I'm trying to build the search for a Django site I am building, and in that search, I am searching in three different models. And to get pagination on the search result list, I would like to use a generic object_list view to display the results. But to do that, I have to merge three querysets into one.
How can I do that? I've tried this:
result_list = []
page_list = Page.objects.filter(
Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term))
article_list = Article.objects.filter(
Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term) |
Q(tags__icontains=cleaned_search_term))
post_list = Post.objects.filter(
Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term) |
Q(tags__icontains=cleaned_search_term))
for x in page_list:
result_list.append(x)
for x in article_list:
result_list.append(x)
for x in post_list:
result_list.append(x)
return object_list(
request,
queryset=result_list,
template_object_name='result',
paginate_by=10,
extra_context={
'search_term': search_term},
template_name="search/result_list.html")
But this doesn't work. I get an error when I try to use that list in the generic view. The list is missing the clone attribute.
How can I merge the three lists, page_list, article_list and post_list?
Concatenating the querysets into a list is the simplest approach. If the database will be hit for all querysets anyway (e.g. because the result needs to be sorted), this won't add further cost.
from itertools import chain
result_list = list(chain(page_list, article_list, post_list))
Using itertools.chain is faster than looping each list and appending elements one by one, since itertools is implemented in C. It also consumes less memory than converting each queryset into a list before concatenating.
Now it's possible to sort the resulting list e.g. by date (as requested in hasen j's comment to another answer). The sorted() function conveniently accepts a generator and returns a list:
result_list = sorted(
chain(page_list, article_list, post_list),
key=lambda instance: instance.date_created)
If you're using Python 2.4 or later, you can use attrgetter instead of a lambda. I remember reading about it being faster, but I didn't see a noticeable speed difference for a million item list.
from operator import attrgetter
result_list = sorted(
chain(page_list, article_list, post_list),
key=attrgetter('date_created'))
Try this:
matches = pages | articles | posts
It retains all the functions of the querysets which is nice if you want to order_by or similar.
Please note: this doesn't work on querysets from two different models.
Related, for mixing querysets from the same model, or for similar fields from a few models, starting with Django 1.11 a QuerySet.union() method is also available:
union()
union(*other_qs, all=False)
New in Django 1.11. Uses SQL’s UNION operator to combine the results of two or more QuerySets. For example:
>>> qs1.union(qs2, qs3)
The UNION operator selects only distinct values by default. To allow duplicate values, use the all=True
argument.
union(), intersection(), and difference() return model instances of
the type of the first QuerySet even if the arguments are QuerySets of
other models. Passing different models works as long as the SELECT
list is the same in all QuerySets (at least the types, the names don’t
matter as long as the types in the same order).
In addition, only LIMIT, OFFSET, and ORDER BY (i.e. slicing and
order_by()) are allowed on the resulting QuerySet. Further, databases
place restrictions on what operations are allowed in the combined
queries. For example, most databases don’t allow LIMIT or OFFSET in
the combined queries.
You can use the QuerySetChain class below. When using it with Django's paginator, it should only hit the database with COUNT(*) queries for all querysets and SELECT() queries only for those querysets whose records are displayed on the current page.
Note that you need to specify template_name= if using a QuerySetChain with generic views, even if the chained querysets all use the same model.
from itertools import islice, chain
class QuerySetChain(object):
"""
Chains multiple subquerysets (possibly of different models) and behaves as
one queryset. Supports minimal methods needed for use with
django.core.paginator.
"""
def __init__(self, *subquerysets):
self.querysets = subquerysets
def count(self):
"""
Performs a .count() for all subquerysets and returns the number of
records as an integer.
"""
return sum(qs.count() for qs in self.querysets)
def _clone(self):
"Returns a clone of this queryset chain"
return self.__class__(*self.querysets)
def _all(self):
"Iterates records in all subquerysets"
return chain(*self.querysets)
def __getitem__(self, ndx):
"""
Retrieves an item or slice from the chained set of results from all
subquerysets.
"""
if type(ndx) is slice:
return list(islice(self._all(), ndx.start, ndx.stop, ndx.step or 1))
else:
return islice(self._all(), ndx, ndx+1).next()
In your example, the usage would be:
pages = Page.objects.filter(Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term))
articles = Article.objects.filter(Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term) |
Q(tags__icontains=cleaned_search_term))
posts = Post.objects.filter(Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term) |
Q(tags__icontains=cleaned_search_term))
matches = QuerySetChain(pages, articles, posts)
Then use matches with the paginator like you used result_list in your example.
The itertools module was introduced in Python 2.3, so it should be available in all Python versions Django runs on.
