如果Django-reversion中没有任何变化，不要创建新版本。

Question

我想保存新的对象版本，只有当一些东西已经改变，在django-reversion中。我翻阅了文件，却没有发现任何关于它的东西。我怎样才能做到这一点？

Answer 1

您可以使用选项。不幸的是

它不遵循关系，因为这可能会很快变得昂贵和缓慢。

如果您真的想忽略后续关系的重复，您有两种可能性：

1. 执行分叉并禁用限制

在这里删除and explicit https://github.com/etianen/django-reversion/blob/master/reversion/revisions.py#L199

默认情况下将ignore_duplicates设置为True https://github.com/etianen/django-reversion/blob/master/reversion/revisions.py#L368

要小心，正如上面提到的，它可能是缓慢的。

1. 侦听修订后提交信号并手动删除重复版本

将ignore_duplicates设置为False并添加信号接收器：

from django.db import transaction
from django.dispatch import receiver
from reversion.models import Revision, Version
from reversion.signals import post_revision_commit


def clear_versions(versions, revision):
    count = 0
    for version in versions:
        previous_version = Version.objects.filter(
            object_id=version.object_id,
            content_type_id=version.content_type_id,
            db=version.db,
            id__lt=version.id,
        ).first()
        if not previous_version:
            continue
        if previous_version._local_field_dict == version._local_field_dict:
            version.delete()
            count += 1
        if len(versions_ids) == count:
            revision.delete()


@receiver(post_revision_commit)
def post_revision_commit_receiver(sender, revision, versions, **kwargs):
    transaction.on_commit(lambda: clear_versions(versions, revision))

它也可能很慢，但您可以异步完成(例如，在芹菜任务中)：

# tasks.py

@celery.task(time_limit=60, ignore_result=True)
def clear_versions(revision_id, versions_ids):
    count = 0
    if versions_ids:
        for version in Version.objects.filter(id__in=versions_ids):
            previous_version = Version.objects.filter(
                object_id=version.object_id,
                content_type_id=version.content_type_id,
                db=version.db,
                id__lt=version.id,
            ).first()
            if not previous_version:
                continue
            if previous_version._local_field_dict == version._local_field_dict:
                version.delete()
                count += 1
    if len(versions_ids) == count:
        Revision.objects.only('id').get(id=revision_id).delete()

# signals.py

@receiver(post_revision_commit)
def post_revision_commit_receiver(sender, revision, versions, **kwargs):
    transaction.on_commit(
        lambda: clear_versions.delay(revision.id, [v.id for v in versions])
    )