如何在 Django 1.9 中加入多个表？

Question

我需要帮助。我的问题是我试图在 Django 1.9 中加入多个表/模型（（Question, Answer, User)），以获得由某个用户提出的一组问题的答案。

我已经有了我想要的 sql 查询：

 SELECT q.id, q.title, COUNT(a.id) AS total_answers 
 FROM review_question AS q
 JOIN review_answer AS a ON q.id = a.question_id
 JOIN users_user AS u ON q.user_id = u.id
 WHERE q.user_id = 1
 GROUP BY q.id, q.title;

这是我的模型：

review/models.py

[问题]

class Question(models.Model):
    user = models.ForeignKey(User, db_index=True, null=True, blank=True)
    tag = models.ManyToManyField(Tag)
    title = models.CharField(max_length=200)

[回答]

class Answer(models.Model):
    question = models.ForeignKey(Question)
    user = models.ForeignKey(User, db_index=True, null=True, blank=True)

users/models.py

[用户]

class User(models.Model):
    username = models.CharField(max_length=100, unique=True)

顺便说一句，在我的文件 users/views.py 我有下一个：

class UserDetailView(DetailView):
    model = User

    def get_context_data(self, **kwargs):
        # calling the get_context_data parent here
        questions = Question.objects.filter(user = context['object']).order_by('created')

但是，我只得到用户提出的所有问题。

我一直在尝试寻找如何将上面的查询转换为 django-orm 的方法，但是我仍然无法做到这一点。任何帮助将不胜感激。

Answer 1

终于可以解决我的问题了。

我做的是下一个：

在我的文件 users/views.py

class UserDetailView(DetailView):
    model = User

    def get_context_data(self, **kwargs):
        # Calling the get_context_data parent
        questions = Question.objects.filter(user = context['object']).order_by('created')

        tags = [ question.tag.all() for question in questions ]

        total_answers = self.get_total(questions) # Calling the function that return the total answer by question of a specific user
        context['question_tags'] = zip(questions, tags, total_answers) #Zipping the lists with results to use it in the template

        return context

    def get_total(self, questions):
        #Return the total answers that a question has (Here is the trick!)
        return [
            Answer.objects.filter(question__id=question.id).annotate(num_answers=Count('question')).count() for question in questions]

这就是我所做的一切。 最后，我要特别感谢@PauloAlmeida 和@MikeVelazco 的帮助！