如何在ListView Django中进行过滤和分页

Question

当我想用django_filter创建的过滤器分页时，我遇到了一个问题，在我的模板中，它向我显示了查询集和过滤器，但分页不起作用，我想知道为什么会发生这种情况，如果你能帮我的话。

我将插入我的代码片段，以便您可以看到。

这是我的观点.py

PD：我有所有必需的进口商品。

@method_decorator(staff_member_required, name='dispatch')
class EmployeeListView(ListView):
model = Employee
paginate_by = 4
def dispatch(self, request, *args, **kwargs):
if not request.user.has_perm('employee.view_employee'):
return redirect(reverse_lazy('home'))
return super(EmployeeListView, self).dispatch(request, *args, **kwargs)

def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['filter'] = EmployeeFilter(self.request.GET, queryset = self.get_queryset())
return context

过滤器.py

import django_filters
from .models import Employee, Accident
class EmployeeFilter(django_filters.FilterSet):
class Meta:
model = Employee
fields = {
'rutEmployee' : ['startswith']
}

Answer 1

您应该覆盖get_queryset。这意味着你必须像这样把你的过滤器放在get_queryset中：

@method_decorator(staff_member_required, name='dispatch')
class EmployeeListView(ListView):
model = Employee
paginate_by = 4
def dispatch(self, request, *args, **kwargs):
if not request.user.has_perm('employee.view_employee'):
return redirect(reverse_lazy('home'))
return super(EmployeeListView, self).dispatch(request, *args, **kwargs)
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['filter'] = EmployeeFilter(self.request.GET, queryset = self.get_queryset())
return context

def get_queryset(self):
queryset = super().get_queryset()
return EmployeeFilter(self.request.GET, queryset=queryset).qs

并在employee_list.html中使用object_list而不是filter，如下所示：

{% for employee in object_list|dictsort:"id" reversed %}

Answer 2

您也可以尝试以下操作：(我的源代码片段(

class ModelListView(ListView):
model = YourModel
paginate_by = 4 # Change this if you don't intend to paginate by 4
ordering = model_field_to_order_by
# variable used to know if a match was found for the search made using django_filters
no_search_result = False
def get_queryset(self, **kwargs):
search_results = YourDjangoFiltersForm(self.request.GET, self.queryset)
self.no_search_result = True if not search_results.qs else False
# Returns the default queryset if an empty queryset is returned by the django_filters
# You could as well return just the search result's queryset if you want to
return search_results.qs.distinct() or self.model.objects.all()
def get_query_string(self):
query_string = self.request.META.get("QUERY_STRING", "")
# Get all queries excluding pages from the request's meta
validated_query_string = "&".join([x for x in re.findall(
r"(w*=w{1,})", query_string) if not "page=" in x])
# Avoid passing the query path to template if no search result is found using the previous query
return "&" + validated_query_string.lower() if (validated_query_string and not self.no_search_result) else ""
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
# Pass to template if you want to do something whenever an empty queryset is return by django_filters
context["no_search_result"] = self.no_search_result
# This is the query string which should be appended to the current page in your template for pagination, very critical
context["query_string"] = self.get_query_string()
context['filter'] = YourDjangoFiltersForm()
return context

在html模板中，您需要附加从视图传递到模板的查询字符串，下面显示了一个示例

{% for i in page_obj.paginator.page_range %}
{% if page_obj.number == i %}
<li class="page-item active" aria-current="page">
<span class="page-link">{{ i }}<span class="sr-only">(current)</span></span>
</li>
{% elif i > page_obj.number|add:'-5' and i < page_obj.number|add:'5' %} <li class="page-item"><a
class="page-link" href="?page={{ i }}{{ query_string }}">{{ i }}</a></li>
{% endif %}
{% endfor %}