带有表单的基于 Django 类的视图 ListView

Question

主视图是一个简单的分页 ListView，我想给它添加一个搜索表单。

我认为这样的事情可以解决问题：

class MyListView(ListView, FormView):
    form_class = MySearchForm
    success_url = 'my-sucess-url'
    model = MyModel
    # ...

但显然我弄错了..我在官方文档中找不到如何做到这一点。

建议？

Answer 1

在 Django 2.2 中，您可以这样做（至少使用 get-request 可以正常工作）：

from django.views.generic import ListView
from django.views.generic.edit import FormMixin

from .models import MyModel
from .forms import MySearchForm

class ListPageView(FormMixin, ListView):
    template_name = 'property_list.html'
    model = MyModel
    form_class = MySearchForm
    queryset = MyModel.objects.all()

在ListView 之前使用FormMixin。如果你想在TemplateView 中使用SearchForm，你可以这样做：

from django.views.generic.base import TemplateView
from django.views.generic.edit import FormMixin

from .models import MyModel
from .forms import MySearchForm

class HomePageView(FormMixin, TemplateView):
    template_name = 'home.html'
    form_class = MySearchForm

Answer 2

the official documentation 中介绍了使用 mixins 向索引和列表视图添加表单。

文档通常建议不要使用这种方法。它建议改为简单地编写更多 python，并手动编写视图。

Answer 3

我也一直在寻找合适的解决方案。但我找不到任何东西，所以只好自己想出一个。

views.py

class VocationsListView(ListView):

    context_object_name = "vocations"
    template_name = "vocations/vocations.html"
    paginate_by = 10

    def get_queryset(self):
        get = self.request.GET.copy()
        if(len(get)):
            get.pop('page')
        self.baseurl = urlencode(get)
        model = Vocation
        self.form = SearchForm(self.request.GET)
        filters = model.get_queryset(self.request.GET)
        if len(filters):
            model = model.objects.filter(filters)
        else:
            model = model.objects.all()
        return model



def get_context_data(self):
    context = super(VocationsListView, self).get_context_data()
    context['form'] = self.form
    context['baseurl']= self.baseurl
    return context

models.py

class Vocation(models.Model):
    title = models.CharField(max_length = 255)
    intro = models.TextField()
    description = models.TextField(blank = True)
    date_created = models.DateTimeField(auto_now_add = True)
    date_modified = models.DateTimeField(auto_now = True)
    created_by = models.ForeignKey(User, related_name = "vocation_created")
    modified_by = models.ForeignKey(User, related_name = "vocation_modified")

    class Meta:
        db_table = "vocation"

    @property
    def slug(self):
        return defaultfilters.slugify(self.title)

    def __unicode__(self):
        return self.title

    @staticmethod
    def get_queryset(params):

        date_created = params.get('date_created')
        keyword = params.get('keyword')
        qset = Q(pk__gt = 0)
        if keyword:
            qset &= Q(title__icontains = keyword)
        if date_created:
            qset &= Q(date_created__gte = date_created)
        return qset

所以基本上我将这段代码添加到我想要实现搜索功能的每个模型类中。这是因为必须明确准备每个模型的过滤器

@staticmethod
def get_queryset(params):

    date_created = params.get('date_created')
    keyword = params.get('keyword')
    qset = Q(pk__gt = 0)
    if keyword:
        qset &= Q(title__icontains = keyword)
    if date_created
        qset &= Q(date_created__gte = date_created)
    return qset

它准备了我用来从模型中检索数据的 qset 过滤器

Answer 4

根据以前的答案，这是我对用于在与 ListView 相同的页面上显示表单的视图的看法：

class IndexView(FormMixin, ListView):
    ''' Homepage: displays list of links, and a form used to create them '''
    template_name = "links/index.html"
    context_object_name = "links"
    form_class = LinkForm

    def get_queryset(self):
        return Links.objects.all()

def add_link(request):
    # Sole job of this function is to process the form when POSTed. 
    if request.method == "POST":
        form = LinkForm(request.POST)

        if form.is_valid():
            Links.objects.create(address=form.cleaned_data['address'])

        return HttpResponseRedirect('/')

然后，最后一件事是将 add_link 视图函数绑定到表单的操作 url，我认为你可以开始了。

Answer 5

这些答案极大地帮助了我朝着正确的方向前进。谢谢各位。

对于我的实现，我需要一个在获取和发布时都返回 ListView 的表单视图。 我不喜欢重复 get 函数的内容，但它需要一些更改。该表单现在也可以从 get_queryset 使用 self.form 获得。

from django.http import Http404
from django.utils.translation import ugettext as _
from django.views.generic.edit import FormMixin
from django.views.generic.list import ListView

class FormListView(FormMixin, ListView):
    def get(self, request, *args, **kwargs):
        # From ProcessFormMixin
        form_class = self.get_form_class()
        self.form = self.get_form(form_class)

        # From BaseListView
        self.object_list = self.get_queryset()
        allow_empty = self.get_allow_empty()
        if not allow_empty and len(self.object_list) == 0:
            raise Http404(_(u"Empty list and '%(class_name)s.allow_empty' is False.")
                          % {'class_name': self.__class__.__name__})

        context = self.get_context_data(object_list=self.object_list, form=self.form)
        return self.render_to_response(context)

    def post(self, request, *args, **kwargs):
        return self.get(request, *args, **kwargs)


class MyListView(FormListView):
    form_class = MySearchForm
    model = MyModel
    # ...