심지어 <a href="https://docs.djangoproject.com/en/1.11/ref/models/relations/" rel="nofollow noreferrer">doc</a></p> <p>models.py에게 다음과 같은 후 내가 <code>ManyToManyField</code> 객체를 추가 할 수 없습니다입니다 장고

Question

에 ManyToManyField 객체를 추가 할 수 없습니다

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") 

그리고

>>> 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 

이 오류가 발생합니다

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' 

파이썬 3.6에서 장고 1.11을 사용하고 있습니다.

Answer 1

개체가 아닌 (object, created)의 튜플을 반환하는 get_or_create을 사용하고 있습니다.

그래서 l1 및 l2은 Label 개체가 아니라 터플입니다. 이것을 many-to-many 매니저에게 넘기는 것은 작동하지 않을 것이다. 올바른

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])