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我只想在 django-reversion 中保存新的对象版本。我浏览了文档并没有找到任何关于它的信息。我怎样才能实现它?

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1 回答 1

2

您可以使用ignore_duplicates选项。很遗憾

它不遵循关系,因为这会很快变得昂贵和缓慢。

如果您真的想忽略跟随关系的重复项,您有两种可能性:

  1. 执行分叉并禁用限制

and explicit在此处删除https://github.com/etianen/django-reversion/blob/master/reversion/revisions.py#L199

默认设置ignore_duplicateshttps://github.com/etianen/django-reversion/blob/master/reversion/revisions.py#L368True

小心,如上所述,它可能很慢。

  1. 监听修订后提交信号并手动删除重复版本

设置ignore_duplicatesFalse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,但您可以异步执行(例如,在 Celery 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])
    )
于 2018-05-17T11:34:30.540 回答