Sort the Django queryset by priority

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Sort the Django queryset by priority

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#1 by (-1 votes)

3

I'm creating a search engine on my platform. I do a search for content title, description and tags inside it. Here is the code:

questions = Question.objects.filter(reduce(lambda x, y: x | y, [(Q(name__icontains=word) | Q(description__icontains=word) | Q(tags__name__icontains=word)) for word in words]))

That way, I can break the words in the text and do a pretty cool search. The problem is, I would like to sort by a priority that I create, in case I wanted to do in the following order:

Exact result

Result that begins with the word / phrase

Result that ends with the word / phrase

Result containing the word / phrase

In this way I will first display the result that is exactly what the user is looking for, then results that begin with what he searches for, finishes and contains.

But with my code I can not see a way to do this other than by doing the pure sql using case when .

django filter

asked by anonymous 14.08.2017 / 17:57

1 answer

Put directory path in a php variable Grid with auto refresh html + angularJS

score -1 · Answer 1

I found a solution but I do not know if it is the best. To tell you the truth, it should not be because I have to 2 queries to bring up a list .

I create 2 querysets, the first for the exact match and the second for the possible matches (as I'm doing in the example I put in the question). Here is the code:

from functools import reduce    

words = text.split()
q1 = Question.objects.filter(name__istartswith=text)[:1]
if q1.count() > 0:
    q1 = [q1[0]]
else:
    q1 = []

q2 = Question.objects.filter(reduce(lambda x, y: x | y, [Q(name__icontains=word) for word in words]))[:4]
if q1:
    q2.exclude(id=q1[0].id)

if q2.count() > 0:
    q2 = [q for q in q2]
else:
    q2 = []

questions = list(q1) + list(q2)

At the end it returns me the 5 results being the 1 perfect match.