django queryset按字段排序

发布于 2025-02-07 09:57:51 字数 904 浏览 0 评论 0原文

我有这样的模型：

class ProgrammingQuestionAnswer(models.Model):
    programming_question = models.ForeignKey(ProgrammingQuestion, on_delete=models.CASCADE, related_name='programming_question_a', null=True, blank=True)
    time = models.DateTimeField(default=timezone.now)
    score = models.IntegerField(null=True, blank=True, default=0)
    user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='writer_answer_programming_question', null=True, blank=True)
    accept = models.BooleanField(default=False)
    file = models.FileField(upload_to=content_file_name)
    result = models.TextField(null=True, blank=True)
    file_name = models.CharField(max_length=500, null=True, blank=True)
    max_score = models.IntegerField(null=True, blank=True, default=0)

我想为解决更多问题的用户编写一个查询（有更多的接受= true），而那些平等的人则按时间（时间字段）进行排序（时间字段）谢谢

原文

i have this model:

class ProgrammingQuestionAnswer(models.Model):
    programming_question = models.ForeignKey(ProgrammingQuestion, on_delete=models.CASCADE, related_name='programming_question_a', null=True, blank=True)
    time = models.DateTimeField(default=timezone.now)
    score = models.IntegerField(null=True, blank=True, default=0)
    user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='writer_answer_programming_question', null=True, blank=True)
    accept = models.BooleanField(default=False)
    file = models.FileField(upload_to=content_file_name)
    result = models.TextField(null=True, blank=True)
    file_name = models.CharField(max_length=500, null=True, blank=True)
    max_score = models.IntegerField(null=True, blank=True, default=0)

I want to write a query for get users who solve more questions(have more accept=True) and those who are equal are sorted by time(time field)
Thanks

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舂唻埖巳落 2025-02-14 09:57:51

尝试

accepted = Q( 'writer_answer_programming_question__accept=True')

User.objects.all.annotate(
    accepted_count = Count('writer_answer_programming_question', filter = accepted    
    ).order_by('accepted_count', 'time'
    )

DOC：
contregation函数，在注释上过滤

Try

accepted = Q( 'writer_answer_programming_question__accept=True')

User.objects.all.annotate(
    accepted_count = Count('writer_answer_programming_question', filter = accepted    
    ).order_by('accepted_count', 'time'
    )

Doc:
Aggregation functions, Filtering on annotations

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