Question

我正在动态构建一些Django过滤器查询using this example：

kwargs = { 'deleted_datetime__isnull': True }
args = ( Q( title__icontains = 'Foo' ) | Q( title__icontains = 'Bar' ) )
entries = Entry.objects.filter( *args, **kwargs )

我只是不确定如何构建args的条目。说我有这个数组：

strings = ['Foo', 'Bar']

我如何从那里到：

args = ( Q( title__icontains = 'Foo' ) | Q( title__icontains = 'Bar' )

我能得到的最接近的是：

for s in strings:
    q_construct = Q( title__icontains = %s) % s
    args.append(s)

但我不知道如何设置|条件。

Answer 1

你有Q类对象的列表，

args_list = [Q1,Q2,Q3]   # Q1 = Q(title__icontains='Foo') or Q1 = Q(**{'title':'value'})  
args = Q()  #defining args as empty Q class object to handle empty args_list
for each_args in args_list :
    args = args | each_args

query_set= query_set.filter(*(args,) ) # will excute, query_set.filter(Q1 | Q2 | Q3)
# comma , in last after args is mandatory to pass as args here

Answer 2

您可以使用kwarg格式直接迭代它（我不知道正确的术语）

argument_list = [] #keep this blank, just decalring it for later
fields = ('title') #any fields in your model you'd like to search against
query_string = 'Foo Bar' #search terms, you'll probably populate this from some source

for query in query_string.split(' '):  #breaks query_string into 'Foo' and 'Bar'
    for field in fields:
        argument_list.append( Q(**{field+'__icontains':query_object} ) ) 

query = Entry.objects.filter( reduce(operator.or_, argument_list) )

# --UPDATE-- here's an args example for completeness

order = ['publish_date','title'] #create a list, possibly from GET or POST data
ordered_query = query.order_by(*orders()) # Yay, you're ordered now!

这将在query_string中的每个字段fields中查找每个字符串，并在结果中查找

我希望我仍然拥有我的原始资源，但这是根据我使用的代码改编的。

Answer 3

firstQ = [
    Q(...),
    Q(...),
    Q(...)
]
import functools
functools.reduce(lambda a, b: a & b, Qrelationship)

或者就我而言，我需要对不同的过滤器集进行“与”操作：

firstQ = [
    Q(...),
    Q(...),
    Q(...)
]
secondQ = [
    Q(...),
    Q(...),
    Q(...)
]
import functools
combined = functools.reduce(lambda a, b: a | b, [
    functools.reduce(lambda a, b: a & b, firstQ),
    functools.reduce(lambda a, b: a & b, secondQ)
])
myqueryset = Model.objects.filter(combined)
# Make sure you apply the Q's first (BEFORE any other filter) or it will fail silently

使用args和kwargs动态构造Django过滤器查询

3 个答案: