11

我的一个 Django 模型有这个奇怪的问题，我能够修复它，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()])

每当我制作一个新的（空的）Team对象并想从中获取players时，我都会得到一个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并仅生成名称来修复它，但我认为它应该工作的方式是只返回名称，因为", ".join...这将是一个空列表。相反，会发生一些我不明白的递归。

4

1 回答 1

20

问题是当实例尚未保存到数据库时，您无法访问该team.players字段。Team尝试这样做会引发ValueError.

但是，在尝试引发 thisValueError时，代码将尝试获取您的team对象的表示形式，该表示形式将间接调用unicode(team). 这将尝试访问self.players，它将尝试ValueError在第一个被提出之前提出另一个。这一直持续到达到最大递归深度，但仍然没有ValueError抛出。因此，您只会看到RuntimeError.

如果您执行以下任一操作，同样会（应该？）发生：

>>> team
>>> repr(team)
>>> unicode(team)
于 2013-07-27T16:05:31.433 回答