Efficient way to update multiple fields of Django model object

Question

I'm trying to update user in Django database.

Fetched data is as follows :

fetched_data = {
     'id': 1,
     'first_name': 'John',
     'last_name': 'Doe',
     'phone': '+32 12',
     'mobile_phone': '+32 13',
     'email': '[email protected]',
     'username': 'myusername'
}

I get the user with this id as follows :

old_user = User.objects.get(pk=fetched_data['id'])

If I update the user as follows :

old_user.username = fetched_data['username']
old_user.first_name = fetched_data['first_name']
......
old_user.save()

it works fine, but I do not want to do it for every record, thus I tried something like :

for fetched_data_key in fetched_data:
    old_user.fetched_data_key = fetched_data['fetched_data_key']
    //old_user[fetched_data_key] = fetched_data['fetched_data_key'] --- I tried this way to
    old_user.save()

But that doesn't work. Any idea how can I update the user without doing it for every record?

Community · Accepted Answer · 2017-05-23 11:45:46Z

73

You can update a row in the database without fetching and deserializing it; update() can do it. E.g.:

User.objects.filter(id=data['id']).update(email=data['email'], phone=data['phone'])

This will issue one SQL update statement, and is much faster than the code in your post. It will never fetch the data or waste time creating a User object.

You cannot, though, send a whole bunch of update data to the SQL database and ask it to map it to different rows in one go. If you need a massive update like that done very quickly, your best bet is probably inserting the data into a separate table and then update it form a select on that table. Django ORM does not support this, as far as I can tell.

edited May 23, 2017 at 11:45

CommunityBot

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answered Jan 19, 2017 at 14:32

9000

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3 Comments

DjangoBlockchain Over a year ago

When creating an object I know you can do something like e = User(**data). Would this work for update?

9000 Over a year ago

@DjangoBlockchain: It may, but it will defeat the purpose of this answer. The idea is to not create an ORM model object for each DB record you want to update. It really kills performance.

DjangoBlockchain Over a year ago

I ended up doing User.objects.filter(id=data['id']).update(**data), so it just updates every field instead of specifying individual fields. Sorry if my question was confusing.

Moinuddin Quadri · Accepted Answer · 2017-01-19 14:51:25Z

34

Even simpler if you use .update() method of QuerySet object as:

my_id = fetched_data.pop('id')  # remove 'id' from `fetch_data`
                                # as it is not supposed to be updated

User.objects.filter(pk=my_id).update(**fetched_data)
#          unpack the content of `dict` ^

It will unpack the content of fetched_data dict and will update the records in User object whose columns are present as key to the fetched_data dict. Since you are calling filter on pk, it will always return single record.

edited Jan 19, 2017 at 14:51

answered Jan 19, 2017 at 14:26

Moinuddin Quadri

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6 Comments

brainLoop Over a year ago

what if I need to update only one object from the queryset?

Moinuddin Quadri Over a year ago

"pk" is the unique key in your Django model. On filtering based on it will return you a QuerySet containing only one object.

brainLoop Over a year ago

the problem is I am not using 'pk'

Moinuddin Quadri Over a year ago

@brainLoop : By default "pk" references the primary key column of your Django model.

UJJAVAL SHAH Over a year ago

@Anonymous I did exactly the same as you explained. The record is updated but my updated_at (auto_now) field is not getting updated. Maybe because we are not doing Obj.save().

|

Arghya Saha · Accepted Answer · 2018-11-27 12:35:54Z

19

If you use .update then there is an issue, since it won't raise post_save or pre_save signal, so if you want to use any signal on any change of data then User.objects.filter(pk=fetched_data['id']).update(**kwargs) won't work.

So you can use setattr() to update the fields and then .save would save the row which would update the row also raise an signal.

old_user = User.objects.get(pk=fetched_data['id'])
for key, value in fetched_data.iteritems():
    setattr(old_user, key, value)
old_user.save()

answered Nov 27, 2018 at 12:35

Arghya Saha

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1 Comment

Chris Over a year ago

This is by far the best way to go. Signals are all over the place in all medium and large Django projects; you wouldn't want to tinker with them

Alex Hall · Accepted Answer · 2017-01-19 14:23:38Z

6

setattr(old_user, fetched_data_key, fetched_data['fetched_data_key'])

answered Jan 19, 2017 at 14:23

Alex Hall

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Comments

Simran Thakur · Accepted Answer · 2020-12-01 07:22:06Z

-1

Blackquotes

def update_block(request, id):

a = RentedNumberBlock.objects.get(id=id)
b = RentedNumber.objects.filter(user=request.user)
c = a.rented_number.values_list('id', flat=True)
if request.POST:
    update = RentedNumberBlock.objects.get(id=id)
    update.name = request.POST['name']
    z = dict(request.POST)['rented_number']
    y = RentedNumber.objects.filter(id__in=z)
    update.rented_number.set(z)
    update.save()
    return HttpResponseRedirect('/block/')
return render(request, "block/update.html", {"a": a , "b": b, "c": c})

answered Dec 1, 2020 at 7:22

Simran Thakur

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1 Comment

Jeanne Dark Over a year ago

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