假设active_on是一个日期对象，将其增加1天，然后执行range

next_day = active_on + datetime.timedelta(1)
queryset = queryset.filter(date_created__range=(active_on, next_day) )

Model.objects.filter(datetime__year=2011, datetime__month=2, datetime__day=30)

很简单，如果你有一个datetimefield你可以使用datetime。date。today()

context['now'] = Mymodel.objects.filter(date_time_field=datetime.date.today())

在Django 1.7.6中工作:

MyObject.objects.filter(datetime_attr__startswith=datetime.date(2009,8,22))

下面是我使用ipython的timeit函数得到的结果:

from datetime import date
today = date.today()

timeit[Model.objects.filter(date_created__year=today.year, date_created__month=today.month, date_created__day=today.day)]
1000 loops, best of 3: 652 us per loop

timeit[Model.objects.filter(date_created__gte=today)]
1000 loops, best of 3: 631 us per loop

timeit[Model.objects.filter(date_created__startswith=today)]
1000 loops, best of 3: 541 us per loop

timeit[Model.objects.filter(date_created__contains=today)]
1000 loops, best of 3: 536 us per loop

Contains似乎更快。

请参阅文章Django文档

ur_data_model.objects.filter(ur_date_field__gte=datetime(2009, 8, 22), ur_date_field__lt=datetime(2009, 8, 23))