将Django Model对象转换为所有字段完整的dict [英] Convert Django Model object to dict with all of the fields intact
问题描述
如何将django Model对象转换为其字段中全部的dict?所有理想情况下都包括外键和可编辑= False的字段。
让我详细说明一下。假设我有一个django模型,如下所示:
从django.db导入模型
类OtherModel(models.Model):pass
class SomeModel(models.Model):
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)
在终端中,我已经完成了以下操作:
other_model = OtherModel()
other_model.save()
instance = SomeModel()
instance.value = 1
实例。 value2 = 2
instance.reference1 = other_model
instance.save()
instance.reference2.add(other_model)
instance.save()
我想将其转换为以下字典:
code> {'created':datetime.dat etime(2015,3,16,21,34,14,926738,tzinfo =< UTC),
u'id':1,
'reference1':1,
' reference2':[1],
'value':1,
'value2':2}
不满意答案的问题:
我如何将Django Model对象转换成字典并仍然拥有外键?
有许多方法可以将实例转换为字典,具有不同程度的角位处理和逼近所需的结果。
1。 实例.__ dict __
实例.__ dict__
$
$ b pre $ < _reference1_cache':< OtherModel:OtherModel object>
'_state':< django.db.models.base.ModelState at 0x1f63310>
'created':datetime.datetime(2014,2, ,
'id':1L,
'reference1_id':1L,
'value':1,
'value2':2}
这是迄今为止最简单的,但是缺少reference2,reference1没有名字,它有两个额外的东西。
2。 import_to_dict
model_to_dict(instance)
$ c $($)
$ b $ $ $ $ $ $ $ $ c>
其中返回
{u'id ':1L,'reference1':1L,'reference2':[1L],'value':1}
这是唯一具有reference2的内容,但是缺少不可编辑的字段。
3。 model_to_dict with fields
from django.forms.models import model_to_dict
model_to_dict(instance,fields = [field.name for字段in instance._meta.fields])
其中返回
{u'id':1L,'reference1':1L,'value':1}
这比标准的model_to_dict调用严格得多。
4 。 query_set.values()
SomeModel.objects.filter(id = instance.id).values()[0]
其中返回
{'created':datetime.datetime(2014,2,21,4,38,51,tzinfo =< UTC),
u'id':1L,
'reference1_id' :1L,
'value':1L,
'value2':2L}
这是与实例相同的输出.__ dict __
但没有额外的字段。
5。自定义函数
django的model_to_dict的代码有很多答案。它显式删除不可编辑的字段,因此删除该检查会导致以下代码,其行为如下:
from django.db .models.fields.related import ManyToManyField
def to_dict(instance):
opts = instance._meta
data = {}
for opt in opts.concrete_fields + opts.many_to_many:
如果isinstance(f,ManyToManyField):
如果instance.pk是None:
data [f.name] = []
else:
data [f.name] = list(f.value_from_object(instance).values_list('pk',flat = True))
else:
data [f.name] = f.value_from_object(instance)
返回数据
虽然这是最复杂的选项,但调用 to_dict(instance)
给我们准确的结果:
{'created':datetime.datetime (2015,3,16,21,34,14,926738,tzinfo =< UTC),
u'id':1,
'reference1':1,
'reference2':[1],
'value':1,
'value2':2}
奖金回合
如果你想要一个具有更好的python命令行显示的django模型,让你的模型子类如下:
from django.db导入模型
pre>
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 opt in cons.concrete_fields + opts .many_to_many:
如果isinstance(f,ManyToManyField):
如果self.pk为无:
data [f.name] = []
else:
数据[f.name] = list(f.value_from_object(self).values_list('pk',flat = True))
else:
data [f.name] = f.value_from_object(self)
返回数据
class Meta:
abstract = True
所以,例如,如果我们定义我们的模型:
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)
现在调用
SomeModel.objects.first()
给出这样的输出:{'created':datetime.datetime(2015,3,16,21,34, 926738,tzinfo =< UTC),
'value':1,'value2':2,'reference1':1,u'id':1,'reference2':[1]}
How does one convert a django Model object to a dict with all of its fields? All ideally includes foreign keys and fields with editable=False.
