Django使用dict参数过滤OR条件

时间:2016-06-30 19:17:40

标签: python django

我在我的Django应用程序上有一个函数,我在其中执行一些Queryset操作并将其结果设置为Memcache。由于它是一种功能,因此必须具有一般用途。因此,为了使其可重复使用,我将dict作为filterexclude操作的参数传递。这是功能:

def cached_query(key, model, my_filter=None, exclude=None, order_by=None, sliced=50):
    """
    :param key: string used as key reference to store on Memcached
    :param model: model reference on which 'filter' will be called
    :param my_filter: dictionary containing the filter parameters (eg.: {'title': 'foo', 'category': 'bar'}
    :param sliced: integer limit of results from the query. The lower the better, since for some reason Django Memcached
        won't store thousands of entries in memory
    :param exclude: dictionary containing the exclude parameters (eg.: {'title': 'foo', 'category': 'bar'}
    :param order_by: tuple containing the list of fields upon which the model will be ordered.
    :return: list of models. Not a QuerySet, since it was sliced.
    """
    result = cache.get(key, None)
    if not result:
        if my_filter:
            result = model.objects.filter(**my_filter)
        if exclude:
            result = result.exclude(**exclude)
        if order_by:
            result = result.order_by(*order_by)
        else:
            result = model.objects.all()
        result = result[:sliced]
        cache.set(key, result, cache_timeout)
    return result

如果我使用像{'title': 'foo', 'name': 'bar'}这样的简单dict过滤查询集,它的工作原理非常好。然而,并非总是如此。我需要使用django.db.models.Q实用程序执行过滤器,以处理需要OR条件的更复杂查询。

那么,我如何将这些参数作为字典传递给过滤器。对此有什么办法吗?

3 个答案:

答案 0 :(得分:8)

您可以将字典重组为单个键值字典列表,并在dict表达式中的每个Q上使用解压缩,如下所示:

from functools import reduce
import operator

from django.db.models import Q

# your dict is my_filter
q = model.objects.filter(reduce(operator.or_, 
                                (Q(**d) for d in [dict([i]) for i in my_filter.items()])))
reduce上的

or_在OR上加入Q个表达式。

您还可以使用生成器表达式,其中listdict

q = model.objects.filter(reduce(operator.or_, 
                                (Q(**d) for d in (dict([i]) for i in my_filter.items()))))

答案 1 :(得分:4)

您可以使用按位|运算符。

my_filter = Q()

# Or the Q object with the ones remaining in the list
my_or_filters = {'some_field__gte':3.5, 'another_field':'Dick Perch'}

for item in my_or_filters:
    my_filter |= Q(**{item:my_or_filters[item]})

model.objects.filter(my_filter)
# unpacks to model.objects.filter(Q(some_field__gte=3.5) | Q(another_field='Dick Perch'))

考虑到这一点,您可能希望将my_filter中存储的所有查询加载到Q个对象中。然后,您可以通过上面使用按位&的相同方法加入所有非OR查询:     my_filter& = ...

答案 2 :(得分:0)

根据@Moses Koledoye的回答,我可以解决这个问题。这就是我现在的功能:

{{1}}