Question

我正在尝试为定义用户访问权限的外部群组创建API。我在django中设置了一个易于管理的设置，但我希望其他团队的输出相当简单。

这是我正在寻找的输出：

{
    "user_list": {
        "user": {
            "username": "username1",
            "top_accesses": ["top_access_1", "top_access_2", "top_access_5"],
            "middle_accesses": ["middle_access_1", "middle_access_2", "middle_access_7"],
            "lower_accesses": ["lower_access_1", "lower_access_2", "lower_access_22"],
        },
        "user": {
            "username": "username2",
            "top_accesses": ["top_access_1", "top_access_2", "top_access_8"],
            "middle_accesses": ["middle_access_3", "middle_access_5", "middle_access_6"],
            "lower_accesses": ["lower_access_21", "lower_access_33", "lower_access_36"],
        }
    }
}

但是，我在使用ORM内置的django来从我的模型中提出这些设置时遇到了麻烦。我可以想一想如何在SQL中做到这一点，但这不是一个特别干净的方法。我知道必须有更好的方法来实现它，因为使用TabularInline显示了我想在管理页面中看到的内容

以下是我的模特：

class TopAccess(models.Model):
    name = models.CharField(max_length=100)
    site_user_access = models.ManyToManyField(User, blank=True)
    site_group_access = models.ManyToManyField(Group, blank=True)

    class Meta:
        verbose_name_plural = "Top Access"

    def __str__(self):
        return self.name


class MiddleAccess(models.Model):
    name = models.CharField(max_length=100)
    site_user_access = models.ManyToManyField(User, blank=True)
    site_group_access = models.ManyToManyField(Group, blank=True)

    class Meta:
        verbose_name_plural = "Middle Access"

    def __str__(self):
        return self.name


class LowerAccess(models.Model):
    name = models.CharField(max_length=100)
    site_user_access = models.ManyToManyField(User, blank=True)
    site_group_access = models.ManyToManyField(Group, blank=True)

    class Meta:
        verbose_name_plural = "Lower Access"

    def __str__(self):
        return self.name

理想情况下，我最终可以返回一个与django-rest-framework一起运行良好的查询对象，因为我喜欢以任何形式请求返回相同数据的方式

修改：这就是我认为接近解决方案的原因，但我知道我没有错误地使用类继承

class MaybeThisCouldWork(User):
    t_user = TopAccess.objects.filter(site_user_access=User)
    m_user = MiddleAccess.objects.filter(site_user_access=User)
    l_user = LowerAccess.objects.filter(site_user_access=User)
    user_groups = User.objects.filter(id=User)
    for user_group in user_groups:
        t_group = TopAccess.objects.filter(
                site_group_access=user_groups
                )
        m_group = MiddleAccess.objects.filter(
                site_group_access=user_groups
                )
        l_group = LowerAccess.objects.filter(
                site_group_access=user_groups
                )
        t_user = t_user | t_group
        m_user = m_user | m_group
        l_user = l_user | l_group

Answer 1

您可以使用序列化程序，可能是这样的，

class TopAccessSerializer(serializers.ModelSerializer):
    class Meta:
        model = TopAccess
        fields = ['name']

class MiddleAccessSerializer(serializers.ModelSerializer):
    class Meta:
        model = MiddleAccess
        fields = ['name']

class LowerAccessSerializer(serializers.ModelSerializer):
    class Meta:
        model = LowerAccess
        fields = ['name']


class UserSerializer(serializers.ModelSerializer):
    topaccess = TopAccessSerializer(source='topaccess_set', many=True)
    middleaccess = MiddleAccessSerializer(source='middleaccess_set', many=True)
    loweraccess = LowerAccessSerializer(source='loweraccess_set', many=True)

    class Meta:
        model = User
        fields = ['username', 'topaccess', 'middleaccess', 'loweraccess']

试图制作一个看起来不像我的模型的django rest api

1 个答案: