Question

在我的应用程序中，我需要在base.html中添加一个表单，我已经完成了。为此我使用了context_processors，现在我的问题是每次我试图发布，我得到一个空白页面并且出现此错误：Method Not Allowed (POST)

在这种形式中，我只想添加一个按钮，将所有当前用户的通知标记为已读。

我知道你可以使用像这样的context_processors：

def my_context(request):
    data = dict()
    if request.user.is_authenticated:
        data['notifications'] = Notification.objects.filter(user=request.user, read=False)
        data['form'] = NotificationForm()
        return data

但不是添加表格我需要这些行：

def my_context(request):
    data = dict()
    if request.user.is_authenticated:
        data['notifications'] = Notification.objects.filter(user=request.user, read=False)
    if request.method == 'POST':
        if 'read-notifications' in request.POST:
            for notification in data['notifications']:
                notification.read = True
                notification.save()
        next = request.POST.get('next', '/')
        return redirect(next)
    return data

base.html中的表单：

<form action="" method="POST">{% csrf_token %}
    <input type="hidden" name="next" value="{{ request.path }}">
    <button type="submit" class="btn m-btn--square  btn-primary" name="read-notifications">Mark as read</button>
</form>

urls.py

url(r'^orders/create/$', views.admin_order_document, name='create_order'),
url(r'^orders/waiting/$', views.OrdersWaitingListView.as_view(), name='order_waiting_list'),
url(r'^orders/unallocated/$', views.OrdersUnallocatedListView.as_view(), name='order_unallocated_list'),
url(r'^orders/working/$', views.OrdersWorkingListView.as_view(), name='order_working_list'),

如何在不出现上述错误的情况下在每个页面上显示此表单？

Answer 1

所以，我选择了不同的方法，我最终得到了这个解决方案：

base.html文件

<form action="{% url 'mark_read' %}" method="POST">{% csrf_token %}
    <input type="hidden" name="next" value="{{ request.path }}">
    <button type="submit" class="btn m-btn--square  btn-primary" name="action" value="mark_read">Marchează ca citite</button>
</form>

views.py

@login_required()
def read_notifications(request):
    if request.method == 'POST':
        if 'action' in request.POST:
            notifications = Notification.objects.filter(user=request.user, read=False)
            for notification in notifications:
                notification.read = True
                notification.save()
        next = request.POST.get('next', '/')
        return redirect(next)

urls.py

url(r'^mark-read/$', views.read_notifications, name='mark_read'),

基本上，表单操作会将我重定向到视图功能，它会执行我必须执行的操作，它会将我重定向回同一页面。

Django在每个页面上添加表单

1 个答案: