python - Django 嵌套事务和异常

Question

为了简单起见，我的代码中有一个嵌套的原子事务，基本上，如果将人员添加到电影失败(请参阅第一个 for 循环)，我想显示一个异常，专门说明此操作失败并回滚 movie.save()，与工作室相同(请参阅第二个 for 循环)。但由于某种原因，当其中任何一个失败 movie.save() 都不会回滚时，我会在数据库中保存一部电影，而没有任何人员或工作室附加到它。

people = ['Mark', 'John', 'Mark']
studios = ['Universal Studios', 'Longcross Studios']

try:
    with transaction.atomic():
        movie = Movie(title=title, description=summary, pub_date=pub_date)
        movie.save()

        # iterate through people list, get the pk from database and add it to the movie
        try:
            with transaction.atomic():
                for p in people:
                    person = Person.objects.get(name=p)
                    movie.people.add(person)

        except Person.DoesNotExist:
            self.stdout.write(self.style.ERROR('One or more performers from %s does not '
                                               'exist in database, please add them and '
                                               'rerun the command.' % str(people)))
            continue

        # iterate through studio list, get the pk from database and add it to the movie
        try:
            with transaction.atomic():
                for s in studios:
                    studio = Studio.objects.get(name=s)
                    movie.studios.add(studio)
        except Studio.DoesNotExist:
            self.stdout.write(self.style.ERROR('One or more studios from %s does not '
                                               'exist in database, please add them and '
                                               'rerun the command.' % str(studios)))
            continue

        movie.save()
        self.stdout.write(self.style.NOTICE('Successfully added movie %s' % title))

except Exception as e:
    self.stdout.write(self.style.ERROR('Failed to add movie to the database.'))

Answer 1

当atomic block 因异常退出时会发生回滚。但是您正在捕获异常并抑制它们，因此回滚不会发生。要让异常在记录消息后传播，您需要使用 raise，而不是 continue。