[英]django-filter use paginations
我正在使用django-filter
包在我的列表視圖上提供搜索功能。
現在我也想為該視圖添加一個分頁。
我正在嘗試將分頁與過濾的查詢集結合起來,但我不知道如何繼續。
到目前為止,我在views.py
上嘗試了以下操作:
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})
這對我有用:
在我的模板中而不是使用這個
<li><a href="?page={{ i }}">{{ i }}</a></li>
我寫了這個:
{% 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 %}
我希望它有幫助:)
要使用 Django 過濾器並對過濾后的結果進行分頁,您可以執行以下操作:
為您的模型創建一個過濾器類:
在my_project/my_app/filters.py
上:
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', ...]
每個FilterSet
對象都有一個.qs
屬性,其中包含過濾后的查詢集,您甚至可以根據需要覆蓋它。
我們將對MyModelFilter
的.qs
屬性進行分頁:
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} )
你有它!
PS_2:如果您要使用 DRF,我已經編寫了一個示例,說明如何在基於函數的視圖中使用分頁,您可以輕松地將其與FilterSet
結合使用:
@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)
為了增加答案,我也使用 html 表以及 django-filters 和 Paginator 來完成。 以下是我的視圖和模板文件。 需要模板標簽來確保將正確的參數傳遞給分頁 url。
搜索視圖.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 %}">« 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 »</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()
這里最重要的部分是您如何在模板中構建 URL 。
你可能有
{% if pages.has_previous %}
<li><a href="?page={{ pages.previous_page_number }}">Prev</a></li>
{% endif %}
如果您僅使用它在初始分頁結果之間切換,那是非常好的。
但棘手的部分是當您使用django-fitler
過濾器時,查詢字符串(在'?'之后的部分)會獲得全新的鍵值對,而忽略您的?page=2
或類似的。
因此,要使用過濾后的結果進行分頁,當您單擊“下一步”或“上一個”按鈕時 - 在django-fitler
的鍵值中,您還需要將&page=5
作為對傳遞。
正如@stathoula 所提到的,您需要檢查查詢字符串中是否已經存在至少一個過濾器字段。 如果是,那么您需要使用已經存在的鍵值對,然后是新的&page=3
對。
看起來很簡單,但是當用戶單擊箭頭時,我不得不做一些小技巧,不要在查詢字符串中一遍又一遍地重復&page=1
。
就我而言,我將“標題”作為過濾器,所以我需要檢查它是否已經存在。
這是我為我的項目所做的完美工作的片段。
模板/分頁.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>
這是視圖,以防萬一:
應用程序/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,
})
這是一個工作示例:
.
我花了一些時間找到 DRYer 和更清潔的解決方案來解決這個問題,我認為最好的解決方案是使用模板標簽的解決方案。
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
並在您的模板中
<a href="{% relative_url i 'page' request.GET.urlencode %}">{{ i }}</a>
資料來源: 處理 QueryString 參數
我對分頁結果“記住過濾器/查詢 URL 參數”的方法:將當前 URL 參數作為上下文變量傳遞:
# 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>
這個和我 100% 合作
視圖.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
並在 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>
據我了解,您的目標是對過濾后的查詢集進行分頁。 如果是這樣,您可以將 PublicationFilter 對象的“qs”屬性傳遞給 Paginator 構造函數:
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包含過濾的 QuerySet
url_filter.queryset包含未過濾的 QuerySet
pip install filter-and-pagination
pip install filter-and-pagination
安裝包from filter_and_pagination import FilterPagination
在 view.py 中導入 FilterPaginationqueryset = FilterPagination.filter_and_pagination(request, Customer)
serialize_data = CustomerSerializer(queryset['queryset'], many=True).data
resultset = {'dataset': serialize_data, 'pagination': queryset['pagination']}
Customer
是 Django 模型 &CustomerSerializer
是一個 DRF 序列化器類如果您仍然遇到任何困難,請聯系我 :)
在get_context_data()
函數中:
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,加載類似
<a href="{{ cleaned_full_path }}page={{ page_obj.paginator.num_pages }}"
除了@stathoula 和對@Benbb96 的回應,我設法用正則表達式刪除了額外的page
參數,覆蓋了基於類的視圖中的setup
方法:
import re
...
class MyView(ListView):
...
def setup(self, request, *args, **kwargs) -> None:
request.GET.get("page")
request.META["QUERY_STRING"] = re.sub("(&|\?)page=(.)*", "", request.META.get("QUERY_STRING", ""))
return super().setup(request, *args, **kwargs)
希望它可以幫助任何人!
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