“pagination de Django” Réponses codées

gérer les requêtes dans Listview django

class ProfileList(ListView):
    template_name = 'your_template.html'
    model = Profile

    def get_queryset(self):
        query = self.request.GET.get('q')
        if query:
            object_list = self.model.objects.filter(name__icontains=query)
        else:
            object_list = self.model.objects.none()
        return object_list
TitanVentura006

modèle de pagination appropriée

{% if is_paginated %}
    {% load proper_paginate %}
    {% load url_replace %}
    <ul class="pagination">
        {% if page_obj.number == 1 %}
            <li class="disabled"><span>⇤</span></li>
        {% else %}
            <li><a class="page-link" href="?{% url_replace request 'page' 1 %}">⇤</a></li>
        {% endif %}
        {% if page_obj.has_previous %}
            <li><a class="page-link" href="?{% url_replace request 'page' page_obj.previous_page_number %}">«</a></li>
        {% else %}
            <li class="disabled"><span>«</span></li>
        {% endif %}
        {% for i in paginator|proper_paginate:page_obj.number %}
            {% if page_obj.number == i %}
                <li class="active"><span>{{ i }} <span class="sr-only">(current)</span></span></li>
            {% else %}
                <li><a class="page-link" href="?{% url_replace request 'page' i %}">{{ i }}</a></li>
            {% endif %}
        {% endfor %}
        {% if page_obj.has_next %}
            <li><a class="page-link" href="?{% url_replace request 'page' page_obj.next_page_number %}">»</a></li>
        {% else %}
            <li class="disabled"><span>»</span></li>
        {% endif %}
        {% if page_obj.number == paginator.num_pages %}
            <li class="disabled"><span>⇥</span></li>
        {% else %}
            <li><a class="page-link" href="?{% url_replace request 'page' paginator.num_pages %}">⇥</a></li>
        {% endif %}
    </ul>
{% endif %}
WarlordOmar

pagination de Django

{% if page_obj.has_previous %}
  <a href="?page={{ page_obj.previous_page_number }}">« Previous page</a>

  {% if page_obj.number > 3 %}
    <a href="?page=1">1</a>
    {% if page_obj.number > 4 %}
      <span>...</span>
    {% endif %}
  {% endif %}
{% endif %}

{% for num in page_obj.paginator.page_range %}
  {% if page_obj.number == num %}
    <a href="?page={{ num }}">{{ num }}</a>
  {% elif num > page_obj.number|add:'-3' and num < page_obj.number|add:'3' %}
    <a href="?page={{ num }}">{{ num }}</a>
  {% endif %}
{% endfor %}

{% if page_obj.has_next %}
  {% if page_obj.number < page_obj.paginator.num_pages|add:'-3' %}
    <span>...</span>
    <a href="?page={{ page_obj.paginator.num_pages }}">{{ page_obj.paginator.num_pages }}</a>
  {% elif page_obj.number < page_obj.paginator.num_pages|add:'-2' %}
    <a href="?page={{ page_obj.paginator.num_pages }}">{{ page_obj.paginator.num_pages }}</a>
  {% endif %}

  <a href="?page={{ page_obj.next_page_number }}">Next Page »</a>
{% endif %}
Adventurous Armadillo

Pagination de recherche de Django

from django.core.paginator import Paginator

def search(request):
    if 'results' in request.GET and request.GET['results']:
        page = request.GET.get('page', 1)

        results = request.GET['results']
        word = words.objects.filter(title__icontains = results).order_by('title')
        paginator = Paginator(word, 25) # Show 25 contacts per page
        word = paginator.page(page)
        return render_to_response('myapp/search.html',
                 {'word': word, 'query': results })
    else:
        return render(request, 'myapp/search.html')
Natthawut Ponkrut

Réponses similaires à “pagination de Django”

Questions similaires à “pagination de Django”

Plus de réponses similaires à “pagination de Django” dans Python

Parcourir les réponses de code populaires par langue

Parcourir d'autres langages de code