django - 在保存期间标记 django 模型实例以供以后处理

Question

我在使用标志来标记模型实例以供将来处理时遇到问题。我有一个类:

class MyModel(models.Model):
    processed = models.BooleanField(default=False)
    changed = models.DateTimeField(auto_now=True)
    # More fields.

    def save(self):
        self.processed = False
        super().save(*args, **kwargs)

然后我有一个管理命令:

class Command(BaseCommand):

    def handle(self, *args: Any, **kwargs: Any) -> None:
        models = MyModel.objects.filter(processed=False).order_by("changed")[:200]
        for model in models:
            # Do some processing
            model.processed = True
            model.save()

现在，很明显，当模型被保存时，它只是将实例重新标记为未处理。
我是 Django 的新手，所以我对模型生命周期和可用方法的了解非常有限。我一直在阅读文档，到目前为止还没有找到任何解决方案。
关于我如何解决这个问题的任何想法？

Answer 1

解决这个问题的最好方法可能是添加一个参数来忽略标记:

class MyModel(models.Model):
    processed = models.BooleanField(default=False)
    changed = models.DateTimeField(auto_now=True)
    # More fields.

    def save(self, *args, unset_processed=True, **kwargs):
        if unset_processed:
            self.processed = False
        super().save(*args, **kwargs)

然后在您的基本命令中，我们可以使用:

class Command(BaseCommand):

    def handle(self, *args: Any, **kwargs: Any) -> None:
        models = MyModel.objects.filter(processed=False).order_by("changed")[:200]
        for model in models:
            # Do some processing
            model.processed = True
            model.save(unset_processed=False)

但请注意，批量创建、更新等将绕过调用 .save(…)在模型上。因此，如果您例如使用:

MyModel.objects.filter(pk__in=[1,4,2,5]).update(some_field=some_value)

例如不会调用 .save(…)方法。 documentation on .update(…) [Django-doc]例如声明:

Finally, realize that update() does an update at the SQL level and,thus, does not call any save() methods on your models, nor doesit emit the pre_save or post_save signals

(…)

If you want to update a bunch of records for a model that has a customsave() method, loop over them and call save(), like this:

for e in Entry.objects.filter(pub_date__year=2010):
    e.comments_on = False
    e.save()