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)

结果是这样的查询:

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

这种方式不需要处理或运营商，减少的等。

有多种方法。

1. 直接使用管道|操作符。

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实现快照

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Q实现

这可能有用https://docs.djangoproject.com/en/dev/topics/db/queries/#spanning-multi-valued-relationships

基本上听起来就像手术室

类似于以前的答案，但更简单一点，没有lambda…

使用OR来过滤这两个条件:

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

要以编程方式得到相同的结果:

filter_kwargs = {
    'field_a': 123,
    'field_b__in': (3, 4, 5, ),
}
list_of_Q = [Q(**{key: val}) for key, val in filter_kwargs.items()]
Item.objects.filter(reduce(operator.or_, list_of_Q))

操作符在标准库:import操作符 从文档字符串:

or_(a, b)——和a | b一样。

对于Python3, reduce不再是内置的，但仍然在标准库中:from functools import reduce

P.S.

不要忘记确保list_of_Q不是空的- reduce()会阻塞在空列表中，它至少需要一个元素。

Item.objects.filter(field_name__startswith='yourkeyword')

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)

结果是这样的查询:

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

这种方式不需要处理或运营商，减少的等。