如何将复杂的 django 查询构建为字符串答案

【问题标题】：How to build complex django query as a string如何将复杂的 django 查询构建为字符串
【发布时间】：2014-07-31 06:46:19
【问题描述】：

我正在动态生成具有多个参数的查询字符串。我试图在我的字符串中包含对象名称（'nut'、'jam'）。查询必须是“OR”查询。我的代码如下，我得到如下所示的错误。 here、here 和 here 的解决方案对我不起作用。

from viewer.models import Model1
from django.db.models import Q
list1 = [
    {'nut' : 'peanut', 'jam' : 'blueberry'},
    {'nut' : 'almond', 'jam' : 'strawberry'}
]
query_string = ""
for x in list1:
    if len(query_string) == 0:
        query_string = "Q(nut='%s', jam='%s')" % (x["nut"], x["jam"])
    else:
        query_string = "%s | Q(nut='%s', jam='%s')" % (query_string, x["nut"], x["jam"])
print query_string # correctly prints Q(nut='peanut', jam='blueberry') | Q(nut='almond', jam='strawberry')
query_results = Model1.objects.filter(query_string)

Error: 
#truncated
File "/Library/Python/2.7/site-packages/Django-1.5.4-py2.7.egg/django/db/models/manager.py", line 155, in filter
    return self.get_query_set().filter(*args, **kwargs)
  File "/Library/Python/2.7/site-packages/Django-1.5.4-py2.7.egg/django/db/models/query.py", line 669, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/Django-1.5.4-py2.7.egg/django/db/models/query.py", line 687, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/Library/Python/2.7/site-packages/Django-1.5.4-py2.7.egg/django/db/models/sql/query.py", line 1271, in add_q
can_reuse=used_aliases, force_having=force_having)
  File "/Library/Python/2.7/site-packages/Django-1.5.4-py2.7.egg/django/db/models/sql/query.py", line 1066, in add_filter
arg, value = filter_expr
ValueError: too many values to unpack

【问题讨论】：

标签： django models django-queryset

【解决方案1】：

构造一个Q对象并在filter()中使用：

from viewer.models import Model1
from django.db.models import Q

list1 = [
    {'nut' : 'peanut', 'jam' : 'blueberry'},
    {'nut' : 'almond', 'jam' : 'strawberry'}
]

q = Q()
for x in list1:
    q.add(Q(**x), Q.OR)

query_results = Model1.objects.filter(q)

或者，您可以使用operator.or_ 加入Q 对象列表：

import operator
from viewer.models import Model1
from django.db.models import Q

list1 = [
    {'nut' : 'peanut', 'jam' : 'blueberry'},
    {'nut' : 'almond', 'jam' : 'strawberry'}
]

query_results = Model1.objects.filter(reduce(operator.or_, 
                                             [Q(**x) for x in list1]))

【讨论】：