Django分页:在分页/非分页ListView之间切换

Question

我正在详细阐述一种聪明的方法，在分页模板和非分页模板之间切换。

我已经有了一个工作分页器，我正在考虑在它旁边添加一个按钮，上面写着“显示所有结果”，该按钮链接到一个非分页列表，然后将有另一个按钮返回到分页列表。

1)简单解决方案

使用2 ListViews，使用属性paginate_by的不同分配(django默认设置分页)，但是由于我的项目中有许多列表，这将不方便(也不太明智)。

2)解决方案我被困在上了

编写Mixin (稍后将由我的ListViews进行扩展)，根据条件设置变量paginate_by，然后向上下文中添加一些有用的变量：

class PaginationMixin:
    no_pagination = False
    no_pagination_url = ''

    def get_paginate_by(self, queryset):
     # overwrite django method
        if self.no_pagination:
            return None
        else:
            return super().get_paginate_by(queryset)

    def get_no_pagination_url(self):
        return self.no_pagination_url

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['no_pagination'] = self.no_pagination
        context['no_pagination_url'] = self.get_no_pagination_url()
        return context

 class MyListView(PaginationMixin, ListView):
     #...
     def get_no_pagination_url(self):
         return reverse('mylist_urlname')

问题:我不知道如何从模板中设置no_pagination变量。有什么办法可以做到吗？

谢谢你的帮助。

更新的解决方案(由@hi-lan解决方案编辑)：通过这种方式，它将显示所有结果，如果存在，也会保留urlparams (来自过滤器或其他)。

class PaginationMixin:

    toggle_pagination = False
    toggle_pagination_url = ''
    no_pagination = False
    view_name = ''
    urlparams_dict = {}

    def get(self, request, page=None, *args, **kwargs):
        #store current GET params and pop 'page' key
        self.urlparams_dict = request.GET            
        self.urlparams_dict.pop('page', None)

        page = page or request.GET.get('page', '1')
        if page == 'all':
            page = self.paginate_by = None
            self.no_pagination = True
        return super().get(request, page=page, *args, **kwargs)

    def get_paginate_by(self, queryset):
        if self.no_pagination:
            return None
        else:
            return super().get_paginate_by(queryset)

    def get_toggle_pagination_url(self):
        # variables to set in view to toggle this mixin
        if self.toggle_pagination and self.view_name:
            if not self.no_pagination:
                extra = {'page': 'all'}
                self.urlparams_dict.update(extra)
            else:
                self.urlparams_dict.pop('page', None)
            # url keeps track of urlparams adds page=all if toggled
            self.toggle_pagination_url = reverse(self.view_name) + '?' + urlencode(self.urlparams_dict)
        return self.toggle_pagination_url

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['toggle_pagination_url'] = self.get_toggle_pagination_url()
        context['toggle_pagination'] = self.toggle_pagination
        return context

Answer 1

问题在于从用户返回的数据流以指示非分页。我唯一能想到的方法就是使用特殊的页码。有两个选项，取决于您配置urls.py的方式。

• 在path('objects/page<int:page>/', PaginatedView.as_view()),情况下，特殊编号为0(因为普通页码从1开始)。
• 在/objects/?page=3的情况下，特殊号码可以是all。

在这两种情况下，我们都需要覆盖get方法，因为它是检索用户选择的位置。

class PaginationMixin:
    no_pagination = False
    view_name = ''

    def get(self, request, page=None, *args, **kwargs):
        page = page or request.GET.get('page', '1')
        if page in ['0', 'all']:
            page = self.paginate_by = None
        else: pass
        return super().get(request, page=page, *args, **kwargs)

    def get_paginate_by(self, queryset):
        # overwrite django method
        if self.no_pagination:
            return None
        else:
            return super().get_paginate_by(queryset)

    def get_no_pagination_url(self):
        # For using path
        extra = {'page': '0'}
        no_pagination_url = reverse(self.view_name, kwargs=extra)
        # For using query params
        extra = {'page': 'all'}
        no_pagination_url = reverse(self.view_name) + '?' + urlencode(extra)
        return no_pagination_url

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['no_pagination'] = self.no_pagination
        context['no_pagination_url'] = self.get_no_pagination_url()
        return context


class MyListView(PaginationMixin, ListView):
    view_name = 'mylist_urlname'
    #...

Answer 2

我正在尝试使用以下视图更新gccallie解决方案：

class StageTempList(PaginationMixin, LoginRequiredMixin, SingleTableMixin, FilterView):
    view_name = 'stagetemp-list'
    table_class = StageTempTable
    model = StageTemp
    filterset_class = StageTempFilter
    template_name = 'stage/stagetemp_list.html'
    paginate_by = 30
    strict = False

但是当get_paginate_by返回None时，我得到25行。Django版本2.1.2

更新:我使用的PaginationMixin类

class PaginationMixin:
    no_pagination = False
    view_name = ''

    def get(self, request, page=None, *args, **kwargs):
        page = page or request.GET.get('page', '1')
        if page in ['0', 'all']:
            page = self.paginate_by = None
            self.no_pagination = True
        else: pass
        return super().get(request, page=page, *args, **kwargs)

    def get_paginate_by(self, queryset):
        # overwrite django method
        if self.no_pagination:
            return None
        else:
            return super().get_paginate_by(queryset)

    def get_no_pagination_url(self):
        extra = {'page': 'all'}
        no_pagination_url = reverse(self.view_name) + '?' + urlencode(extra)
        return no_pagination_url

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['no_pagination'] = self.no_pagination
        context['no_pagination_url'] = self.get_no_pagination_url()
        return context