Fastest way to get the first object from a queryset in django?

Question

Use the convenience methods .first() and .last():

MyModel.objects.filter(blah=blah).first()

They both swallow the resulting exception and return None if the queryset returns no objects.

These were added in Django 1.6, which was released in Nov 2013.

You can use array slicing:

Entry.objects.all()[:1].get()

Which can be used with .filter():

Entry.objects.filter()[:1].get()

You wouldn't want to first turn it into a list because that would force a full database call of all the records. Just do the above and it will only pull the first. You could even use .order_by() to ensure you get the first you want.

Be sure to add the .get() or else you will get a QuerySet back and not an object.

r = list(qs[:1])
if r:
  return r[0]
return None

Now, in Django 1.9 you have first() method for querysets.

YourModel.objects.all().first()

This is a better way than .get() or [0] because it does not throw an exception if queryset is empty, Therafore, you don't need to check using exists()

If you plan to get first element often - you can extend QuerySet in this direction:

class FirstQuerySet(models.query.QuerySet):
    def first(self):
        return self[0]


class ManagerWithFirstQuery(models.Manager):
    def get_query_set(self):
        return FirstQuerySet(self.model)

Define model like this:

class MyModel(models.Model):
    objects = ManagerWithFirstQuery()

And use it like this:

 first_object = MyModel.objects.filter(x=100).first()

This could work as well:

def get_first_element(MyModel):
    my_query = MyModel.objects.all()
    return my_query[:1]

if it's empty, then returns an empty list, otherwise it returns the first element inside a list.

It can be like this

obj = model.objects.filter(id=emp_id)[0]

or

obj = model.objects.latest('id')