django-filter 使用分页

Question

我正在使用 django-filter 包在我的列表视图中提供搜索功能。

现在我也想为该视图添加分页。
我正在尝试将分页与过滤查询集相结合，但我不知道如何继续。

到目前为止，我已经在 views.py 上尝试了以下操作：

def search(request):
    qs = local_url.objects.filter(global_url__id=1).all()
    paginator = Paginator(qs, 25)
    page = request.GET.get('page')
    try:
        pub = paginator.page(page)
    except PageNotAnInteger:
        pub = paginator.page(1)
    except EmptyPage:
       pub = paginator.page(paginator.num_pages)
    url_filter = PublicationFilter(request.GET, queryset=qs)
    return render(request, 'ingester/search_list.html', {'filter': url_filter, 'publication':pub})

Answer 1

据我了解，您的目标是对过滤后的查询集进行分页。如果是这样，您可以将 PublicationFilter 对象的 "qs" 属性传递给 Paginator 构造函数：

def search(request):
    qs = local_url.objects.filter(global_url__id=1).all()
    url_filter = PublicationFilter(request.GET, queryset=qs)
    paginator = Paginator(url_filter.qs, 25)
    page = request.GET.get('page')
    try:
        pub = paginator.page(page)
    except PageNotAnInteger:
        pub = paginator.page(1)
    except EmptyPage:
        pub = paginator.page(paginator.num_pages)
    url_filter = PublicationFilter(request.GET, queryset=qs)
    return render(request, 'ingester/search_list.html', {'publication':pub})

url_filter.qs 包含筛选的查询集
url_filter.queryset 包含未过滤的查询集

Answer 2

要使用 Django 过滤器并对过滤结果进行分页，您可以执行以下操作：

为您的模型创建过滤器 class：

在 my_project/my_app/filters.py:

import django_filters

class MyModelFilter(django_filters.FilterSet):
    class Meta:
        model = MyModel
        # Declare all your model fields by which you will filter
        # your queryset here:
        fields = ['field_1', 'field_2', ...]

每个 FilterSet 对象都有一个 .qs 属性其中包含 过滤查询集 你甚至可以 override it if you want.

我们将分页 .qs 属性的 MyModelFilter:

在 my_project/my_app/views.py:

from . import filters
from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger

def my_view(request):
    # BTW you do not need .all() after a .filter() 
    # local_url.objects.filter(global_url__id=1) will do
    filtered_qs = filters.MyModelFilter(
                      request.GET, 
                      queryset=MyModel.objects.all()
                  ).qs
    paginator = Paginator(filtered_qs, YOUR_PAGE_SIZE)

    page = request.GET.get('page')
    try:
        response = paginator.page(page)
    except PageNotAnInteger:
        response = paginator.page(1)
    except EmptyPage:
        response = paginator.page(paginator.num_pages)

    return render(
        request, 
        'your_template.html', 
        {'response': response}
    )

给你！

PS_1: 我的 Django 过滤器，"plays" better with Django Rest Framework.

PS_2: 如果你打算使用 DRF，我已经写了一个关于如何在基于函数的视图中使用分页的例子，你可以很容易地结合一个 FilterSet:

@api_view(['GET',])
def my_function_based_list_view(request):
    paginator = PageNumberPagination()
    filtered_set = filters.MyModelFilter(
                       request.GET, 
                       queryset=MyModel.objects.all()
                   ).qs
    context = paginator.paginate_queryset(filtered_set, request)
    serializer = MyModelSerializer(context, many=True)
    return paginator.get_paginated_response(serializer.data)

Answer 3

这对我有用：

在我的模板中而不是使用这个

<li><a href="?page={{ i }}">{{ i }}</a></li>

我写了这个：

{% if 'whatever_parameter_you_use_to_filter' in request.get_full_path %}
   <li><a href="{{ request.get_full_path }}&page={{ i }}"{{ i }}</a></li>
{% else %}
   <li><a href="?page={{ i }}">{{ i }}</a></li>
{% endif %}

希望对您有所帮助:)

Answer 4

这里最重要的部分是如何在模板中构建 URL。

你可能有

{% if pages.has_previous %}
<li><a href="?page={{ pages.previous_page_number }}">Prev</a></li>
{% endif %}

如果您仅使用它在初始分页结果之间切换，这非常好。

但棘手的部分是当您使用 django-fitler 过滤器时，查询字符串（'?' 之后的那部分）变得全新 key-values对，忽略你的 ?page=2 或类似的。

因此，要使分页与过滤结果一起使用，当您单击 "Next" 或 "Prev" 按钮时 - 在 django-fitler 的 key-values 中，您还需要通过&page=5 一对。

如@stathoula 所述，您需要检查查询字符串中是否已存在至少一个过滤字段。如果是，那么您需要使用已经存在的 key-value 对，然后是新的 &page=3 对。

