在Django中加入两个包含相同外键的表 [英] JOIN two tables containing same foreign key in Django

问题描述

所以我有两个模型包含主键'User'

  class Reviews（models.Model）：
 subject_user = models.ForeignKey（User，related_name ='submitted_user'）
 actor = models.ForeignKey（User）## irrelavent 
 text = models.TextField（）
 
 class友谊（models.Model）：
 head_user = models.ForeignKey（User，related_name ='follow'）
 tail_user = models.ForeignKey（User）## irrelavent 
  

现在我想查找也在 Friendsip.head_user 中的用户的所有评论。例如，说用户'1'为用户'3'和用户'5'

$ b $写了一个评论b

 评论：
 subject_user_id actor_id文本
 3 1 LoremIpsum1 
 5 1 LoremIpsum2 
 
友谊： 
 head_user tail_user 
 3 9 
 8 9 ### Irrelavent 
  

现在我输出应该是对应于'LoremIpsum1'的评论对象，即subject_user也存在于 Friendship.head_user 中的评论对象。我尝试了以下

  reviews = Reviews.objects.all（）。select_related（'friendship__head_user'）
  

但它是给我所有的评论

编辑：

这是我如何插入评论：

  review，created = Reviews.object.get_or_create（subject_user = target，actor = actor）
 review.text = NewReviewText 
 review.save（）
  / pre> 
 
解决方案
如果我正确理解你，这应该是有效的：
  heads =（f.head_user for f in Friendship.objects.all（））
评论= Reviews.objects.all（）。filter（subject_user__in= heads） 
  
首先，您抓住友谊中的所有用户 s，然后找到所有的评论，这似乎符合您的要求：
 
 
我想查找所有的用户评论在友谊中。 head_user  
 
 
你不能使用的原因  select_related  它是否希望在两种模式之间存在明确的关系，在这种情况下，这种模式不是。在评论和友谊之间没有定义外键，所以在这种情况下最合适的方式可能是我已经提出了。
 
 
 
然而，直到你已经在愤怒中进行了测试，并做了一些分析来确认是系统中一个关键的慢点，我注意到Donald Knuth的话：
 
 
我们应该忘记小的效率，约97％的时间：过早优化是所有的根源
 
 
 
So I have two models containing primary key 'User'
class Reviews(models.Model):
    subject_user = models.ForeignKey(User,related_name='reviewed_user')
    actor = models.ForeignKey(User)                         ## irrelavent 
    text = models.TextField()

class Friendship(models.Model):
    head_user = models.ForeignKey(User,related_name='followed')
    tail_user = models.ForeignKey(User)                     ## irrelavent 
Now I want to find all the reviews of users who are also in the Friendsip.head_user. For example, say User '1' wrote a review each for User '3' and User '5'
Reviews:
subject_user_id     actor_id     text
3                   1            LoremIpsum1
5                   1            LoremIpsum2

Friendship:
head_user           tail_user
3                   9
8                   9      ### Irrelavent
Now I output should be reviews object corresponding to 'LoremIpsum1' i.e. the review object whose subject_user is also present in Friendship.head_user. I tried the following
reviews = Reviews.objects.all().select_related('friendship__head_user')
But it is giving me all the reviews

Edit:

This is how I insert the review:
review, created = Reviews.object.get_or_create(subject_user=target, actor=actor)
review.text = NewReviewText
review.save()

解决方案
If I understand you correctly, this should work:
heads = (f.head_user for f in Friendship.objects.all())
reviews = Reviews.objects.all().filter("subject_user__in"=heads)
First you grab all the head users in Friendships and then find all their reviews, which seems to match your requirement:

  
I want to find all the reviews of users who are also in the Friendship.head_user
The reason you can't use select_related is that it expects that there will be an explicit relationship between the two models, which in your case there is not. There is no foreign key defined between a Review and a Friendship, so in this case the most optimal way is probably what I've proposed.

However, until you've tested this in anger and done some analysis to confirm is is a critical slow point in your system, I'd heed the words of Donald Knuth:

  
We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil.


                        
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