Question

如何在slug中添加id，使其显示为

＆＃34;这-是-A-交易-1＆＃34;

其中＆＃34;这是一笔交易＆＃34;是标题和＆＃34; 1＆＃34;是身份证号码。目前使用当前代码，它显示为"this-is-a-deal-None"

models.py ... slug = models.SlugField（）

    def save(self, *args, **kwargs):
        if not self.id:
            # Newly created object, so set slug
            self.slug = '%s-%s' %(
                            slugify(self.deal_title),
                            str(self.id))

views.py

def deal_page(request, deal_title_slug):

    # Create a context dictionary which we can pass to the template rendering engine.
    context_dict = {}

    try:
        deal = SubmitDeal.objects.get(slug=deal_title_slug)
        context_dict['deal_title'] = deal.deal_title

        context_dict['deal'] = deal
    except SubmitDeal.DoesNotExist:
        # We get here if we didn't find the specified category.
        # Don't do anything - the template displays the "no category" message for us.
        pass

    return render(request, 'deal_page.html', context_dict)

urls.py

url(r'^(?P<deal_title_slug>[\w\-]+)/$', 'submit_deals.views.deal_page', name='deal_page'),

admin.py

class SubmitModelAdmin(admin.ModelAdmin):
    prepopulated_fields = {'slug':('deal_title',)} #prepopulate slug with deal_title
    class meta:
        model = SubmitDeal

admin.site.register(SubmitDeal, SubmitModelAdmin)

Answer 1

你不应该改变数据库中的slu ..在网址中分隔slug和id：

url(r'^(?P<slug>[\w\-]+)-(?P<pk>\d+)/$', 'submit_deals.views.deal_page',
                                         name='deal_page'),

然后使用pk获取对象并忽略slug：

def deal_page(request, slug, pk):
    try:
        deal = SubmitDeal.objects.get(pk=pk)
        context_dict = {'deal_title': deal.deal_title, 'deal': deal}
    except SubmitDeal.DoesNotExist:
        context_dict = {}
   return render(request, 'deal_page.html', context_dict)

在Django中如何在slug标题中添加id

1 个答案: