Anh*_*yen 14 python django pagination django-filter
我正在使用该django-filter包在我的列表视图上提供搜索功能.
现在我想为该视图添加一个分页.
我正在尝试将分页与过滤的查询集结合起来,但我不知道如何继续.
到目前为止,我已经尝试了以下内容views.py:
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})
Eli*_*adi 10
为了添加答案，我也使用 html 表以及 django-filters 和 Paginator 来完成。以下是我的视图和模板文件。需要模板标签来确保将正确的参数传递给分页 URL。
搜索视图.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})
状态模板.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()
小智 7
我花了一些时间才找到更干燥和更清洁的解决方案来解决此问题，在我看来，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
并在你的模板中
<a href="{% relative_url i 'page' request.GET.urlencode %}">{{ i }}</a>
来源：处理查询字符串参数
要使用Django过滤器并对过滤后的结果进行分页，您可以执行以下操作：
为模型创建一个过滤器类：
开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', ...]
我们将对我们的.qs财产进行分页MyModelFilter：
开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}
)
在那里，您拥有了！
PS_2：如果您将要使用DRF，我已经写了一个示例，说明如何在基于函数的视图中使用分页，可以轻松地将其与结合使用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)
小智 6
这对我有用:
在我的模板中而不是使用它
<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 %}
我希望它有帮助:)
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