views.py

Question

我正在设置期刊订阅购物车，并希望用户能够根据与当前日期的距离订购订阅。

每个订阅包含3个问题。我想让人们选择从当前期刊开始1年或从下一期开始1年。

我的模型很简单：

class Subscription(models.Model):
  start_date = models.DateField(max_length=10)
  end_date = models.DateField(max_length=10)
  date = models.DateTimeField(auto_now_add=True, blank=True)
  def __unicode__(self):
      return unicode(self.start_date)

理想情况下，“从当前开始的订单1年”会在过去找Subscription个start_date，但过去最近的。{/ p>

“从下一个开始的订单1年”将返回Subscription，其中start_date最接近当前日期。

我如何在视图中实现这种逻辑？

Answer 1

views.py

class NextSubscription(View):
    def get(self, request):
        try:
            return Subscription.objects.filter(start_date__gt=date.today()).order_by("start_date")[0]
        except IndexError:
            raise Exception("No Next Subscription Found")

class CurrentSubscription(View):
    def get(self, request):
        try:
            return Subscription.objects.filter(start_date__lt=date.today()).order_by("-start_date")[0]
        except IndexError:
            raise Exception("No Current Subscription Found")

Answer 2

我认为最好将其放在forms逻辑中而不是views中。我认为还有一个Issue模型，如下所示：

class Issue(models.Model):
    issue_date = models.DateField()
    ....

保存Subscription后，您的save()应该执行以下操作：

forms.py

from models import Subscription
from django import forms
class SubscriptionForm(forms.ModelForm):
    def save(self, commit=True):
        instance = super(SubscriptionForm, self).save(commit=False)
        if instance.start_from_next:  # You should add a start_from_next boolean field to know if user wants to start from previous or next issue
            the_issue = Issue.objects.filter(issue_date__gt=date.today()).order_by("start_date")[0]
        else:
            the_issue = Issue.objects.filter(issue_date__lt=date.today()).order_by("-start_date")[0]
        instance.start_date = the_issue.start_date
        if commit:
            instance.save()
        return instance

    class Meta:
        model = Subscription

Django使用timedelta过滤模板中的结果

2 个答案:

views.py

forms.py