[Django]-Fastest way to get the first object from a queryset in django?

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.

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()

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

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.

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()

It can be like this

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

or

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

You should use django methods, like exists. Its there for you to use it.

if qs.exists():
    return qs[0]
return None

You can get the first object with these ways as shown below:

MyModel.objects.first()
MyModel.objects.all().first()
MyModel.objects.all()[0]
MyModel.objects.filter().first()
MyModel.objects.filter()[0]

In addition, you can get the last object with these ways as shown below:

MyModel.objects.last()
MyModel.objects.all().last()
MyModel.objects.filter().last()