成功时如何重定向UpdateView? [英] How to redirect an UpdateView upon success?

问题描述

我创建了一个小的Django应用程序来管理适合简单模型的数据.现在，我只需要两个视图:一个视图列出所有记录，另一个视图以通用形式编辑记录.除成功更新后从编辑视图重定向之外，其他所有功能均按预期运行. urls.py 中包含以下内容:

I created a small Django application to manage data that fits a simple a model. For now I only need two views: one to list all records and another to edit a record with a generic form. Everything functions as expected, except the redirection from the edit view upon a successful update. In urls.py are the following contents:

from django.urls import path
from . import views

app_name = 'reqs'
urlpatterns = [
    path('', views.IndexView.as_view(), name='index'),
    path('<int:pk>/', views.ReqUpdateView.as_view(), name='update'),
]

在 forms.py 中:

from django.forms import ModelForm
from .models import Requirement


class RequirementForm(ModelForm):
    class Meta:
        model = Requirement
        fields = ['name', 'priority', 'source' , 'rationale']

还有庙宇 requirement_form.html :

<h1>{{ requirement.id }} - {{ requirement.name }}</h1>

<form method="post" novalidate>
  {% csrf_token %}
    <table>
    {{ form.as_table }}
    <tr><td></td><td><button type="submit">Save</button></td></tr>
    </table>
</form>
    
{% if error_message %}<p><strong>{{ error_message }}</strong></p>{% endif %}

<br><br>
<a href="{% url 'reqs:index' %}">Back to list</a>

最后一次尝试将更新重定向到列表的 views.py :

Finally views.py, on a first attempt to redirect the update to the list:

from django.views.generic import ListView, UpdateView
from django.urls import reverse_lazy
from .models import Requirement
from .forms import RequirementForm

class IndexView(ListView):
    template_name = 'reqs/index.html'
    context_object_name = 'requirements_list'

    def get_queryset(self):
        return Requirement.objects.order_by('subject')

class ReqUpdateView(UpdateView):
    model = Requirement
    form_class = RequirementForm
    success_url = reverse_lazy('/')

使用此公式，保存按钮会产生以下错误:

With this formulation the Save button produces this error:

找不到反向的'/'."/"不是有效的视图函数或模式名称.

Reverse for '/' not found. '/' is not a valid view function or pattern name.

我还尝试使用空字符串作为 reverse_lazy 的参数以及路径名 index ，但是会生成类似的错误消息.

I also tried an empty string as argument to reverse_lazy, as well as the path name index, but a similar error message is produced.

再次尝试重定向到同一页面，重新定义 get_success_url 方法什么也不做:

On a second attempt I tried to redirect to the same page, redefining the get_success_url method to do nothing:

class ReqUpdateView(UpdateView):
    model = Requirement
    form_class = RequirementForm
    context_object_name = 'requirement_update'
    def get_success_url(self):
        pass #return the appropriate success url

这将返回404错误，尝试将浏览器重定向到/reqs/1/None .

This returns a 404 error trying to redirect the browser to /reqs/1/None.

第三次尝试重定向到具有相同记录的表单:

A third attempt to redirect to the form with the same record:

class ReqUpdateView(UpdateView):
    model = Requirement
    form_class = RequirementForm
    context_object_name = 'requirement_update'
    def get_success_url(self):
           pk = self.kwargs["pk"]
           return reverse("update", kwargs={"pk": pk})

哪个抱怨找不到视图:

未找到反向的更新".更新"不是有效的视图函数或模式名称.

Reverse for 'update' not found. 'update' is not a valid view function or pattern name.

如何将成功重定向到有效URL?只要有效，它既可以是项目列表，也可以是项目更新视图.

How can I redirect success to a valid URL? It can either be the items list or the item update view, as long as it works.

推荐答案

reverse/reverse_lazy用于使用视图名称或模式名称获取URL.如果您想直接使用网址，只需输入:

reverse / reverse_lazy are used to get the url using view name or pattern name. If you want to use a url directly just write:

success_url = '/'

对于 return reverse("update"，kwargs = {" pk:pk})而言，由于您设置了 app_name ='reqs'您应该使用 return reverse("reqs:update"，kwargs = {" pk:pk}).

For the case of return reverse("update", kwargs={"pk": pk}) not working since you set app_name = 'reqs' you should be using return reverse("reqs:update", kwargs={"pk": pk}) instead.

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