问题:如何在Django中动态组成OR查询过滤器?

从一个示例中,您可以看到一个多重或查询过滤器:

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

例如,这导致:

[<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))

From an example you can see a multiple OR query filter:

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

For example, this results in:

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

However, I want to create this query filter from a list. How to do that?

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


回答 0

您可以按以下方式链接查询:

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)

You could chain your queries as follows:

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)

回答 1

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.query

To build more complex queries there is also the option to use built in Q() object’s constants Q.OR and Q.AND together with the add() method like so:

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.query

回答 2

使用python的reduce函数编写Dave Webb答案的更短方法:

# 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)  

A shorter way of writing Dave Webb’s answer using python’s reduce function:

# 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)  

回答 3

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))
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))

回答 4

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

如果您确实需要使用动态逻辑进行查询,则可以执行以下操作(丑陋且未经测试):

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

Maybe it’s better to use sql IN statement.

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

See queryset api reference.

If you really need to make queries with dynamic logic, you can do something like this (ugly + not tested):

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

回答 5

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,但这似乎就是您要尝试的方法。

因此,您想要的是:

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

See the docs:

>>> 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([])
{}

Note that this method only works for primary key lookups, but that seems to be what you’re trying to do.

So what you want is:

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

回答 6

如果我们要以编程方式设置要查询的数据库字段:

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))

In case we want to programmatically set what db field we want to query:

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))

回答 7

使用reduceor_运算符按乘法字段进行过滤的解决方案。

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是一种新的格式字符串文字。它是在python 3.6中引入的

Solution which use reduce and or_ operators to filter by multiply fields.

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()))
)

p.s. f is a new format strings literal. It was introduced in python 3.6


回答 8

您可以使用| =运算符来使用Q对象以编程方式更新查询。

You can use the |= operator to programmatically update a query using Q objects.


回答 9

这是动态pk列表:

pk_list = qs.values_list('pk', flat=True)  # i.e [] or [1, 2, 3]

if len(pk_list) == 0:
    Article.objects.none()

else:
    q = None
    for pk in pk_list:
        if q is None:
            q = Q(pk=pk)
        else:
            q = q | Q(pk=pk)

    Article.objects.filter(q)

This one is for dynamic pk list:

pk_list = qs.values_list('pk', flat=True)  # i.e [] or [1, 2, 3]

if len(pk_list) == 0:
    Article.objects.none()

else:
    q = None
    for pk in pk_list:
        if q is None:
            q = Q(pk=pk)
        else:
            q = q | Q(pk=pk)

    Article.objects.filter(q)

回答 10

另一种选择我是不知道的直到最近- QuerySet还覆盖了&|~,等运营商。OR Q对象的其他答案是对该问题的更好解决方案,但是出于兴趣/观点的考虑,您可以执行以下操作:

id_list = [1, 2, 3]
q = Article.objects.filter(pk=id_list[0])
for i in id_list[1:]:
    q |= Article.objects.filter(pk=i)

str(q.query)将返回一个查询,其中包含WHERE子句中的所有过滤器。

Another option I wasn’t aware of until recently – QuerySet also overrides the &, |, ~, etc, operators. The other answers that OR Q objects are a better solution to this question, but for the sake of interest/argument, you can do:

id_list = [1, 2, 3]
q = Article.objects.filter(pk=id_list[0])
for i in id_list[1:]:
    q |= Article.objects.filter(pk=i)

str(q.query) will return one query with all the filters in the WHERE clause.


回答 11

对于循环:

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)

这两个都相当于 Article.objects.filter(pk__in=values)

values空着时要考虑什么很重要。许多Q()以起始值开头的答案将返回所有内容Q(pk__in=[])是一个更好的起点。这是一个始终失败的Q对象,优化器可以很好地处理它(即使对于复杂的方程式)。

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values为空时返回所有内容,则应进行AND操作~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

重要的是要记住,什么都不Q()是,不是一个总是成功的Q对象。涉及此操作的任何操作都会将其完全删除。

For loop:

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

Reduce:

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)

Both of these are equivalent to Article.objects.filter(pk__in=values)

It’s important to consider what you want when values is empty. Many answers with Q() as a starting value will return everything. Q(pk__in=[]) is a better starting value. It’s an always-failing Q object that’s handled nicely by the optimizer (even for complex equations).

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

If you want to return everything when values is empty, you should AND with ~Q(pk__in=[]) to ensure that behaviour:

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

It’s important to remember that Q() is nothing, not an always-succeeding Q object. Any operation involving it will just drop it completely.


回答 12

easy ..
from django.db.models import Q import your model args =(Q(visibility = 1)|(Q(visibility = 0)&Q(user = self.user)))#元组参数= {} #dic顺序=’create_at’限制= 10

Models.objects.filter(*args,**parameters).order_by(order)[:limit]

easy..
from django.db.models import Q import you model args = (Q(visibility=1)|(Q(visibility=0)&Q(user=self.user))) #Tuple parameters={} #dic order = ‘create_at’ limit = 10

Models.objects.filter(*args,**parameters).order_by(order)[:limit]

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