如何将django Model对象转换为包含所有字段的dict ?理想情况下,所有字段都包含外键和editable=False。

让我详细说明一下。假设我有一个django模型,如下所示:

from django.db import models

class OtherModel(models.Model): pass

class SomeModel(models.Model):
    normal_value = models.IntegerField()
    readonly_value = models.IntegerField(editable=False)
    auto_now_add = models.DateTimeField(auto_now_add=True)
    foreign_key = models.ForeignKey(OtherModel, related_name="ref1")
    many_to_many = models.ManyToManyField(OtherModel, related_name="ref2")

在终端中,我做了以下工作:

other_model = OtherModel()
other_model.save()
instance = SomeModel()
instance.normal_value = 1
instance.readonly_value = 2
instance.foreign_key = other_model
instance.save()
instance.many_to_many.add(other_model)
instance.save()

我想把它转换成下面的字典:

{'auto_now_add': datetime.datetime(2015, 3, 16, 21, 34, 14, 926738, tzinfo=<UTC>),
 'foreign_key': 1,
 'id': 1,
 'many_to_many': [1],
 'normal_value': 1,
 'readonly_value': 2}

回答不满意的问题:

Django:将整个Model对象集转换为单个字典

如何将Django Model对象转换为字典,同时还保留外键?


当前回答

将模型转换为字典并保留所有的外键模型关系。我使用了以下方法:

无详细名称

from django.forms.models import model_to_dict

instance = MyModel.objects.get(pk=1) # EXAMPLE

instance_dict = {key: getattr(instance, key) for key in model_to_dict(instance).keys()}

输出

{'foreign_key': [<OtherModel: OtherModel object>],
 'id': 1,
 'many_to_many': [<OtherModel: OtherModel object>],
 'normal_value': 1}

如果你想在模板中为外键关系显示__str__()值,这可能很有用。

将关键字参数fields=和exclude=包含到model_to_dict(instance,[…])中,使您可以过滤特定的字段。

详细名称

from django.forms.models import model_to_dict

instance = MyModel.objects.get(pk=1) # EXAMPLE

instance_dict = {instance._meta.get_field(key).verbose_name if hasattr(instance._meta.get_field(key), 'verbose_name') else key: getattr(instance, key) for key in model_to_dict(instance).keys()}

示例输出(如果给定示例有详细的名称)

{'Other Model:': [<OtherModel: OtherModel object>],
 'id': 1,
 'My Other Model:': [<OtherModel: OtherModel object>],
 'Normal Value:': 1}

其他回答

最简单的方式,

如果你的查询是Model.Objects.get(): Get()将返回单个实例,因此您可以直接从实例中使用__dict__。 model_dict = Model.Objects.get().__dict__ 过滤器()/ (): All ()/filter()将返回实例列表,因此您可以使用values()来获取对象列表。 model_values = Model.Objects.all().values()

当我试图使用django-rest框架将django站点转换为API时,我遇到了这个问题。通常django会从数据库中返回三种类型的对象。它们包括一个查询集、一个模型实例和一个分页器对象。对我来说,这些是需要转换的。

查询集

queryset就像django中的模型对象列表。这是把它转换成字典的代码。

model_data=Model.object.all()# This returns a queryset object
model_to_dict=[model for model in model_data.values()]
return Response(model_to_dict,status=status.HTTP_200_OK)

模型实例

模型实例是模型的单个对象。

model_instance=Model.objects.get(pk=1)# This will return only a single model object
model_to_dict=model_to_dict(model_instance)
return Response(model_to_dict,status=status.HTTP_200_OK)

Paginator对象

分页器对象是一个包含特定页面的模型对象的对象。

model_queryset=Model.objects.all()
paginator = Paginator(model_queryset, 10)
try:
    selected_results = paginator.page(page)
except Exception:
    selected_results=result
paginator_to_dict=list(selected_results.object_list.values())
return Response(selected_results,status=status.HTTP_200_OK)

至少我是这么解决的。

@Zags的解决方案太棒了!

不过,为了使它对JSON友好,我要为datefields添加一个条件。

奖金轮

如果你想要一个更好的python命令行显示的django模型,让你的models子类如下:

from django.db import models
from django.db.models.fields.related import ManyToManyField

class PrintableModel(models.Model):
    def __repr__(self):
        return str(self.to_dict())

    def to_dict(self):
        opts = self._meta
        data = {}
        for f in opts.concrete_fields + opts.many_to_many:
            if isinstance(f, ManyToManyField):
                if self.pk is None:
                    data[f.name] = []
                else:
                    data[f.name] = list(f.value_from_object(self).values_list('pk', flat=True))
            elif isinstance(f, DateTimeField):
                if f.value_from_object(self) is not None:
                    data[f.name] = f.value_from_object(self).timestamp()
            else:
                data[f.name] = None
            else:
                data[f.name] = f.value_from_object(self)
        return data

    class Meta:
        abstract = True

例如,如果我们这样定义我们的模型:

class OtherModel(PrintableModel): pass

class SomeModel(PrintableModel):
    value = models.IntegerField()
    value2 = models.IntegerField(editable=False)
    created = models.DateTimeField(auto_now_add=True)
    reference1 = models.ForeignKey(OtherModel, related_name="ref1")
    reference2 = models.ManyToManyField(OtherModel, related_name="ref2")

调用sommodel .objects.first()现在给出如下输出:

{'created': 1426552454.926738,
'value': 1, 'value2': 2, 'reference1': 1, u'id': 1, 'reference2': [1]}

我创建了一个小片段,利用django的model_to_dict,但遍历对象的关系。 对于循环依赖项,它终止递归并放入引用依赖项对象的字符串。您可以将其扩展为包含不可编辑字段。

我在测试期间使用它来创建模型快照。

from itertools import chain

from django.db.models.fields.files import FileField, ImageField
from django.forms.models import model_to_dict


def get_instance_dict(instance, already_passed=frozenset()):
    """Creates a nested dict version of a django model instance
    Follows relationships recursively, circular relationships are terminated by putting
    a model identificator `{model_name}:{instance.id}`.
    Ignores image and file fields."""
    instance_dict = model_to_dict(
        instance,
        fields=[
            f
            for f in instance._meta.concrete_fields
            if not isinstance(f, (ImageField, FileField))
        ],
    )

    already_passed = already_passed.union(
        frozenset((f"{instance.__class__.__name__}:{instance.id}",))
    )
    # Go through possible relationships
    for field in chain(instance._meta.related_objects, instance._meta.concrete_fields):
        if (
            (field.one_to_one or field.many_to_one)
            and hasattr(instance, field.name)
            and (relation := getattr(instance, field.name))
        ):
            if (
                model_id := f"{relation.__class__.__name__}:{relation.id}"
            ) in already_passed:
                instance_dict[field.name] = model_id
            else:
                instance_dict[field.name] = get_instance_dict(relation, already_passed)

        if field.one_to_many or field.many_to_many:
            relations = []
            for relation in getattr(instance, field.get_accessor_name()).all():
                if (
                    model_id := f"{relation.__class__.__name__}:{relation.id}"
                ) in already_passed:
                    relations.append(model_id)
                else:
                    relations.append(get_instance_dict(relation, already_passed))
            instance_dict[field.get_accessor_name()] = relations

    return instance_dict

更简单的方法是使用pprint,它是基于Python的

import pprint
item = MyDjangoModel.objects.get(name = 'foo')
pprint.pprint(item.__dict__, indent = 4)

这将提供类似于json.dumps(…, indent = 4),但它正确地处理了可能嵌入在模型实例中的奇怪的数据类型,如ModelState和UUID等。

在Python 3.7上测试