我的Django模型之一存在这个奇怪的问题,我能够解决它,但不明白发生了什么。
这些是模型:
class Player(models.Model):
facebook_name = models.CharField(max_length=100)
nickname = models.CharField(max_length=40, blank=True)
def __unicode__(self):
return self.nickname if self.nickname else self.facebook_name
class Team(models.Model):
name = models.CharField(max_length=50, blank=True)
players = models.ManyToManyField(Player)
def __unicode__(self):
name = '(' + self.name + ') ' if self.name else ''
return name + ", ".join([unicode(player) for player in self.players.all()])
每当我创建一个新的(空的)Team
对象并希望从中获取players
时,我得到了RuntimeError: maximum recursion depth exceeded
。
例如:
>>> team = Team()
>>> team.players
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/fields/related.py", line 897, in __get__
through=self.field.rel.through,
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/fields/related.py", line 586, in __init__
(instance, source_field_name))
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/base.py", line 421, in __repr__
u = six.text_type(self)
File "/Users/walkman/Projects/fociadmin/fociadmin/models.py", line 69, in __unicode__
return name + ", ".join([unicode(player) for player in self.players.all()])
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/fields/related.py", line 897, in __get__
through=self.field.rel.through,
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/fields/related.py", line 586, in __init__
(instance, source_field_name))
File "/Users/walkman/Projects/fociadmin/venv/lib/python2.7/site-packages/django/db/models/base.py", line 421, in __repr__
u = six.text_type(self)
File "/Users/walkman/Projects/fociadmin/fociadmin/models.py", line 69, in __unicode__
return name + ", ".join([unicode(player) for player in self.players.all()])
...
为什么会这样?我能够通过检查pk
来修复它,然后只生成名称,但我认为它应该起作用的方式只返回名称,因为", ".join...
将是一个空列表。相反,发生了一些我不理解的递归。
答案 0 :(得分:18)
问题是,当team.players
实例尚未保存到数据库时,您无法访问Team
字段。尝试这样做会引发ValueError
。
但是,在尝试引发此ValueError
时,代码将尝试获取将team
对象的表示,该对象将间接调用unicode(team)
。这将尝试访问self.players
,它会在第一个引发之前尝试引发另一个ValueError
。这一直持续到达到最大递归深度,但仍然没有抛出ValueError
。因此,您只会看到RuntimeError
。
如果您执行以下任一操作,那么(应该?)会发生同样的情况:
>>> team
>>> repr(team)
>>> unicode(team)