带分页的 Django 过滤器：按下一步后显示同一页面 url

Question

当我搜索某些东西（比如 tom）时，它在第一页上效果很好。但是如果我点击下一页url，它显示相同的结果，没有任何变化，但是在url，它变成了http://127.0.0.1:8000/search/?caption=tom to http://127.0.0.1:8000/search/?caption=tom&?page=2

filters.py:

class VideoFilter(django_filters.FilterSet):
    class Meta:
        model = Video
        fields = ['caption']
        filter_overrides = {
             models.CharField: {
                 'filter_class': django_filters.CharFilter,
                 'extra': lambda f: {
                     'lookup_expr': 'icontains',
                 },
             },
         }

views.py:

def search(request):
    queryset = Video.objects.all()
    filterset = VideoFilter(request.GET, queryset=queryset)
    if filterset.is_valid():
        queryset = filterset.qs

    paginator = Paginator(queryset, 2)
    page_number = request.GET.get('page')
    print(page_number)# always prints **none**
    queryset = paginator.get_page(page_number)
    
    return render(request, 'search.html',{
        'result': queryset,
        'caption': request.GET['caption'],
    })

search.html:

{% extends 'navbar.html' %}
{% block body %}

<!-- more code -->
{% if result.has_next %}
    <a href="?caption={{caption}}&?page={{result.next_page_number}}"><button>See more results</button></a>
{% endif %}

{% endblock body %}

navbar.html:

<!-- more code -->
<form action="/search/" method="GET"> <!-- working without csrf_token -->
            <input type="text" placeholder="Search" id="search" name="caption" required />
            <button type="submit">Search</button>
        </form>

问题出在哪里？我如何访问下一页？

Answer 1

您在按钮的 URL 中输入错误，query string [wiki] 中的 key-value 对由符号分隔（ &)，而不是带问号的符号 (~~&?~~)。通过这样写，Django 会将其解释为：

>>> QueryDict('caption=tom&?page=2')
<QueryDict: {'caption': ['tom'], '?page': ['2']}>

那么参数就不是~~page~~，而是?page.

我们可以通过删除 URL 中标题和页面之间的问号来解决此问题：

<a href="?caption={{ caption|urlencode }}<b>&</b>page={{ result.next_page_number }}">

您还应该使用 |urlencode template filter [Django-doc] to prevent a wrong querystring, for example when the caption would have a questionmark, number sign (#), etc. The |urlencode filter will transform the text in percent encoding [wiki]。

带分页的 Django 过滤器：按下一步后显示同一页面 url

django filters with pagination: showing same page after pressing next url

python

html

django

django-filter

django-pagination