You can also do that using the ORM instead of SQL, just instead of doing
the calls on the instanced model use a manager and do the calls on all the
objects in the database, something like this:
class Objectmanager(models.Manager):
def SumAB(self):
return sum([(obj.a + obj.b) for obj
Thanks, the best bet is the raw query, because I actually need the Sum of
the method result for all the instances in the query...
2014-11-24 3:52 GMT-05:00 monoBOT :
> You can create a model method (with no parameter) that can solve that, so
> you can call it from the
You can create a model method (with no parameter) that can solve that, so
you can call it from the templates.
something like:
class MyObject(models.Model):
a = models.IntegerField()
b = models.Integerfield()
def miOperation(self):
return self.a + self.b
def
related: https://groups.google.com/forum/#!topic/django-users/s9qgXC4TNrA
Short answer: you can't. A raw query would be your best bet(i.e. "SELECT
SUM(a + b) AS sum FROM app_X")
On Sunday, November 23, 2014 4:42:48 PM UTC-6, Jorge Andrés Vergara Ebratt
wrote:
>
> Hello everyone,
>
> Well, the
What do you mean by the sum or average of the method? As in the sum or
average of the method applied to a list of objects "X"?
On Sun, Nov 23, 2014 at 7:41 PM, Jorge Andrés Vergara Ebratt <
javebr...@gmail.com> wrote:
> Hello everyone,
>
> Well, the tittle says it all:
>
> I have a model X
>
>
Hello everyone,
Well, the tittle says it all:
I have a model X
class X(models.Model)
a = models.IntegerField()
b = models.IntegerField()
def getC(self):
return a + b
So, when I'm inside a template I can call {{x.getC}} and it gets the method
for the current instance it's in, but
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