Django1.2+ South 0.7 + Django -恼人的AutoOneToOneField导致TypeError：'LegacyConnection‘对象不可迭代
Django1.2+ South 0.7 + Django -恼人的AutoOneToOneField导致TypeError：'LegacyConnection‘对象不可迭代
提问于 2010-05-06 20:52:45
我使用的是Django1.2主干和South 0.7，还有一个从Django复制过来的AutoOneToOneField --很烦人。South抱怨说，字段没有定义规则，并且South的新版本不再具有自动字段类型解析器。因此，我阅读了南方文档，并编写了以下定义(基本上是OneToOneField规则的精确副本)：
rules = [
(
(AutoOneToOneField),
[],
{
"to": ["rel.to", {}],
"to_field": ["rel.field_name", {"default_attr": "rel.to._meta.pk.name"}],
"related_name": ["rel.related_name", {"default": None}],
"db_index": ["db_index", {"default": True}],
},
)
]
from south.modelsinspector import add_introspection_rules
add_introspection_rules(rules, ["^myapp"]) 现在，当我进行模式迁移时，South会引发以下错误。
Traceback (most recent call last):
File "manage.py", line 11, in <module>
execute_manager(settings)
File "django/core/management/__init__.py", line 438, in execute_manager
utility.execute()
File "django/core/management/__init__.py", line 379, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "django/core/management/base.py", line 196, in run_from_argv
self.execute(*args, **options.__dict__)
File "django/core/management/base.py", line 223, in execute
output = self.handle(*args, **options)
File "South-0.7-py2.6.egg/south/management/commands/schemamigration.py", line 92, in handle
(k, v) for k, v in freezer.freeze_apps([migrations.app_label()]).items()
File "South-0.7-py2.6.egg/south/creator/freezer.py", line 33, in freeze_apps
model_defs[model_key(model)] = prep_for_freeze(model)
File "South-0.7-py2.6.egg/south/creator/freezer.py", line 65, in prep_for_freeze
fields = modelsinspector.get_model_fields(model, m2m=True)
File "South-0.7-py2.6.egg/south/modelsinspector.py", line 322, in get_model_fields
args, kwargs = introspector(field)
File "South-0.7-py2.6.egg/south/modelsinspector.py", line 271, in introspector
arg_defs, kwarg_defs = matching_details(field)
File "South-0.7-py2.6.egg/south/modelsinspector.py", line 187, in matching_details
if any([isinstance(field, x) for x in classes]):
TypeError: 'LegacyConnection' object is not iterable这与Django 1.2主干最近的变化有关吗？我该如何解决这个问题？
我按如下方式使用此字段：
class Bar(models.Model):
foo = AutoOneToOneField("foo.Foo", primary_key=True, related_name="bar")参考django-tagging中的字段代码：
class AutoSingleRelatedObjectDescriptor(SingleRelatedObjectDescriptor):
def __get__(self, instance, instance_type=None):
try:
return super(AutoSingleRelatedObjectDescriptor, self).__get__(instance, instance_type)
except self.related.model.DoesNotExist:
obj = self.related.model(**{self.related.field.name: instance})
obj.save()
return obj
class AutoOneToOneField(OneToOneField):
def contribute_to_related_class(self, cls, related):
setattr(cls, related.get_accessor_name(), AutoSingleRelatedObjectDescriptor(related))回答 3
Stack Overflow用户
回答已采纳
发布于 2010-05-06 21:32:27
尝试更改此行
(AutoOneToOneField),要这样做：
(AutoOneToOneField,),像你这样声明的元组是不可迭代的。
Stack Overflow用户
发布于 2010-05-06 21:14:11
通过移除规则并在AutoOneToOneField中添加以下方法，解决了此问题：
def south_field_triple(self):
"Returns a suitable description of this field for South."
from south.modelsinspector import introspector
field_class = OneToOneField.__module__ + "." + OneToOneField.__name__
args, kwargs = introspector(self)
return (field_class, args, kwargs)Stack Overflow用户
发布于 2010-05-06 21:32:40
您的规则有简单的python相关问题..在元组中，如果其中只有一项，则必须添加逗号。
因此将(AutoOneToOneField)更改为(AutoOneToOneField，)，
但老实说，我不知道我可以在字段中使用方法而不是规则。我将应用您的补丁并提交到django- apply存储库。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/2781210
复制相关文章
相似问题
腾讯云开发者
