在Django QuerySet中，如何以多对一关系筛选“不存在”(In a Django QuerySet, how to filter for “not exists” in a many

在Django QuerySet中，如何以多对一关系筛选“不存在”(In a Django QuerySet, how to filter for “not exists” in a many-to-one relationship)

我有两个这样的模型：

class User(models.Model): email = models.EmailField() class Report(models.Model): user = models.ForeignKey(User)

实际上，每个模型都有更多的领域对这个问题无关紧要。

我想过滤所有以'a'开头并且没有报告的电子邮件的用户。 基于其他字段会有更多的.filter()和.exclude()条件。

我想要像这样来处理它：

users = User.objects.filter(email__like = 'a%') users = users.filter(<other filters>) users = ???

我想要 ？？？ 过滤掉没有关联报告的用户。 我将如何做到这一点？ 如果这是不可能的，我已经介绍过了，什么是替代方法？

I have two models like this:

class User(models.Model): email = models.EmailField() class Report(models.Model): user = models.ForeignKey(User)

In reality each model has more fields which are of no consequence to this question.

I want to filter all users who have an email which starts with 'a' and have no reports. There will be more .filter() and .exclude() criteria based on other fields.

I want to approach it like this:

users = User.objects.filter(email__like = 'a%') users = users.filter(<other filters>) users = ???

I would like ??? to filter out users who do not have reports associated with them. How would I do this? If this is not possible as I have presented it, what is an alternate approach?

最满意答案

使用isnull 。

users_without_reports = User.objects.filter(report__isnull=True) users_with_reports = User.objects.filter(report__isnull=False).distinct()

当您使用isnull=False ，需要使用distinct()来防止重复结果。

Use isnull.

users_without_reports = User.objects.filter(report__isnull=True) users_with_reports = User.objects.filter(report__isnull=False).distinct()

When you use isnull=False, the distinct() is required to prevent duplicate results.