IntegrityError:
                context = {'form':form}
                form._errors[''] = ErrorList(['It appears you have already requested a password reset, please \
                check ' + request.POST['email2'] + ' for the reset link.'])
                return render_template(request,'passwordReset/forgot_password.html',context)
            if success_url is None:
                to, args, kwargs = backend.post_forgot_password(request, new_user)
                return redirect(to, *args, **kwargs)
            else:
                return redirect(success_url)
守望孤独 2024-11-14 17:35:59

您必须在表单模板中的

标记之间包含 {% csrf_token %}

<form method="post" action"/wikicamp/{{page_name}}/save/">
    {% csrf_token %}
    <textarea name="content" rows="20" cols="60">{{content}}</textarea><br>
    <input type="submit" value="Save Page"/>
</form>

如果 csrf_token 未呈现到您的表单中,请确保您在视图的响应中提供 RequestContext

from django.shortcuts import render_to_response
from django.template import RequestContext

def app_view(request):
    return render_to_response('app_template.html', 
                              app_data_dictionary, 
                              context_instance=RequestContext(request))

或者,使用此快捷方法:

from django.views.generic.simple import direct_to_template

def app_view(request):             
    return direct_to_template(request, 'app_template.html', app_data_dictionary)

RequestContext 始终可用。

You've got to include {% csrf_token %} in your form's template between your <form> tags.

<form method="post" action"/wikicamp/{{page_name}}/save/">
    {% csrf_token %}
    <textarea name="content" rows="20" cols="60">{{content}}</textarea><br>
    <input type="submit" value="Save Page"/>
</form>

If the csrf_token is not rendered into your form make sure you're providing the RequestContext in the view's response:

from django.shortcuts import render_to_response
from django.template import RequestContext

def app_view(request):
    return render_to_response('app_template.html', 
                              app_data_dictionary, 
                              context_instance=RequestContext(request))

Or, use this shortcut method:

from django.views.generic.simple import direct_to_template

def app_view(request):             
    return direct_to_template(request, 'app_template.html', app_data_dictionary)

The RequestContext is always available when you're using generic views.

难忘№最初的完美 2024-11-14 17:35:59

我猜您错过了表单声明中的符号“=”。

action"/wikicamp/{{page_name}}/save/"

action="/wikicamp/{{page_name}}/save/"

幸运的是,这可能不是一个错误。
因此,如果这不是解决方案,请尝试一些更简单的示例:

# settings.py

TEMPLATE_DIRS = (
    # Here comes something like "C:/www/django/templates"
)

MIDDLEWARE_CLASSES = (
    ...
    'django.middleware.csrf.CsrfViewMiddleware',
    ...
)

# urls.py

urlpatterns = patterns('',
    ('^foo', foo),
)


# views.py
from django.http import HttpResponse
from django.shortcuts import render_to_response
from django.core.context_processors import csrf

def foo(request):
    d = {}
    d.update(csrf(request))
    if 'output' in request.POST:
        d.update({'output':request.POST['output']})
    return render_to_response('foo.html',d)

# foo.html template
<html>
<h1> Foo </h1>
<form action="/foo" method = "post">
    {% csrf_token %}
    <input type="text" name="output"></input>
    <input type="submit" value="go"></input>
</form>
<p> Output: {{ output }} </p>
</html>

希望这会起作用

I guess you've missed the symbol '=' in the form declaration.

action"/wikicamp/{{page_name}}/save/"

action="/wikicamp/{{page_name}}/save/"

Fortunately, it might be not a mistake.
So if it is not a solution, try some more easy example:

# settings.py

TEMPLATE_DIRS = (
    # Here comes something like "C:/www/django/templates"
)

MIDDLEWARE_CLASSES = (
    ...
    'django.middleware.csrf.CsrfViewMiddleware',
    ...
)

# urls.py

urlpatterns = patterns('',
    ('^foo', foo),
)


# views.py
from django.http import HttpResponse
from django.shortcuts import render_to_response
from django.core.context_processors import csrf

def foo(request):
    d = {}
    d.update(csrf(request))
    if 'output' in request.POST:
        d.update({'output':request.POST['output']})
    return render_to_response('foo.html',d)

# foo.html template
<html>
<h1> Foo </h1>
<form action="/foo" method = "post">
    {% csrf_token %}
    <input type="text" name="output"></input>
    <input type="submit" value="go"></input>
</form>
<p> Output: {{ output }} </p>
</html>

Hope this will work

情愿 2024-11-14 17:35:59

上面在第三行中使用“request.POST”而不是“c.POST”

def save_page (request,page_name):
    content = request.POST["content"]

并在“edit_page”中进行更改

-   return render_to_response("edit.html",{"page_name":page_name, "content":content})
+   t = get_template('edit.html')
+   html = t.render(Context({"page_name":page_name, "content":content}))
+   return HttpResponse(html)

- :remove 
+ :add

re above use "request.POST" not "c.POST" in the 3rd line

def save_page (request,page_name):
    content = request.POST["content"]

and change in "edit_page"

-   return render_to_response("edit.html",{"page_name":page_name, "content":content})
+   t = get_template('edit.html')
+   html = t.render(Context({"page_name":page_name, "content":content}))
+   return HttpResponse(html)

- :remove 
+ :add
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