In case you want to chain a lot of querysets, try this:
from itertools import chain
result = list(chain(*docs))
where: docs is a list of querysets
The big downside of your current approach is its inefficiency with large search result sets, as you have to pull down the entire result set from the database each time, even though you only intend to display one page of results.
In order to only pull down the objects you actually need from the database, you have to use pagination on a QuerySet, not a list. If you do this, Django actually slices the QuerySet before the query is executed, so the SQL query will use OFFSET and LIMIT to only get the records you will actually display. But you can't do this unless you can cram your search into a single query somehow.
Given that all three of your models have title and body fields, why not use model inheritance? Just have all three models inherit from a common ancestor that has title and body, and perform the search as a single query on the ancestor model.
This can be achieved by two ways either.
1st way to do this
Use union operator for queryset | to take union of two queryset. If both queryset belongs to same model / single model than it is possible to combine querysets by using union operator.
For an instance
pagelist1 = Page.objects.filter(
Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term))
pagelist2 = Page.objects.filter(
Q(title__icontains=cleaned_search_term) |
Q(body__icontains=cleaned_search_term))
combined_list = pagelist1 | pagelist2 # this would take union of two querysets
2nd way to do this
One other way to achieve combine operation between two queryset is to use itertools chain function.
from itertools import chain
combined_results = list(chain(pagelist1, pagelist2))
You can use Union:
qs = qs1.union(qs2, qs3)
But if you want to apply order_by on the foreign models of the combined queryset... then you need to Select them beforehand this way... otherwise it won't work.
Example
qs = qs1.union(qs2.select_related("foreignModel"), qs3.select_related("foreignModel"))
qs.order_by("foreignModel__prop1")
where prop1 is a property in the foreign model.
DATE_FIELD_MAPPING = {
Model1: 'date',
Model2: 'pubdate',
}
def my_key_func(obj):
return getattr(obj, DATE_FIELD_MAPPING[type(obj)])
And then sorted(chain(Model1.objects.all(), Model2.objects.all()), key=my_key_func)
Quoted from https://groups.google.com/forum/#!topic/django-users/6wUNuJa4jVw. See Alex Gaynor
Requirements:
Django==2.0.2, django-querysetsequence==0.8
In case you want to combine querysets and still come out with a QuerySet, you might want to check out django-queryset-sequence.
But one note about it. It only takes two querysets as it's argument. But with python reduce you can always apply it to multiple querysets.
from functools import reduce
from queryset_sequence import QuerySetSequence
combined_queryset = reduce(QuerySetSequence, list_of_queryset)
And that's it. Below is a situation I ran into and how I employed list comprehension, reduce and django-queryset-sequence
from functools import reduce
from django.shortcuts import render
from queryset_sequence import QuerySetSequence
class People(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
mentor = models.ForeignKey('self', null=True, on_delete=models.SET_NULL, related_name='my_mentees')
class Book(models.Model):
name = models.CharField(max_length=20)
owner = models.ForeignKey(Student, on_delete=models.CASCADE)
# as a mentor, I want to see all the books owned by all my mentees in one view.
def mentee_books(request):
template = "my_mentee_books.html"
mentor = People.objects.get(user=request.user)
my_mentees = mentor.my_mentees.all() # returns QuerySet of all my mentees
mentee_books = reduce(QuerySetSequence, [each.book_set.all() for each in my_mentees])
return render(request, template, {'mentee_books' : mentee_books})
Here's an idea... just pull down one full page of results from each of the three and then throw out the 20 least useful ones... this eliminates the large querysets and that way you only sacrifice a little performance instead of a lot.
The best option is to use the Django built-in methods:
# Union method
result_list = page_list.union(article_list, post_list)
That will return the union of all the objects in those querysets.
If you want to get just the objects that are in the three querysets, you will love the built-in method of querysets, intersection.
