[Django]-Constructing Django filter queries dynamically with args and kwargs

by

13👍

✅

You can iterate it directly using a kwarg format (I don’t know the proper term)

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!

This will look for each string in your query_string in each field in fields and OR the result

I wish I still had my original source for this, but this is adapted from code I use.

👤j_syk

14👍

you have list of Q class objects,

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
👤Roshan

1👍

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

Or in my case, I needed to AND to different sets of filters:

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
👤Ryan

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