Question

每当我按下指向下一页的链接（这是一个get请求）时，过滤器都会被绕过，并且我会获得整个列表的相应页面（没有过滤器）。例如，如果我在第1页上找到过滤列表，然后单击“下一步”，那么我将在第2页上找到整个未过滤列表。在models.py中，

Processing by BuildingsController#update as HTML
  Parameters: {"authenticity_token"=>"p2fFVpKZnw5HroZvy3qGAbguTxf+3oEOjOrhEGTu4ImS7GpVwXrcj78XPt5ENcN6GQx9K9Cn/c3ZvMwOSqVYbQ==", "building"=>{"frequency"=>["", "3"]}, "commit"=>"Update Building", "id"=>"5"}

Building Update (0.8ms)  UPDATE "buildings" SET "frequency" = $1, "updated_at" = $2 WHERE "buildings"."id" = $3  [["frequency", "---\n- ''\n- '3'\n"], ["updated_at", "2020-04-05 17:56:22.142645"], ["id", 5]]
  ↳ app/controllers/buildings_controller.rb:40:in `update'
   (1.4ms)  COMMIT
  ↳ app/controllers/buildings_controller.rb:40:in `update'
Redirected to http://localhost:3000/buildings/5
Completed 302 Found in 14ms (ActiveRecord: 3.7ms | Allocations: 6210)

在filters.py中，

class Person(models.Model):
    name = models.CharField(max_length=50, unique=True)
    gender = models.CharField(max_length=7, choices=GENDER_CHOICES)
    category = models.CharField(max_length=20, choices=get_all_category_choices())

在views.py中，

import django_filters

from .models import Person

class PersonFilter(django_filters.FilterSet):

    class Meta:
        model = Person
        fields = [
            'name',
            'gender',
            'category',
        ]

在app1 / show_all_persons_page.html中，

def show_all_persons_page(request):
    context = {}

    filtered_person_list = PersonFilter(
        request.GET,
        queryset=Person.objects.all()
    )

    context['filtered_person_list'] = filtered_person_list

    paginated_filtered_person_list = Paginator(filtered_person_list.qs, 3)
    page_number = request.GET.get('page')
    person_page_obj = paginated_filtered_person_list.get_page(page_number)

    context['person_page_obj'] = person_page_obj

    return render(request, 'app1/show_all_persons_page.html', context)

Answer 1

已解决。从https://simpleisbetterthancomplex.com/snippet/2016/08/22/dealing-with-querystring-parameters.html实现的概念。为了保留当前的url参数，我们必须定义一个自定义模板标签。在app1下创建一个名为templatetags的新目录。在该目录中创建一个__init__.py。创建另一个.py文件，例如app1_extras。在app1_extras内部，定义一个自定义模板标记，如下所示。 app1 / templatetags / app1_extras.py，

from django import template

register = template.Library()

@register.simple_tag
def relative_url(value, field_name, urlencode=None):
    url = '?{}={}'.format(field_name, value)
    if urlencode:
        querystring = urlencode.split('&')
        filtered_querystring = filter(lambda p: p.split('=')[0] != field_name, querystring)
        encoded_querystring = '&'.join(filtered_querystring)
        url = '{}&{}'.format(url, encoded_querystring)
    return url

这将获取我们当前的网址，并确保参数和页码都在适当的位置。 field_name是字符串“ page”，value是将从模板传递的页码。 urlencode将我们当前的URL表示为字符串。在app1 / show_all_persons.html中，

{% load app1_extras %}

<form method="get">
    {{ filtered_person_list.form.as_p }}
    <button type="submit">Search</button>
</form>

{% for person in person_page_obj %}
    <img src="{{ person.picture.url }}" width="240">
    <h2>
        {{person.name}}
        <br>
        {{person.gender}}
        <br>
        {{person.category}}
        <br>
    </h2>
    <br>
{% endfor %}

<div class="pagination">
    <span class="step-links">
        {% if person_page_obj.has_previous %}
            <a href="{% relative_url 1 'page' request.GET.urlencode %}">&laquo; first</a>
            <a href="{% relative_url person_page_obj.previous_page_number 'page' request.GET.urlencode %}">previous</a>
        {% endif %}

        <span class="current">
            Page {{ person_page_obj.number }} of {{ person_page_obj.paginator.num_pages }}.
        </span>

        {% if person_page_obj.has_next %}
            <a href="{% relative_url person_page_obj.next_page_number 'page' request.GET.urlencode %}">next</a>
            <a href="{% relative_url person_page_obj.paginator.num_pages 'page' request.GET.urlencode %}">last &raquo;</a>
        {% endif %}
    </span>
</div>

{％load app1_extras％}将加载我们的新自定义模板标记。 relative_url函数现在传递给我们页面链接的href。这些参数按顺序传递，其中request.GET.urlencode将包含我们当前的URL，并将传递给urlencode参数。

有什么办法可以将分页与django-filter一起使用吗？

1 个答案: