如何在Django中动态组合OR查询过滤器？

Question

从示例中,您可以看到多个OR查询过滤器:

Article.objects.filter(Q(pk=1) | Q(pk=2) | Q(pk=3))

例如,这会导致:

[<Article: Hello>, <Article: Goodbye>, <Article: Hello and goodbye>]

但是,我想从列表中创建此查询过滤器.怎么做？

例如 [1, 2, 3] -> Article.objects.filter(Q(pk=1) | Q(pk=2) | Q(pk=3))

Answer 1

您可以按如下方式链接查询:

values = [1,2,3]

# Turn list of values into list of Q objects
queries = [Q(pk=value) for value in values]

# Take one Q object from the list
query = queries.pop()

# Or the Q object with the ones remaining in the list
for item in queries:
    query |= item

# Query the model
Article.objects.filter(query)

Answer 2

要构建更复杂的查询,还可以选择使用内置的Q()对象的常量Q.OR和Q.AND以及add()方法,如下所示:

list = [1, 2, 3]
# it gets a bit more complicated if we want to dynamically build
# OR queries with dynamic/unknown db field keys, let's say with a list
# of db fields that can change like the following
# list_with_strings = ['dbfield1', 'dbfield2', 'dbfield3']

# init our q objects variable to use .add() on it
q_objects = Q()

# loop trough the list and create an OR condition for each item
for item in list:
    q_objects.add(Q(pk=item), Q.OR)
    # for our list_with_strings we can do the following
    # q_objects.add(Q(**{item: 1}), Q.OR)

queryset = Article.objects.filter(q_objects)

# sometimes the following is helpful for debugging (returns the SQL statement)
# print queryset.query

Answer 3

使用python的reduce函数编写Dave Webb答案的简短方法:

# For Python 3 only
from functools import reduce

values = [1,2,3]

# Turn list of values into one big Q objects  
query = reduce(lambda q,value: q|Q(pk=value), values, Q())  

# Query the model  
Article.objects.filter(query)  

Answer 4

from functools import reduce
from operator import or_
from django.db.models import Q

values = [1, 2, 3]
query = reduce(or_, (Q(pk=x) for x in values))

Answer 5

也许最好使用sql IN语句.

Article.objects.filter(id__in=[1, 2, 3])

请参阅queryset api参考.

如果你真的需要用动态逻辑进行查询,你可以做这样的事情(丑陋+未经测试):

query = Q(field=1)
for cond in (2, 3):
    query = query | Q(field=cond)
Article.objects.filter(query)

Answer 6

查看文档:

>>> Blog.objects.in_bulk([1])
{1: <Blog: Beatles Blog>}
>>> Blog.objects.in_bulk([1, 2])
{1: <Blog: Beatles Blog>, 2: <Blog: Cheddar Talk>}
>>> Blog.objects.in_bulk([])
{}

请注意,此方法仅适用于主键查找,但这似乎是您尝试执行的操作.

所以你想要的是:

Article.objects.in_bulk([1, 2, 3])

Answer 7

如果我们想以编程方式设置我们想要查询的db字段:

import operator
questions = [('question__contains', 'test'), ('question__gt', 23 )]
q_list = [Q(x) for x in questions]
Poll.objects.filter(reduce(operator.or_, q_list))

Answer 8

For循环

values = [1, 2, 3]
q = Q(pk__in=[]) # generic "always false" value
for val in values:
    q |= Q(pk=val)
Article.objects.filter(q)

减少

from functools import reduce
from operator import or_

values = [1, 2, 3]
q_objects = [Q(pk=val) for val in values]
q = reduce(or_, q_objects, Q(pk__in=[]))
Article.objects.filter(q)

这两者都相当于Article.objects.filter(pk__in=values)

为什么Q()危险

重要的是要考虑当values空的时候你想要什么。许多以Q()为起始值的答案将返回所有内容。Q(pk__in=[])是一个更好的起始值。它是一个总是失败的 Q 对象，优化器可以很好地处理它（即使对于复杂的方程）。

Article.objects.filter(Q(pk__in=[]))  # doesn't hit DB
Article.objects.filter(Q(pk=None))    # hits DB and returns nothing
Article.objects.none()                # doesn't hit DB
Article.objects.filter(Q())           # returns everything

如果您想在为空时返回所有内容values，您应该与~Q(pk__in=[])以确保该行为：

values = []
q = Q()
for val in values:
    q |= Q(pk=val)
Article.objects.filter(q)                     # everything
Article.objects.filter(q | author="Tolkien")  # only Tolkien

q &= ~Q(pk__in=[])
Article.objects.filter(q)                     # everything
Article.objects.filter(q | author="Tolkien")  # everything

Q()是什么也不是，不是一个永远成功的 Q 对象。任何涉及它的操作都会将其彻底删除。

Answer 9

在此处与reduceand or_运算符配合使用解决方案，如何通过乘法字段进行过滤。

from functools import reduce
from operator import or_
from django.db.models import Q

filters = {'field1': [1, 2], 'field2': ['value', 'other_value']}

qs = Article.objects.filter(
   reduce(or_, (Q(**{f'{k}__in': v}) for k, v in filters.items()))
)

ps f是在python3.6中添加的新格式字符串文字