[PATCH 2/2] retag: Properly display the final count

[Finucane, Stephen]

> On Mon, Nov 09, 2015 at 02:37:54AM +0000, Finucane, Stephen wrote:
> > > i == count cannot happen in the loop as i will vary from 0 to count -
> 1.
> > >
> > > Signed-off-by: Damien Lespiau <damien.lespiau at intel.com>
> > > ---
> > >  patchwork/management/commands/retag.py | 3 ++-
> > >  1 file changed, 2 insertions(+), 1 deletion(-)
> > >
> > > diff --git a/patchwork/management/commands/retag.py
> > > b/patchwork/management/commands/retag.py
> > > index e67d099..cb95398 100644
> > > --- a/patchwork/management/commands/retag.py
> > > +++ b/patchwork/management/commands/retag.py
> > > @@ -38,7 +38,8 @@ class Command(BaseCommand):
> > >
> > >          for i, patch in enumerate(query.iterator()):
> > >              patch.refresh_tag_counts()
> > > -            if (i % 10) == 0 or i == count:
> > > +            if (i % 10) == 0:
> >
> > What happens if count == 11?
> 
> I don't where you are going with this question. In the loop we'll print
> something out when i is 0, then 10. Out of the loop we'll print out
> 11/11. But that cannot be what you were asking, am I missing a question
> behind the question?

That's what I get for trying to be less prescriptive heh. It doesn't help
when I give the wrong value also. I meant what happens if count == 12? From
what I can see, it should have looked like this:

    if (i % 10) == 0 or i == count - 1:

i.e. every 10th iteration, or on the last iteration. However, the presence of
the print on the line after this loop handles that last iteration case making
this statement, as you say, unnecessary. Therefore:

Acked-by: Stephen Finucane <stephen.finucane at intel.com>