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Django-TypeError:int()参数必须是字符串或数字,而不是'dict'

[英]Django - TypeError: int() argument must be a string or a number, not 'dict'

I'm trying to save a multiple Heat instances and store it to a list heats then used the list to a bulk_create as data. 我想保存多个Heat实例,并将其存储到一个列表中heats ,然后用列表到bulk_create数据。

In my models.py 在我的models.py中

class Animal(models.Model):

    farm = models.ForeignKey(Farm, related_name='farm_animals', on_delete=models.CASCADE)
    herd = models.ForeignKey(Herd, related_name='animals', on_delete=models.CASCADE)

    name = models.CharField(max_length=25)
    ....

class Heat(models.Model):

    # Relationship Fields
    animal = models.ForeignKey(Animal, related_name='heats', on_delete=models.CASCADE)

    # Fields
    ....

In my views.py 在我的views.py中

@transaction.atomic
def create(self, request):
    heats = []
    for item in request.data:
        animal = Animal(item['animal'])
        heat = Heat(item['heat'])
        heat.animal = animal

        heats.append(heat)

    Heat.objects.bulk_create(heats)

The request.data is a json serialized. request.data是序列化的json。 Here what it looks like. 在这里看起来像什么。

[
    {
        "animal" : {
            "id" : 1,
            ....
        },
        "heat" : {
            ....
        }
    },
    {
        "animal" : {
            "id" : 2,
            ....
        },
        "heat" : {
            ....
        }
    }, 
    {
        "animal" : {
            "id" : 3,
            ....
        },
        "heat" : {
            ....
        }
    }
]

But i got the error: TypeError: int() argument must be a string or a number, not 'dict' 但是我得到了错误: TypeError: int() argument must be a string or a number, not 'dict'


Here is the full traceback: 这是完整的回溯:

Traceback (most recent call last):
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\core\handlers\exception.py", line 39, in inner
    response = get_response(request)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\core\handlers\base.py", line 187, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\core\handlers\base.py", line 185, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\views\decorators\csrf.py", line 58, in wrapped_view
    return view_func(*args, **kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\rest_framework\viewsets.py", line 83, in view
    return self.dispatch(request, *args, **kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\rest_framework\views.py", line 477, in dispatch
    response = self.handle_exception(exc)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\rest_framework\views.py", line 437, in handle_exception
    self.raise_uncaught_exception(exc)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\rest_framework\views.py", line 474, in dispatch
    response = handler(request, *args, **kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\utils\decorators.py", line 185, in inner
    return func(*args, **kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\farm_management\heat\views.py", line 188, in create
    Heat.objects.bulk_create(heats)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\manager.py", line 85, in manager_method
    return getattr(self.get_queryset(), name)(*args, **kwargs)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\query.py", line 449, in bulk_create
    self._batched_insert(objs_with_pk, fields, batch_size)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\query.py", line 1068, in _batched_insert
    self._insert(item, fields=fields, using=self.db)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\query.py", line 1045, in _insert
    return query.get_compiler(using=using).execute_sql(return_id)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\sql\compiler.py", line 1053, in execute_sql
    for sql, params in self.as_sql():
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\sql\compiler.py", line 1006, in as_sql
    for obj in self.query.objs
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\sql\compiler.py", line 945, in prepare_value
    value = field.get_db_prep_save(value, connection=self.connection)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\fields\__init__.py", line 755, in get_db_prep_save
    prepared=False)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\fields\__init__.py", line 938, in get_db_prep_value
    value = self.get_prep_value(value)
  File "C:\Users\Web\Desktop\PyDev\Envs\djangular\lib\site-packages\django\db\models\fields\__init__.py", line 946, in get_prep_value
    return int(value)
TypeError: int() argument must be a string or a number, not 'dict'

I can't fully understand the situation since i am new to django development. 由于我不熟悉Django开发,因此我无法完全了解情况。 Please help. 请帮忙。

You should extract already existing animals from the database instead of creating new ones. 您应该从数据库中提取已经存在的动物,而不是创建新的动物。 And it's a good idea to extract the animals "in bulk" too to increase performance. 最好“大量”提取动物以提高性能。

The code: 编码:

@transaction.atomic
def create(self, request):
    # Extracting animals
    animal_id_list = [item['animal']['id'] for item in request.data]
    animals = Animal.objects.in_bulk(animal_id_list)
    animals_dict = {animal.id: animal for animal in animals}

    # Creating heats
    heats = []
    for item in request.data:
        heat = Heat(**item['heat'])
        heat.animal = animals_dict[item['animal']['id']]
        heats.append(heat)
    Heat.objects.bulk_create(heats)

This is because of the foreign key relationship between models. 这是因为模型之间存在外键关系。 If your model has a ForeignKey called animal, then your data_dict should have an animal_id field. 如果您的模型有一个名为animal的ForeignKey,则您的data_dict应该有一个animal_id字段。 But django.forms.model_to_dict() returns a dict with an animal field. 但是django.forms.model_to_dict()返回带有动物字段的字典。
So you can't do MyModel(**model_to_dict(my_instance)); 所以你不能做MyModel(**model_to_dict(my_instance)); you have to rename the animal field to animal_id 您必须将动物字段重命名为animal_id

Something like this should work: 这样的事情应该起作用:
heat.animal_id = animal.id

In your create method, you should pass item['heat'] as kwargs and save animal before using it in heat. 在您的create方法中,应将item['heat']传递为kwargs并保存动物,然后再将其用于加热。 Eg 例如

@transaction.atomic
def create(self, request):
    heats = []
    for item in request.data:
        animal = Animal.objects.get(id=item['animal']['id'])
        heat = Heat(animal=animal, **item['heat'])

        heats.append(heat)

    Heat.objects.bulk_create(heats) 

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