我想分頁我的基於類的視圖。這是我的觀點的外觀:基於類的基於Django的視圖
class IssuesByTitleView(ListView):
context_object_name = "issue_list"
def issues(request):
issue_list = Issue.objects.all()
###### Commented out does not work ######
# paginator = Paginator(issue_list, 24)
# try:
# page = int(request.GET.get('page', '1'))
# except ValueError:
# page = 1
# try:
# issues = paginator.page(page)
# except (EmptyPage, InvalidPage):
# issues = paginator.page(paginator.num_pages)
def get_queryset(self):
self.title = get_object_or_404(Title, slug=self.kwargs['title_slug'])
return Issue.objects.filter(title=self.title).order_by('-number')
def get_context_data(self, **kwargs):
context = super(IssuesByTitleView, self).get_context_data(**kwargs)
context['title'] = self.title
return context
這裏是我的模型的某些方面的例子:
class Title(models.Model):
CATEGORY_CHOICES = (
('Ongoing', 'Ongoing'),
('Ongoing - Canceled', 'Ongoing - Canceled'),
('Limited Series', 'Limited Series'),
('One-shot', 'One-shot'),
('Other', 'Other'),
)
title = models.CharField(max_length=64)
vol = models.IntegerField(blank=True, null=True, max_length=3)
year = models.CharField(blank=True, null=True, max_length=20, help_text="Ex) 1980 - present, 1980 - 1989.")
category = models.CharField(max_length=30, choices=CATEGORY_CHOICES)
is_current = models.BooleanField(help_text="Check if the title is being published where Emma makes regular appearances.")
slug = models.SlugField()
class Meta:
ordering = ['title']
def get_absolute_url(self):
return "/titles/%s" % self.slug
def __unicode__(self):
class Issue(models.Model):
title = models.ForeignKey(Title)
number = models.CharField(max_length=20, help_text="Do not include the '#'.")
...
當然,按照Django文檔,定義視圖時分頁系統的工作原理通過這樣的事情:def view(request):
我也想知道如何可以拉出下一個和以前的對象。
我需要一個鏈接到「下一期(與名稱和問題編號的上下文)」,然後是「上一期」鏈接。請注意,僅僅更改問題的下一個或前一個號碼的模板鏈接是行不通的。
所以,如果任何人都可以幫助我,那就太好了。
這工作,但我怎麼現在通過它的模板?例如:{{issue.paginator.num_pages}}的{Page {{issue.number}}。 {%if issue.has_previous%} « Previous {%endif%} {%if issues.has_next%} | Next » {%endif%}'不起作用。 – AAA 2011-05-16 15:00:13
「page_obj」上下文變量將具有您需要的信息。即{{page_obj.paginator.num_pages}}',page_obj.has_previous'的頁面{{page_obj.number}}。還有'is_paginated'上下文變量來檢查是否有分頁。 – 2011-05-16 17:00:10
是的,這是回答這個問題時的關鍵丟失細節;) – defbyte 2012-04-20 17:52:40