我有一个工作订单模型,其中包含需要工作订单的字段。要获得工作单列表以及早期需要的工作单,我会这样做:
wo = Work_Order.objects.order_by('dateWORequired')
这很好用,但只有在该字段中确实有值。如果没有必需的日期,则值为None
。然后,工作单列表中的所有None
位于顶部,然后剩余的工单按正确顺序排列。
如何在底部获取None
?
答案 0 :(得分:38)
Django 1.11将此添加为本机功能。这有点令人费解。 It is documented
只有一个字段,升序:
wo = Work_Order.objects.order_by(F('dateWORequired').asc(nulls_last=True))
使用两个字段进行排序,均为降序:
wo = Work_Order.objects.order_by(F('dateWORequired').desc(nulls_last=True), F('anotherfield').desc(nulls_last=True))
答案 1 :(得分:37)
q = q.extra(select={
'date_is_null': 'dateWORequired IS NULL',
},
order_by=['date_is_null','dateWORequired'],
)
你可能需要在order_by部分的date_is_null之前 - 但这就是你如何控制行为。
答案 2 :(得分:8)
当问到这个问题时,这是不可用的,但是自从Django 1.8以来我认为这是最好的解决方案:
from django.db.models import Coalesce, Value
long_ago = datetime.datetime(year=1980, month=1, day=1)
Work_Order.objects.order_by('dateWORequired')
MyModel.objects.annotate(date_null=
Coalesce('dateWORequired', Value(long_ago))).order_by('date_null')
Coalesce
选择第一个非空值,因此您可以创建一个值date_null
来订购,其中只有dateWORequired但null
替换为很久以前的日期。
答案 3 :(得分:4)
要求: Python 3.4,Django 10.2,PostgreSQL 9.5.4
变体1
解决方案:
class IsNull(models.Func):
template = "%(expressions)s IS NULL"
用法(总是最新的):
In [1]: a = User.polls_manager.users_as_voters()
In [4]: from django.db import models
In [5]: class IsNull(models.Func):
...: template = "%(expressions)s IS NULL"
...:
In [7]: a = a.annotate(date_latest_voting_isnull=IsNull('date_latest_voting'))
In [9]: for i in a.order_by('date_latest_voting_isnull', 'date_latest_voting'):
...: print(i.date_latest_voting)
...:
2016-07-30 01:48:11.872911+00:00
2016-08-31 13:13:47.240085+00:00
2016-09-16 00:04:23.042142+00:00
2016-09-18 19:45:54.958573+00:00
2016-09-26 07:27:34.301295+00:00
2016-10-03 14:01:08.377417+00:00
2016-10-21 16:07:42.881526+00:00
2016-10-23 11:10:02.342791+00:00
2016-10-31 04:09:03.726765+00:00
None
In [10]: for i in a.order_by('date_latest_voting_isnull', '-date_latest_voting'):
...: print(i.date_latest_voting)
...:
2016-10-31 04:09:03.726765+00:00
2016-10-23 11:10:02.342791+00:00
2016-10-21 16:07:42.881526+00:00
2016-10-03 14:01:08.377417+00:00
2016-09-26 07:27:34.301295+00:00
2016-09-18 19:45:54.958573+00:00
2016-09-16 00:04:23.042142+00:00
2016-08-31 13:13:47.240085+00:00
2016-07-30 01:48:11.872911+00:00
None
备注的
变体2
解决方案:
from django.db import models
from django.db import connections
from django.db.models.sql.compiler import SQLCompiler
class NullsLastCompiler(SQLCompiler):
# source code https://github.com/django/django/blob/master/django/db/models/sql/compiler.py
def get_order_by(self):
result = super(NullsLastCompiler, self).get_order_by()
# if result exists and backend is PostgreSQl
if result and self.connection.vendor == 'postgresql':
# modified raw SQL code to ending on NULLS LAST after ORDER BY
# more info https://www.postgresql.org/docs/9.5/static/queries-order.html
result = [
(expression, (sql + ' NULLS LAST', params, is_ref))
for expression, (sql, params, is_ref) in result
]
return result
class NullsLastQuery(models.sql.Query):
# source code https://github.com/django/django/blob/master/django/db/models/sql/query.py
def get_compiler(self, using=None, connection=None):
if using is None and connection is None:
raise ValueError("Need either using or connection")
if using:
connection = connections[using]
# return own compiler
return NullsLastCompiler(self, connection, using)
class NullsLastQuerySet(models.QuerySet):
# source code https://github.com/django/django/blob/master/django/db/models/query.py
def __init__(self, model=None, query=None, using=None, hints=None):
super(NullsLastQuerySet, self).__init__(model, query, using, hints)
# replace on own Query
self.query = query or NullsLastQuery(model)
用法:的
# instead of models.QuerySet use NullsLastQuerySet
class UserQuestionQuerySet(NullsLastQuerySet):
def users_with_date_latest_question(self):
return self.annotate(date_latest_question=models.Max('questions__created'))
#connect to a model as a manager
class User(AbstractBaseUser, PermissionsMixin):
.....
questions_manager = UserQuestionQuerySet().as_manager()
结果(总是最新的):
In [2]: qs = User.questions_manager.users_with_date_latest_question()
In [3]: for i in qs:
...: print(i.date_latest_question)
...:
None
None
None
2016-10-28 20:48:49.005593+00:00
2016-10-04 19:01:38.820993+00:00
2016-09-26 00:35:07.839646+00:00
None
2016-07-27 04:33:58.508083+00:00
2016-09-14 10:40:44.660677+00:00
None
In [4]: for i in qs.order_by('date_latest_question'):
...: print(i.date_latest_question)
...:
2016-07-27 04:33:58.508083+00:00
2016-09-14 10:40:44.660677+00:00
2016-09-26 00:35:07.839646+00:00
2016-10-04 19:01:38.820993+00:00
2016-10-28 20:48:49.005593+00:00
None
None
None
None
None
In [5]: for i in qs.order_by('-date_latest_question'):
...: print(i.date_latest_question)
...:
2016-10-28 20:48:49.005593+00:00
2016-10-04 19:01:38.820993+00:00
2016-09-26 00:35:07.839646+00:00
2016-09-14 10:40:44.660677+00:00
2016-07-27 04:33:58.508083+00:00
None
None
None
None
None
注意:
基于Django: Adding "NULLS LAST" to query和Django的源代码
全局模型的所有领域(同时有利有弊)
没有不必要的字段
缺点 - 仅在PostgreSQL上测试
答案 4 :(得分:0)
我努力使用纯Django,而不是放入SQL。
F()表达式函数可以与order_by一起使用,所以我尝试编写一种创建表达式的方法,该表达式将所有数字设置为相同的值,但是将所有NULL设置为另一个特定值。
MySQL将按照升序排序0之前的NULL,反之亦然。
这样可行:
order_by( (0 * F('field')).asc() ) # Nulls first
# or:
order_by( (0 * F('field')).desc() ) # Nulls last
然后,您可以在该表达式之前或之后将任何其他字段传递给相同的order_by调用。
我用日期尝试过,同样的事情发生了。 e.g:
SELECT 0*CURRENT_TIMESTAMP;
评估为0。