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Home/ Questions/Q 7167675
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T14:37:29+00:00 2026-05-28T14:37:29+00:00

I have two models like this: class Store(models.Model): name = models.CharField(max_length=255) class Order(models.Model): store

  • 0

I have two models like this:

class Store(models.Model):
    name = models.CharField(max_length=255)

class Order(models.Model):
    store = models.ForeignKey(Store)
    date = models.DateTimeField(auto_now_add=True)
    success = models.BooleanField()

I would like to filter all the records from the Store model whose latest order was successful i.e. success == True.

Although it looks very simple, I’m having issues figuring how I would accomplish this using the ORM query system.

Any help? Thanks.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T14:37:30+00:00Added an answer on May 28, 2026 at 2:37 pm

    My approach is this: do 2 lists, first one with (id_store, last_success_date) tuples and second one with (id_store, last_date) tuples:

    l_succ = stores.objects.filter( 
                           order__success = True 
                      ).annotate(
                           last_success=Max('order__date')
                      ).value_list (
                           'id', 'last_success'
                      )
    #l_succ = [ (1, '1/1/2011'), (2, '31/12/2010'), ... ] <-l_succ result
    
    l_last = stores.objects.annotate(
                           last_date=Max('order__date')
                      ).value_list (
                           'id', 'last_date'
                      )
    #l_last = [ (1, '1/1/2011'), (2, '3/1/2011'), ... ]   <-l_last result
    

    Then take store ids for stores that last data and last success date are equals, and you have the query:

    store_success_ids =  [ k[0] for k in l_succ if k in l_last ]
    #store_success_ids = [1, 5, ... ]          <-store_success_ids result
    #Cast l_last to dictionary to do lookups if you have a lot of stores.
    
    result = Store.objects.filter( pk__in = store_success_ids)        
    

    It seems an elegant solution, only four lines of code for a complex query (but with a simple requeriment). Disclaimer, it is not tested.

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