选择为整数的Django模型

我试图在Django中制作一个反馈应用程序，但我无法进行评估

在我的models.py中，我有5种选择，从非常差到优秀。但我希望它们可以作为数字使用，这样我就可以评估整体价值

阅读后，我将我的评分从VERYBAD=‘非常糟糕’改为VERYBAD=1，但不，我甚至无法保存我的值/表格

Feedback(models.Model):
    event = models.ForeignKey(Event)
    # Choice
    VERYBAD = 'Very bad'            #old
    BAD = 2                         #new
    OKAY = 3
    GOOD = 4
    EXCELLENT = 5
    RATING = (
        (VERYBAD, 'Very bad'),
        (BAD, 'Bad'),
        (OKAY, 'Okay'),
        (GOOD, 'Good'),
        (EXCELLENT, 'Excellent'),
    )

    # The ratings
    organisation = models.CharField(
        max_length=9,
        choices=RATING,
        default=OKAY,
    )
   ....

所以在我看来，我认为我可以做数学，但我做不到

def rating(request, event_id):
    myevent = Event.objects.get(id=event_id)
    feedback_items = Feedback.objects.filter(event=myevent)
    num_of_items = len(feedback_items)

    def evaluate(feedback_items):
       # The overall rating
       organisation = 0
       for item in feedback_items: 
            organisation += item.organisation

       organisation /= num_of_items

    context = {'feedback_items':feedback_items,
           'num_of_items': num_of_items,
           'myevent': myevent,}

return render(request, 'feedback/rating.html', context)

---

`

显然，您必须将

组织

设置为

整数域

，否则您将无法希望执行整数运算。如果您已经在生产中使用了该字段，请使用一组迁移来添加新的IntegerField，从现有的CharField值填充它（使用dict将旧字符串值映射到新的int值），最后删除原始字段

此外，您可能希望了解如何正确使用Django的ORM功能，如获取相关模型实例和在DB级别执行聚合，即：

from django.db.models import Sum, Count, Avg

def rating(request, event_id):
    myevent = Event.objects.get(id=event_id)

    # this is how you get related models...
    feedback_items = myevent.feedback_set.all()

    # and this is how you use SQL aggregation functions:
    values = feedback_items.aggregate(
        sum=Sum('organisation'),
        count=Count('pk'),
        avg=Avg('organisation')
        )

    # Now 'values' should be a dict with 'sum', 'count' and 'avg'
    # keys, and values['avg'] should be equal to 
    # float(values['sum']) / values['count']
    # FWIW you probably don't need the 'sum' field at all
    # I mentionned it so you can check whether 'avg' value
    # is correct...

     context = {
       'feedback_items':feedback_items,
       'num_of_items': values['count'],
       'avg': values['avg'],
       'myevent': myevent,
        }

     # etc

---

显然，您必须将

组织

设置为一个

整数域

，否则您就不能希望执行整数运算。如果您已经在生产中使用了该字段，请使用一组迁移来添加新的IntegerField，从现有的CharField值填充它（使用dict将旧字符串值映射到新的int值），最后删除原始字段

此外，您可能希望了解如何正确使用Django的ORM功能，如获取相关模型实例和在DB级别执行聚合，即：

from django.db.models import Sum, Count, Avg

def rating(request, event_id):
    myevent = Event.objects.get(id=event_id)

    # this is how you get related models...
    feedback_items = myevent.feedback_set.all()

    # and this is how you use SQL aggregation functions:
    values = feedback_items.aggregate(
        sum=Sum('organisation'),
        count=Count('pk'),
        avg=Avg('organisation')
        )

    # Now 'values' should be a dict with 'sum', 'count' and 'avg'
    # keys, and values['avg'] should be equal to 
    # float(values['sum']) / values['count']
    # FWIW you probably don't need the 'sum' field at all
    # I mentionned it so you can check whether 'avg' value
    # is correct...

     context = {
       'feedback_items':feedback_items,
       'num_of_items': values['count'],
       'avg': values['avg'],
       'myevent': myevent,
        }

     # etc

---

非常感谢你。在重读了making queries一章之后，我还设法进行了一个查询，该查询只显示了事件类别中我尚未注册的战斗机。竖起大拇指非常感谢你。在重读了making queries一章之后，我还设法进行了一个查询，该查询只显示了事件类别中我尚未注册的战斗机。竖起大拇指