在 Django ListView 中实现查询表单的最佳方式是什么？

Question

上周我在 Django 中遇到了一个与 listView 和 Forms 相关的任务，我想知道在 ListView 中实现搜索表单的最佳方式（+Pythonic）是什么，在我阅读 1 and 2 之后我明白了一个主要想法，所以我实施了第一个解决方案，我想收到您的反馈。这里的目标是通过代码字段执行查询并保留查询集以便与分页同步。

forms.py

class InscriptionQueryForm(forms.Form):
    query_inscription = forms.CharField(label=_('Code'), required=False)

models.py

class Inscription(models.Model):
    code = models.CharField(max_length=10, unique=True)
    start_on = models.DateField()
    finish_on = models.DateField()
    active = models.BooleanField(default=False)

views.py

class InscriptionListView(ListView, FormMixin):
    model = Inscription
    paginate_by = 4
    context_object_name = 'inscriptions'
    form_class = InscriptionQueryForm
    form = None
    object_list = None
    search = False

    def get_queryset(self):
        form = self.form_class(self.request.POST)

        if form.is_valid() and self.request.method == 'POST':
            self.request.session['query_inscription'] = \
                form.cleaned_data['query_inscription']
            return self.model.objects.filter(
                code__icontains=form.cleaned_data['query_inscription']).\
                order_by('-active')

        if self.request.method == 'GET' and \
                'query_inscription' in self.request.session:

            return self.model.objects.filter(
                code__icontains=self.request.session.get(
                    'query_inscription', '')).order_by('-active')

        return self.model.objects.all().order_by('-active')

    def get(self, request, *args, **kwargs):
        # From ProcessFormMixin
        self.form = self.get_form(self.form_class)

        # From BaseListView
        if self.request.GET.get('page', False) or self.search:
            self.object_list = self.get_queryset()
        else:
            self.search = False
            self.object_list = self.model.objects.all().order_by('-active')
            if 'query_inscription' in self.request.session:
                del self.request.session['query_inscription']

        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):
        self.search = True
        return self.get(request, *args, **kwargs)

大家怎么看？我相信还有很多其他更好的方法。

Answer 1

上周我遇到了类似的问题。我的模型是一个普通的 django 用户。然而，这里可以使用相同的方法。

我猜您想使用搜索字段搜索铭文并对结果进行分页。当您浏览这些页面时，您希望看到搜索查询的结果。

您的 models.py 保持原样。我们将修改 forms.py 因为它将能够初始化搜索请求

class InscriptionQueryForm(forms.Form):
query_inscription = forms.CharField(label='Code', required=False)

def __init__(self, query_inscription):
    super(InscriptionQueryForm, self).__init__()
    self.fields['query_inscription'].initial = query_inscription

现在让我们看看views.py。您不需要像 schillingt 在他的评论中提到的那样在会话中存储您的请求值。您只需使用搜索请求初始化您的表单。

class InscriptionListView(ListView):
model = Inscription
paginate_by = 4
context_object_name = 'inscriptions'
query_inscription = ''

def get_context_data(self, **kwargs):
    context = super(InscriptionListView, self).get_context_data(**kwargs)
    context['form'] = InscriptionQueryForm(self.query_inscription)
    # this parameter goes for the right pagination
    context['search_request'] = ('query_inscription=' +
                                 unicode(self.query_inscription))
    return context

def get(self, request, *args, **kwargs):
    self.query_inscription = request.GET.get('query_inscription', '')
    return super(InscriptionListView, self).get(request, *args, **kwargs)

def get_queryset(self):
    if not self.query_inscription:
        inscription_list = Inscription.objects.all()
    else:
        inscription_list = Inscription.objects.filter(
            code__icontains=self.query_inscription)

    return inscription_list.order_by('-active')

最后要提到的是分页 inscription_list.html

<form action="">
{{ form.as_p }}
<input type="submit" value="search"/>
</form>
<hr/>
{% if inscriptions %}
    {% for inscription in inscriptions %}
        {{ inscription.code }} {{ inscription.start_on }} {{ inscription.finish_on }}
        <hr/>
    {% endfor %}
    {% if is_paginated %}
        <div class="pagination">
            <span class="page-links">
                {% if page_obj.has_previous %}
                    <a href="?{{ search_request }}&page={{ page_obj.previous_page_number }}">previous</a>
                {% endif %}
                <span class="page-current">
                    Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}
                </span>
                {% if page_obj.has_next %}
                    <a href="?{{ search_request }}&page={{ page_obj.next_page_number }}">next</a>
                {% endif %}
            </span>
        </div>
    {% endif %}
{% endif %}

就是这样！

在 Django ListView 中实现查询表单的最佳方式是什么？

What is the best way to implement Query Form in Django ListView?

django

listview

django-class-based-views

django-1.7