django通过字段对object_list排序

Question

说我有很好的老人/团体/会员关系

class Person(models.Model):
    name = models.CharField(max_length=128)

    def __unicode__(self):
        return self.name

class Group(models.Model):
    name = models.CharField(max_length=128)
    members = models.ManyToManyField(Person, through='Membership')

    def __unicode__(self):
        return self.name

class Membership(models.Model):
    person = models.ForeignKey(Person)
    group = models.ForeignKey(Group)
    date_joined = models.DateField()
    invite_reason = models.CharField(max_length=64)

我想要一个通用视图，显示属于某个组(在URL中指定)的人员列表，该列表按Membership类的date_joined字段排序。我想我可以做一个这样的视图

def group_list(request, pagina):
    return list_detail.object_list(
        request,
        queryset = Person.objects.filter(group__name = slug),
        template_object_name = 'person',
        extra_context = { 'membership' : Membership.objects.all() }, 

        )

然后只需通过membership__date_joined访问模板中的date_joined，但没有成功

Answer 1

您实际要做的是从Membership对象开始。对于特定的Person和Group组合，这是唯一独一无二的事情-如果您从Person开始，就像您正在做的那样，访问他们的成员资格将为您提供所有成员，而不是您想要的成员。所以：

return list_detail.object_list(
    request,
    queryset = Membership.objects.filter(group__name = slug),
    template_object_name = 'membership',
)

现在在您的模板中：

{% for member in membership %}
    {{ member.person.name }}: {{ member.date_joined }}
{% endif %}

Answer 2

一种选择是直接在中介模型上过滤：

members = [member.person for member in Membership.objects.filter(group__name=slug).order_by('-date_joined')]

或者仅仅是members = Membership.objects.filter(group__name=slug).order_by('-date_joined')

然后在模板中迭代此列表，如下所示：

{% for member in members %}
    {{ member.person.name }} : {{ member.date_joined }}
{% endfor %}