# intersection method
result_list = page_list.intersection(article_list, post_list)
This will do the work without using any other libraries:
result_list = page_list | article_list | post_list
You can use "|"(bitwise or) to combine the querysets of the same model as shown below:
# "store/views.py"
from .models import Food
from django.http import HttpResponse
def test(request):
# ↓ Bitwise or
result = Food.objects.filter(name='Apple') | Food.objects.filter(name='Orange')
print(result)
return HttpResponse("Test")
Output on console:
<QuerySet [<Food: Apple>, <Food: Orange>]>
[22/Jan/2023 12:51:44] "GET /store/test/ HTTP/1.1" 200 9
And, you can use |= to add the queryset of the same model as shown below:
# "store/views.py"
from .models import Food
from django.http import HttpResponse
def test(request):
result = Food.objects.filter(name='Apple')
# ↓↓ Here
result |= Food.objects.filter(name='Orange')
print(result)
return HttpResponse("Test")
Output on console:
<QuerySet [<Food: Apple>, <Food: Orange>]>
[22/Jan/2023 12:51:44] "GET /store/test/ HTTP/1.1" 200 9
Be careful, if adding the queryset of a different model as shown below:
# "store/views.py"
from .models import Food, Drink
from django.http import HttpResponse
def test(request):
# "Food" model # "Drink" model
result = Food.objects.filter(name='Apple') | Drink.objects.filter(name='Milk')
print(result)
return HttpResponse("Test")
There is an error below:
AssertionError: Cannot combine queries on two different base models.
[22/Jan/2023 13:40:54] "GET /store/test/ HTTP/1.1" 500 96025
But, if adding the empty queryset of a different model as shown below:
# "store/views.py"
from .models import Food, Drink
from django.http import HttpResponse
def test(request):
# "Food" model # Empty queryset of "Drink" model
result = Food.objects.filter(name='Apple') | Drink.objects.none()
print(result)
return HttpResponse("Test")
There is no error below:
<QuerySet [<Food: Apple>]>
[22/Jan/2023 13:51:09] "GET /store/test/ HTTP/1.1" 200 9
Again be careful, if adding the object by get() as shown below:
# "store/views.py"
from .models import Food
from django.http import HttpResponse
def test(request):
result = Food.objects.filter(name='Apple')
# ↓↓ Object
result |= Food.objects.get(name='Orange')
print(result)
return HttpResponse("Test")
There is an error below:
AttributeError: 'Food' object has no attribute '_known_related_objects'
[22/Jan/2023 13:55:57] "GET /store/test/ HTTP/1.1" 500 95748
This recursive function concatenates array of querysets into one queryset.
def merge_query(ar):
if len(ar) ==0:
return [ar]
while len(ar)>1:
tmp=ar[0] | ar[1]
ar[0]=tmp
ar.pop(1)
return ar
suppose
class Msg(models.Model):
...
likes = models.ManyToManyField(User,...)
channelname = models.CharField(...)
Now my queryset is
queryset = Msg.objects.filter(channelname='home')
What should i do after this to get somelike
[{id:xyz,liked=true},{id:tuv,liked=true},{id:abc,liked:false}]
You can annotate an Exists() subquery using the through model of your many to many field:
from django.db.models import Exists, OuterRef
liked_subquery = Msg.likes.through.objects.filter(
msg=OuterRef('pk'), # Filter for outer queries Msg instance
user=some_user_instance # Filter for user whose like we are checking for
)
queryset = Msg.objects.filter(
channelname='home'
).annotate(
liked=Exists(liked_subquery)
).values('id', 'liked')
print(list(queryset))
I am writing queryset that will return this type
date
total_shipping_fee
2021-04-16
5,000
2021-04-17
100,000
where
class Payments(models.Model):
created = models.DateTimeField()
.....
SELECT DATE(created) from payment_payment .... group by 'created' -> outputs a correct query
My question is how to query/ or cast
Payment.objects.filter(created_range=("2021-05-14", "2021-05-14")).values('created').annotate(total_shipping_fee=Sum('total_shipping_fee'))
so that I can have queryset in above raw sql. I think that is my problem to CAST DATE(created) in django queryset. Thanks
You can work with:
from django.db.models import F, Sum
Payment.objects.filter(
created__date_range=('2021-05-14', '2021-05-14')
).values(create_date=F('created__date')).annotate(
total_shipping_fee=Sum('total_shipping_fee')
).order_by('create_date')
here we thus first takes as values the truncated date of the DateTimeField, and then we use .order_by(…) [Django-doc] to enforce grouping by that date.
The output is a QuerySet of dictionaries, each with two items: { 'create_date': …, 'total_shipping_fee': … }.
we have Project as main model, which contains 2 fields of M2M relationship.
class First(models.Model):
first_results_M2M = models.ManyToManyField(First_Results)
class Second(models.Model):
second_results_M2M = models.ManyToManyField(Second_Results)
class Project(models.Model):
project_first_M2M = models.ManyToManyField(First)
project_second_M2M = models.ManyToManyField(Second)
I m trying to count all the objects present in first_results_M2M of all the project_first_M2M objects within each Project object.
Here's the below example to count all the objects of first_results_M2M for Project object 1.
total_first_all = First_Results.objects.filter(first__project__id=1).count()
I want to render the total count of total_first_all and total_second_all in the template.
