问题描述
有没有办法获取django查询表达式的列表(例如 Q(first_name =Jordan)
,其中 Q
是 django.db.models.Q
)和按位或它们在一起? 换句话说我有这样的东西:
search_string =各种搜索词
/ pre>
我想这样做:
search_params = [Q(description__icontains = term)for re.split(r'\W',search_string)]
search_params = something_magical(search_params)
results = Record.objects.filter(search_params)
其中
search_params
现在相当于Q(description__icontains =各种)| Q(description__icontains =search| Q(description__icontains =terms
我知道这可能是这样的功能: / p>
def something_magical(lst):
result = lst [0]
for l in lst [1: ]
result | = l
return result
所以我想知道这个功能已经在Python中构建了(我假设它比我的函数更优化)。
尽管我对这个应用感兴趣, m / p>
解决方案你可能想要
<$ python 3
search_params = reduce(operator.or_,search_params,Q())
这将会在< |
code> search_params ,从空的条件开始 Q()
。
Is there a way to take a list of django query expresses (e.g. Q(first_name="Jordan")
, where Q
is django.db.models.Q
) and bitwise OR them together?
In other words, I have something like this:
search_string = "various search terms"
And I want to do this:
search_params = [Q(description__icontains=term) for term in re.split(r'\W', search_string)]
search_params = something_magical(search_params)
results = Record.objects.filter(search_params)
where search_params
now is equivalent to Q(description__icontains="various") | Q(description__icontains="search" | Q(description__icontains="terms"
I know it would be possible with a function like this:
def something_magical(lst):
result = lst[0]
for l in lst[1:]
result |= l
return result
So I'm wondering if this functionality is already built into Python (and I'm assuming it's more optimized than my function here).
Although I'm interested in it for this application, I'm also just interested in it theoretically.
You probably want
import operator
from functools import reduce # Python 3
search_params = reduce(operator.or_, search_params, Q())
This will place a bit-wise or (|
) between all the items in search_params
, starting with an empty condition Q()
.
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