Update all models at once in Django

Question

I am trying to update position field for all objects in specific order at once in Django (python).

This is how I've done it now, but the problem is that it makes loads of queries.

    servers = frontend_models.Server.objects.all().order_by('-vote_count')

    i = 1
    for server in servers:
        server.last_rank = i
        server.save()
        i += 1

Is there a way to update with

Model.objects.all().order_by('some_field').update(position=some_number_that_changes_for each_object)

Thank you!

Solution

You can use the F() expression from django.db.models to do the same:

Model.objects.all().order_by('some_field').update(position=F(some_field)+1)

which will generate a single SQL query to update all of the column values, so it is efficient to the database too.

Comments

[dani herrera]

django query api don't has windows functions (I think). You should execute custom sql directly if your RDBMS has this feature.

[Daniel Roseman]

This doesn't answer the OP's question, as F() can only calculate based on the value of a field in the same row, rather than incrementing across rows as he wants.

[lprsd]

For the specific stated problem, it can be refactored to work on the same row too; no?

[Byteme]

this could work only if i could use previous row with F(same_field_of_previous_row) + 1

[Andy]

"Instances of F() act as a reference to a model field within a query. These references can then be used in query filters to compare the values of two different fields on the same model instance." Hence this wouldn't accomplish what the OP needs to do.

[Skylude]

Also we should have some quotes inside that F('some_field') just to be really explicit.