Django自递归外键过滤器查询所有子代 [英] Django self-recursive foreignkey filter query for all childs
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问题描述
我有一个自引用外键关系的模型:
class Person(TimeStampedModel):
pre>
name = models.CharField(max_length = 32)
parent = models.ForeignKey('self',null = True,blank = True,related_name ='children')
现在我想让一个人的所有多层次的孩子。我如何编写Django查询?
解决方案您可以随时向模型添加递归函数:
编辑:根据SeomGi Han更正
def get_all_children(self,include_self = True)
r = []
如果include_self:
r.append(self)
for Person.objects.filter(parent = self):
_r = c。 get_all_children(include_self = True)
如果0< len(_r):
r.extend(_r)
return r
(如果您有很多递归或数据,请不要使用此...)
仍然建议由errx建议的mptt。
I have this model with a self referencing Foreign Key relation:
class Person(TimeStampedModel): name = models.CharField(max_length=32) parent = models.ForeignKey('self', null=True, blank=True, related_name='children')
Now I want to get all the multi level children for a person. How do I write a Django query for it? It needs to behave like recursive function.
解决方案You can always add a recursive function to your model:
EDIT: Corrected according to SeomGi Han
def get_all_children(self, include_self=True): r = [] if include_self: r.append(self) for c in Person.objects.filter(parent=self): _r = c.get_all_children(include_self=True) if 0 < len(_r): r.extend(_r) return r
(Don't use this if you have a lot of recursion or data ...)
Still recommending mptt as suggested by errx.
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