Question

即使遵循doc

，我也无法添加ManyToManyField个对象

models.py

class Label(models.Model):
    ...
    name = models.CharField(blank=False, max_length=100)

class Template(models.Model):
    ...
    labels = models.ManyToManyField(Label, blank=True, related_name="labels")

然后

>>> from content.models import Label, Template
>>> l1 = Label.objects.get_or_create(name='one') # saves in db
>>> l2 = Label.objects.get_or_create(name='two') # saves in db
>>> t1 = Template.objects.get(pk=1)           # loads existing
>>> t1.labels.set([l1,l2])                       # fails

抛出此错误

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/related_descriptors.py", line 1007, in set 
    self.add(*new_objs)
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/related_descriptors.py", line 934, in add 
    self._add_items(self.source_field_name, self.target_field_name, *objs)
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/related_descriptors.py", line 1083, in _add_items
    '%s__in' % target_field_name: new_ids,
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/query.py", line 784, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/query.py", line 802, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/sql/query.py", line 1250, in add_q
    clause, _ = self._add_q(q_object, self.used_aliases)
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/sql/query.py", line 1276, in _add_q
    allow_joins=allow_joins, split_subq=split_subq,
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/sql/query.py", line 1206, in build_filter
    condition = lookup_class(lhs, value)
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/lookups.py", line 24, in __init__
    self.rhs = self.get_prep_lookup()
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/related_lookups.py", line 56, in get_prep_lookup
    self.rhs = [target_field.get_prep_value(v) for v in self.rhs]
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/related_lookups.py", line 56, in <listcomp>
    self.rhs = [target_field.get_prep_value(v) for v in self.rhs]
  File "/path/env3tt/lib/python3.6/site-packages/django/db/models/fields/__init__.py", line 966, in get_prep_value
    return int(value)
TypeError: int() argument must be a string, a bytes-like object or a number, not 'Label'

我在Python 3.6上使用Django 1.11。

Answer 1

您正在使用get_or_create，它返回(object, created)元组，而不仅仅是一个对象。

因此，l1和l2不是Label个对象，而是元组。将此传递给多对多经理是行不通的。

按如下方式更改您的代码：

from content.models import Label, Template
# Ignore the second item returned by get_or_create
l1, _ = Label.objects.get_or_create(name='one') 
l2, _ = Label.objects.get_or_create(name='two') #
t1 = Template.objects.get(pk=1)
t1.labels.set([l1,l2])

无法在Django中添加ManyToManyField对象

1 个答案: