Re: [SQLObject] Indexed access to a SelectResults

[Ian B. <ia...@co...>]

On Wed, 2003-07-16 at 08:37, Matt Goodall wrote:
> If you have code like this
> 
>     all = Entity.select('all')
>     for i in range(len(all)):
>        print all[i]
> 
> you get a whole stream of SQL calls. For a table with n rows you 
> actually get 2n+1 SQL calls: 1 initial COUNT(*) and then a COUNT(*) and 
> SELECT for each row.
> 
> It makes sense for a SelectResults objects to cache the COUNT result on 
> the first len() so that future len() calls don't go back to the database?

Huh... I don't know why it's creating lots of COUNT calls.  Is it
really?  It would, though, create a bunch of SELECTs, like SELECT blah
LIMIT 1 OFFSET 0, SELECT blah LIMIT 1 OFFSET 1, etc.

The problem is the way you're using it.  You should iterate directly
over the result, or force getting everything at once by doing list(all).
Like either:

all = list(Entity.select())
for i in range(len(all)):
    print all[i]

# or:
all = Entity.select()
for i, obj in enumerate(all):
    print obj

Of course, the second is better as you won't pull all the objects into
memory at once.  enumerate isn't defined in Python 2.2, so you can use:

from __future__ import generators

def enumerate(iterable):
    i = 0
    for obj in iterable:
        yield i, obj
        i += 1