这看起来很简单，但我不得不做一些小技巧，不要在用户点击箭头时在查询字符串中一遍又一遍地重复 &page=1。

在我的例子中，我将 'title' 作为过滤器，所以我需要检查它是否已经存在。

这是我为我的项目所做的完美工作的片段。

templates/pagination.html

<div class="paginator">

    {% with request.get_full_path as querystring %}
        <ul class="pagination nav navbar-nav">

            <!-- Previous page section -->
            {% if pages.has_previous %}
                {% if 'title' in querystring %}
                    {% if 'page' in querystring %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring|slice:":-7" }}&page={{ pages.previous_page_number }}">Prev</a>
                        </li>
                    {% else %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring }}&page={{ pages.previous_page_number }}">Prev</a>
                        </li>
                    {% endif %}
                {% else %}
                    <li class="paginator {% if pages.number == page %}active{% endif %}">
                        <a href="?page={{ pages.previous_page_number }}">Prev</a>
                    </li>
                {% endif %}
            {% endif %}

            <!-- All pages section -->
            {% for page in pages.paginator.page_range %}
                {% if 'title' in querystring %}
                    {% if 'page' in querystring %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring|slice:":-7" }}&page={{ page }}">{{ page }}</a>
                        </li>
                    {% else %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring }}&page={{ page }}">{{ page }}</a>
                        </li>
                    {% endif %}
                {% else %}
                    <li class="paginator {% if pages.number == page %}active{% endif %}">
                        <a href="?page={{ page }}">{{ page }}</a>
                    </li>
                {% endif %}
            {% endfor %}

            <!-- Next page section -->
            {% if pages.has_next %}
                {% if 'title' in querystring %}
                    {% if 'page' in querystring %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring|slice:":-7" }}&page={{ pages.next_page_number }}">Next</a>
                        </li>
                    {% else %}
                        <li class="paginator {% if pages.number == page %}active{% endif %}">
                            <a href="{{ querystring }}&page={{ pages.next_page_number }}">Next</a>
                        </li>
                    {% endif %}
                {% else %}
                    <li class="paginator {% if pages.number == page %}active{% endif %}">
                        <a href="?page={{ pages.next_page_number }}">Next</a>
                    </li>
                {% endif %}
            {% endif %}

        </ul>
    {% endwith %}

</div>

这里是视图，以防万一：

app/views.py

def index(request):
    condo_list = Condo.objects.all().order_by('-timestamp_created')
    condo_filter = CondoFilter(request.GET, queryset=condo_list)

    paginator = Paginator(condo_filter.qs, MAX_CONDOS_PER_PAGE)
    page = request.GET.get('page')

    try:
        condos = paginator.page(page)
    except PageNotAnInteger:
        condos = paginator.page(1)
    except EmptyPage:
        condos = paginator.page(paginator.num_pages)


    return render(request, 'app/index.html', {
        'title': 'Home',
        'condos': condos,
        'page': page,
        'condo_filter': condo_filter,
    })

这是一个工作示例：

.

Answer 5

为了增加答案，我还使用了 html 表以及 django-filters 和 Paginator。下面是我的视图和模板文件。需要模板标签以确保将正确的参数传递给分页 url.

search_view.py

from django.shortcuts import render
from app.models.filters_model import ApiStatusFilter
from app.models.api_status import ApiStatus
from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger
from datetime import datetime, timedelta

def status(request):
    all_entries_ordered = ApiStatus.objects.values().order_by('-created_at')[:200]

    for dictionarys in all_entries_ordered:
        dictionarys

    apistatus_list = ApiStatus.objects.values().order_by('-created_at')
    apistatus_filter = ApiStatusFilter(request.GET, queryset=apistatus_list)

    paginator = Paginator(apistatus_filter.qs, 10)
    page = request.GET.get('page')
    try:
        dataqs = paginator.page(page)
    except PageNotAnInteger:
        dataqs = paginator.page(1)
    except EmptyPage:
        dataqs = paginator.page(paginator.num_pages)

    return render(request, 'status_page_template.html', {'dictionarys': dictionarys, 'apistatus_filter': apistatus_filter, 'dataqs': dataqs, 'allobjects': apistatus_list})

status_template.html

{% load static %}
{% load my_templatetags %}

<!DOCTYPE html>
<html lang="en">
    <head>
        <link rel="stylesheet" type="text/css" href="{% static 'css/table_styling.css' %}">
        <meta charset="UTF-8">
        <title>TEST</title>
    </head>

    <body>
         <table>
            <thead>
                <tr>
                    {% for keys in dictionarys.keys %} 
                        <th>{{ keys }}</th>
                    {% endfor %}
                </tr>
            </thead>
                <form method="get">
                    {{ apistatus_filter.form.as_p }}
                    <button type="submit">Search</button>
                        {% for user in dataqs.object_list %}
                        <tr>
                            <td>{{ user.id }}</td>
                            <td>{{ user.date_time }}</td>
                            <td>{{ user.log }}</td>
                        </tr>
                        {% endfor %}
                </form>
            </tbody>
        </table>

