[英]Django filter on generic relationship (unique constraint exception)
I have a model below which points to a generic relationship.我在下面有一个模型,它指向一个通用关系。 This can either be a Post
object or a Reply
object.这可以是Post
对象或Reply
对象。
class ReportedContent(models.Model):
reporter = models.ForeignKey(User, on_delete=models.CASCADE)
# Generic relation for posts and replies
content_type = models.ForeignKey(ContentType, on_delete=models.CASCADE)
object_id = models.PositiveIntegerField()
content_object = GenericForeignKey()
class Meta:
unique_together = ('reporter', 'object_id', 'content_type')
I would like to check if the content_object is already exists before I get a duplicate key value violates unique constraint
exception.我想在获得duplicate key value violates unique constraint
异常之前检查 content_object 是否已经存在。
Django documentation mentioned that: Django 文档提到:
# This will fail
>>> ReportedContent.objects.filter(content_object=content)
# This will also fail
>>> ReportedContent.objects.get(content_object=content)
So how can I filter on generic relation?那么如何过滤通用关系呢? or how can I deal with this exception specifically?或者我该如何处理这个异常?
you can filter by object_id
and content_type
.您可以按object_id
和content_type
进行过滤。 just make sure you do it right, get content_type
this way:只要确保你做对了,就这样获取content_type
:
from django.contrib.contenttypes.models import ContentType
# ...
content_type = ContentType.objects.get(app_label='name_of_your_app', model='model_name')
for handling the exception :用于处理异常:
if ReportedContent.objects.filter(object_id=content.id,content_type=content_type):
raise Exception('your exception message')
I realize this is an old(ish) question, but I thought I'd offer an alternative method in case others run across this post as I did.我意识到这是一个古老的(ish)问题,但我想我会提供一种替代方法,以防其他人像我一样遇到这篇文章。
Instead of doing a separate .get()
on the ContentType model, I just incorporate the app/model names in my filter, like this:我没有在 ContentType 模型上执行单独的.get()
,而是将应用程序/模型名称合并到我的过滤器中,如下所示:
queryset = ReportedContent.objects.filter(
object_id=parent_object.id,
content_type__app_label=app_label,
content_type__model=model_name
)
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