Django过滤器 - 还是？

Question

我如何在django过滤器中执行"或"操作.

基本上,我希望能够列出用户添加的项目(它们被列为创建者)或项目已被批准

所以我基本上需要选择

item.creator = owner or item.moderated = False

我如何在django中执行此操作(最好使用过滤器/查询集)

Answer 1

有些Q对象允许复杂的查找.例:

from django.db.models import Q

Item.objects.filter(Q(creator=owner) | Q(moderated=False))

Answer 2

你可以使用| operator直接组合查询集而不需要Q对象:

result = Item.objects.filter(item.creator = owner) | Item.objects.filter(item.moderated = False)

(编辑 - 我最初不确定这是否会导致额外的查询,但@spookylukey指出懒惰的查询集评估会解决这个问题)

Answer 3

值得注意的是,可以添加Q表达式.

例如:

from django.db.models import Q

query = Q(first_name='mark')
query.add(Q(email='mark@test.com'), Q.OR)
query.add(Q(last_name='doe'), Q.AND)

queryset = User.objects.filter(query)

最终会得到如下查询:

(first_name = 'mark' or email = 'mark@test.com') and last_name = 'doe'

这种方式不需要处理或操作,减少等.

Answer 4

你想让过滤器动态,那么你必须使用Lambda

from django.db.models import Q

brands = ['ABC','DEF' , 'GHI']

queryset = Product.objects.filter(reduce(lambda x, y: x | y, [Q(brand=item) for item in brands]))

reduce(lambda x, y: x | y, [Q(brand=item) for item in brands]) 相当于

Q(brand=brands[0]) | Q(brand=brands[1]) | Q(brand=brands[2]) | .....

Answer 5

有多种方法可以做到这一点。

1.直接使用管道| 操作员。

from django.db.models import Q

Items.objects.filter(Q(field1=value) | Q(field2=value))

2.使用__or__方法。

Items.objects.filter(Q(field1=value).__or__(field2=value))

3. 通过更改默认操作。（注意重置默认行为）

Q.default = Q.OR # Not recommended (Q.AND is default behaviour)
Items.objects.filter(Q(field1=value, field2=value))
Q.default = Q.AND # Reset after use.

4. 通过使用Q类参数_connector。

logic = Q(field1=value, field2=value, field3=value, _connector=Q.OR)
Item.objects.filter(logic)

Q 实施快照

class Q(tree.Node):
    """
    Encapsulate filters as objects that can then be combined logically (using
    `&` and `|`).
    """
    # Connection types
    AND = 'AND'
    OR = 'OR'
    default = AND
    conditional = True

    def __init__(self, *args, _connector=None, _negated=False, **kwargs):
        super().__init__(children=[*args, *sorted(kwargs.items())], connector=_connector, negated=_negated)

    def _combine(self, other, conn):
        if not(isinstance(other, Q) or getattr(other, 'conditional', False) is True):
            raise TypeError(other)

        if not self:
            return other.copy() if hasattr(other, 'copy') else copy.copy(other)
        elif isinstance(other, Q) and not other:
            _, args, kwargs = self.deconstruct()
            return type(self)(*args, **kwargs)

        obj = type(self)()
        obj.connector = conn
        obj.add(self, conn)
        obj.add(other, conn)
        return obj

    def __or__(self, other):
        return self._combine(other, self.OR)

    def __and__(self, other):
        return self._combine(other, self.AND)
    .............

参考号 Q 实施

Answer 6

类似于旧的answera,但有点简单,没有lambda:

filter_kwargs = {
    'field_a': 123,
    'field_b__in': (3, 4, 5, ),
}

要使用OR以下方法过滤这两个条件

Item.objects.filter(Q(field_a=123) | Q(field_b__in=(3, 4, 5, ))

要以编程方式获得相同的结果:

list_of_Q = [Q(**{key: val}) for key, val in filter_kwargs.items()]
Item.objects.filter(reduce(operator.or_, list_of_Q))

(为清楚起见,这里分为两行)

operator在标准库中:import operator
来自docstring:

or_(a,b) - 与|相同 湾

对于Python3,reduce不在标准库中: reduce

PS

不要忘记确保from functools import reduce不是空的 - list_of_Q将在空列表中阻塞,它至少需要一个元素.