        <div class="pagination">
            <span>
                {% if dataqs.has_previous %}
                    <a href="?{% query_transform request page=1 %}">&laquo; first</a>
                    <a href="?{% query_transform request page=dataqs.previous_page_number %}">previous</a>
                {% endif %}

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

                {% if dataqs.has_next %}
                    <a href="?{% query_transform request page=dataqs.next_page_number %}">next</a>
                    <a href="?{% query_transform request page=dataqs.paginator.num_pages %}">last &raquo;</a>
                {% endif %}
            </span>
        </div> 
    </body>
</html>

my_templatetags.py

from django import template

register = template.Library()

@register.simple_tag
def query_transform(request, **kwargs):
    updated = request.GET.copy()
    for k, v in kwargs.items():
        if v is not None:
            updated[k] = v
        else:
            updated.pop(k, 0)

    return updated.urlencode()

Answer 6

我花了一些时间才找到 DRYer 和更清洁的解决方案来解决这个问题，我认为最好的解决方案是使用模板标签的解决方案。

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

并在您的模板中

<a href="{% relative_url i 'page' request.GET.urlencode %}">{{ i }}</a>

来源：Dealing With QueryString Parameters

Answer 7

简单又甜美，
使用这个，pip install filter-and-pagination
https://pypi.org/project/filter-and-pagination/

实施步骤

通过 pip install filter-and-pagination
通过 from filter_and_pagination import FilterPagination 在 view.py
在您的函数中按照以下标准编写代码...

queryset = FilterPagination.filter_and_pagination(request, Customer)
serialize_data = CustomerSerializer(queryset['queryset'], many=True).data
resultset = {'dataset': serialize_data, 'pagination': queryset['pagination']}

在这段代码中Customer是Django模型&
CustomerSerializer 是一个 DRF 序列化器 class

在结果集中它包含数据集和分页数据，采用这种格式（API 响应）link：https://github.com/ashish1997it/filter-pagination-dj#demo
对于 API 请求，请遵循 PostMan collection link: https://github.com/ashish1997it/filter-pagination-dj#postman 在 header 部分，它将采用一个参数并请求您根据您的要求

如果您仍然遇到任何困难，请与我联系:)

Answer 8

我对分页结果“记住 filter/query URL 参数”的方法：将当前 URL 参数作为上下文变量传递：

# views.py

class PublicationFilterView(FilterView):
    model = Publication
    filterset_class = PublicationFilter
    paginate_by = 15

    def get_context_data(self, *args, **kwargs):
        _request_copy = self.request.GET.copy()
        parameters = _request_copy.pop('page', True) and _request_copy.urlencode()
        context = super().get_context_data(*args, **kwargs)
        context['parameters'] = parameters
        return context

# templates/path/to/pagination.html

<a href="?page={{ page_obj.next_page_number }}&{{ parameters }}">
  Next
</a>

Answer 9

这个对我来说 100% 有效

views.py:

def search(request):
    
    category=Category.objects.all()
    try:
        qs=request.GET["qs"]
        products=Product.objects.filter(Q(name__icontains=qs) |Q(details__icontains=qs) | Q(category__name__icontains=qs) | Q(branch__child__icontains=qs) | Q(manufacturer__name__icontains=qs) | Q(color__name__icontains=qs)).distinct()
        print(products)
        search=f"qs={qs}"
    except:
        search=None

并在 HTML

 <ul class="shop-p__pagination">
                                        {% if products.has_provious %}
                                        <li>

                                       <a class="fas fa-angle-left" href="?page={{ products.previous_page_number }}&{search}"></a></li>
                                       {% endif  %}
                                       {% for i in products.paginator.page_range %}
                                       {% if products.number == i %}
                                       
                                       <li class="is-active"><a href="?page={{i}}&{{search}}">{{i}}</a></li>
                                       {% else %}
                                       <li><a href="?page={{i}}&{{search}}">{{i}}</a></li>

                                       {% endif %}
                                     {% endfor %}
                               
                                       {% if products.has_next %}
                                   <li>

                                       <a class="fas fa-angle-right" href="?page={{ products.next_page_number }}&{{search}}"></a></li>
                                       {% endif %}
                                    </ul>

Answer 10

在get_context_data()函数中：

form_submitted = 'csrfmiddlewaretoken' in self.request.GET
context['cleaned_full_path'] = '{}{}'.format(
    self.request.get_full_path().split('&page' if form_submitted else '?page')[0],
    '&' if form_submitted else '?'
)

然后，在您的模板中，加载类似

的内容

<a href="{{ cleaned_full_path }}page={{ page_obj.paginator.num_pages }}"

django-filter 使用分页

django-filter use paginations

python

django

pagination

django-filter

实施步骤