Let me elaborate. Let's say I have a django model like the following:
from django.db import models class OtherModel(models.Model): pass class SomeModel(models.Model): 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")
In the terminal, I have done the following:
other_model = OtherModel() other_model.save() instance = SomeModel() instance.value = 1 instance.value2 = 2 instance.reference1 = other_model instance.save() instance.reference2.add(other_model) instance.save()
I want to convert this to the following dictionary:
{'created': datetime.datetime(2015, 3, 16, 21, 34, 14, 926738, tzinfo=<UTC>), u'id': 1, 'reference1': 1, 'reference2': [1], 'value': 1, 'value2': 2}
Questions with unsatisfactory answers:
Django: Converting an entire set of a Model's objects into a single dictionary
How can I turn Django Model objects into a dictionary and still have their foreign keys?
解决方案There are many ways to convert instance to a dictionary, with varying degrees of corner case handling and closeness to the desired result.
1.
instance.__dict__
instance.__dict__
which returns
{'_reference1_cache': <OtherModel: OtherModel object>, '_state': <django.db.models.base.ModelState at 0x1f63310>, 'created': datetime.datetime(2014, 2, 21, 4, 38, 51, 844795, tzinfo=<UTC>), 'id': 1L, 'reference1_id': 1L, 'value': 1, 'value2': 2}
This is by far the simplest, but is missing reference2, reference1 is misnamed, and it has two extra things in it.
2. model_to_dict
from django.forms.models import model_to_dict model_to_dict(instance)
which returns
{u'id': 1L, 'reference1': 1L, 'reference2': [1L], 'value': 1}
This is the only one with reference2, but is missing the uneditable fields.
3. model_to_dict with fields
from django.forms.models import model_to_dict model_to_dict(instance, fields=[field.name for field in instance._meta.fields])
which returns
{u'id': 1L, 'reference1': 1L, 'value': 1}
This is strictly worse than the standard model_to_dict invocation.
4. query_set.values()
SomeModel.objects.filter(id=instance.id).values()[0]
which returns
{'created': datetime.datetime(2014, 2, 21, 4, 38, 51, tzinfo=<UTC>), u'id': 1L, 'reference1_id': 1L, 'value': 1L, 'value2': 2L}
This is the same output as
instance.__dict__
but without the extra fields.
5. Custom Function
The code for django's model_to_dict had most of the answer. It explicitly removed non-editable fields, so removing that check results in the following code which behaves as desired:
from django.db.models.fields.related import ManyToManyField def to_dict(instance): opts = instance._meta data = {} for f in opts.concrete_fields + opts.many_to_many: if isinstance(f, ManyToManyField): if instance.pk is None: data[f.name] = [] else: data[f.name] = list(f.value_from_object(instance).values_list('pk', flat=True)) else: data[f.name] = f.value_from_object(instance) return data
While this is the most complicated option, calling
to_dict(instance)
gives us exactly the desired result:{'created': datetime.datetime(2015, 3, 16, 21, 34, 14, 926738, tzinfo=<UTC>), u'id': 1, 'reference1': 1, 'reference2': [1], 'value': 1, 'value2': 2}
Bonus Round
If you want a django model that has a better python command-line display, have your models child class the following:
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)) else: data[f.name] = f.value_from_object(self) return data class Meta: abstract = True
So, for example, if we define our models as such:
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")
Calling
SomeModel.objects.first()
now gives output like this:{'created': datetime.datetime(2015, 3, 16, 21, 34, 14, 926738, tzinfo=<UTC>), 'value': 1, 'value2': 2, 'reference1': 1, u'id': 1, 'reference2': [1]}
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