Project_Query = Project.objects.all()
for each_proj in Project_Query:
print(each_proj.total_first_all) ## should print the count the `first_resuls_M2M` for each project obj.
Please let me know how to do achieve it in more effecient/fast way besides annotate.
annotate.total_first_all=Count('project_first_M2M__first_results_M2M')
You .annotate(..) [Django-doc] your queryset, like:
from django.db.models import Count
project_query = Project.objects.annotate(
total_first_all=Count('project_first_M2M__first_results_M2M')
)
for project in project_query:
print(project.total_first_all)
This will not make a query per Project object, but calculate the counts for all Projects in "bulk".
For multiple ones, you can make use of subqueries to reduce the amount of nested JOIN:
from django.db.models import Count, OuterRef, Subquery
project_query = Project.objects.annotate(
total_first_all=Subquery(
First_Results.objects.filter(first__project=OuterRef('pk')).values('first__project').values(cnt=Count('*')).order_by('first__project')
),
total_second_all=Subquery(
Second_Results.objects.filter(second__project=OuterRef('pk')).values('second__project').values(cnt=Count('*')).order_by('second__project')
)
)
I have a list of objects how can I run a query to give the max value of a field:
I'm using this code:
def get_best_argument(self):
try:
arg = self.argument_set.order_by('-rating')[0].details
except IndexError:
return 'no posts'
return arg
rating is an integer
See this. Your code would be something like the following:
from django.db.models import Max
# Generates a "SELECT MAX..." query
Argument.objects.aggregate(Max('rating')) # {'rating__max': 5}
You can also use this on existing querysets:
from django.db.models import Max
args = Argument.objects.filter(name='foo') # or whatever arbitrary queryset
args.aggregate(Max('rating')) # {'rating__max': 5}
If you need the model instance that contains this max value, then the code you posted is probably the best way to do it:
arg = args.order_by('-rating')[0]
Note that this will error if the queryset is empty, i.e. if no arguments match the query (because the [0] part will raise an IndexError). If you want to avoid that behavior and instead simply return None in that case, use .first():
arg = args.order_by('-rating').first() # may return None
Django also has the 'latest(field_name = None)' function that finds the latest (max. value) entry. It not only works with date fields but also with strings and integers.
You can give the field name when calling that function:
max_rated_entry = YourModel.objects.latest('rating')
return max_rated_entry.details
Or you can already give that field name in your models meta data:
from django.db import models
class YourModel(models.Model):
#your class definition
class Meta:
get_latest_by = 'rating'
Now you can call 'latest()' without any parameters:
max_rated_entry = YourModel.objects.latest()
return max_rated_entry.details
I've tested this for my project, it finds the max/min in O(n) time:
from django.db.models import Max
# Find the maximum value of the rating and then get the record with that rating.
# Notice the double underscores in rating__max
max_rating = App.objects.aggregate(Max('rating'))['rating__max']
return App.objects.get(rating=max_rating)
This is guaranteed to get you one of the maximum elements efficiently, rather than sorting the whole table and getting the top (around O(n*logn)).
sol 01:
from .models import MyMODEL
max_rating = MyMODEL.objects.order_by('-rating').first()
sol 02:
from django.db.models import Max
from .models import MyMODEL
max_rating = MyMODEL.objects.aggregate(Max('rating'))
If you also want to get a value other than None in case the table is empty (e.g. 0), combine Max with Coalesce:
from django.db.models import Max, Value
from django.db.models.functions import Coalesce
max_rating = SomeModel.objects.aggregate(
max_rating=Coalesce(Max('rating'), Value(0))
)['max_rating']
To maybe improve on #afahim answer with regards to #Raydel Miranda comment, if you want a random comment. If you want all, then use just the filter
from django.db.models import Max
# Find the maximum value of the rating and then get the record with that rating.
# Notice the double underscores in rating__max
max_rating = App.objects.aggregate(Max('rating'))['rating__max']
return App.objects.filter(rating=max_rating).first()
maybe it will help someone's trouble, in CBV
def get_queryset(self):
sorgu = Sunum.objects.values('id', 'firma', 'projeadi', 'sunumdurum__durum', 'sunumdurum__aciklama'
).annotate(max_rank=Max('sunumdurum__kayittarihi'))
szlk={}
for sor in sorgu :
ana = sor['id'], sor['firma'], sor['projeadi']
dana = sor['sunumdurum__durum'], sor['sunumdurum__aciklama'], sor['max_rank']
szlk.setdefault(ana, dana)
return szlk