如何在Django中动态编写OR查询过滤器？
如何在Django中动态编写OR查询过滤器？
提问于 2009-05-12 12:08:51
从一个示例中，您可以看到一个多OR查询过滤器：
Article.objects.filter(Q(pk=1) | Q(pk=2) | Q(pk=3))例如，这将导致：
[<Article: Hello>, <Article: Goodbye>, <Article: Hello and goodbye>]但是，我希望从列表创建此查询筛选器。如何做到这一点？
例如[1, 2, 3] -> Article.objects.filter(Q(pk=1) | Q(pk=2) | Q(pk=3))
回答 12
Stack Overflow用户
回答已采纳
发布于 2009-05-12 12:21:28
您可以按如下方式链接您的查询：
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)Stack Overflow用户
发布于 2015-03-20 00:26:08
要构建更复杂的查询，还可以选择使用内置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(id__in=[])
# 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.queryStack Overflow用户
发布于 2009-05-22 13:36:13
使用python's reduce function编写Dave Webb答案的一种更简单的方法
# 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) 页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/852414
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