Django：如何跟随ForeignKey（'self'）向后

问题描述

class Achievement（MyBaseModel）：
parent_achievement = models.ForeignKey（'self'，blank = True，null = True，help_text =必须在之前完成的成就这个被实现了）＃从父保留的长名称


我可以做：

  Achievement.objects.get（pk =1）。parent_achievement 
  

这是非常好的。但是如何获得所有的孩子？

  Achievement.objects.get（pk =1）。parent_achievement_set 
  

不工作（可能应该有更多的符号），我没有看到太多搜索时。

可以吗？

解决方案

默认情况下，django将调用相反的模型名称，后跟_set，因此它将be

  Achievement.objects.get（pk =1）。achievement_set 
  

如果这不适合您，请使用 related_name 可选参数 models.ForeignKey ：

  class Achievement（MyBaseModel）：
 parent_achievement = models。 ForeignKey（
'self'，
 blank = True，
 null = True，
 help_text =在此之前必须完成的成就，
 related_name =child_achievement_set
）从父保留的长名称
 
 Achievement.objects.get（pk =1）child_achievement_set 
  

class Achievement(MyBaseModel):
    parent_achievement = models.ForeignKey('self', blank=True, null=True, help_text="An achievement that must be done before this one is achieved") # long name since parent is reserved

I can do :

Achievement.objects.get(pk="1").parent_achievement

which is great. But how do I get all the children?

Achievement.objects.get(pk="1").parent_achievement_set

doesn't work (and probably should have some more notation around it), and I didn't see too much when searching.

Is it possible? Fall into SQL?

解决方案

By default, django will call the reverse the model name, followed by "_set", so it would be

Achievement.objects.get(pk="1").achievement_set

If that doesn't suit you, use the related_name optional argument to models.ForeignKey:

class Achievement(MyBaseModel):
    parent_achievement = models.ForeignKey(
        'self', 
        blank=True, 
        null=True, 
        help_text="An achievement that must be done before this one is achieved",
        related_name="child_achievement_set"
    ) # long name since parent is reserved

Achievement.objects.get(pk="1").child_achievement_set

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