Thank you all for your hints!!!
The easiest way, of course, is to use *orderby *and *limitby*. It works
fine.
But the MAX method is usefull too. I'll use it in my projects.
On Wednesday, September 12, 2012 5:09:07 PM UTC+4, Vladimir Makarov wrote:
>
> So, I need to select data from the table but
Yes, I'm glad you pointed it out.
On Wednesday, September 12, 2012 5:44:58 PM UTC-4, Niphlod wrote:
>
> @anthony, massimo: I surely trust that you know what are you doing.
> My reply was only a specification for "future references" as the title of
> the post could be found by some users and may c
@anthony, massimo: I surely trust that you know what are you doing.
My reply was only a specification for "future references" as the title of
the post could be found by some users and may come handy.
--
On a second look this assumes that no record was deleted. It does not
always select the last 10 records.
On Wednesday, 12 September 2012 11:29:04 UTC-5, Annet wrote:
>
> Hi Vladimir,
>
> Some time ago Anthony provided me with the following solution:
>
> maxID=db(db.node).select(db.node.id.max
Yes, the "max" solution was originally for a different problem that only
needed the single max value and involved only one query. In this case, the
orderby/limitby solution is the way to go.
Anthony
On Wednesday, September 12, 2012 2:36:37 PM UTC-4, Niphlod wrote:
>
> ps: methods are NOT equiva
ps: methods are NOT equivalent. They are if you have "continous" ids.
But, e.g., you remove some rows. You end up with
1,2,3,4,5,6,7,8,9,10,11,13,15,17,20.
Second method (i.e. calc max and go back by ten) leaves you with
20,17,15,13,11,10 (and takes two queries)
First method (i.e. orderby + limi
On Wednesday, September 12, 2012 12:35:41 PM UTC-4, Massimo Di Pierro wrote:
>
> Instead of this
>
> maxID=db(db.node).select(db.node.id.max()).first()['MAX(node.id)']
>
> I would do
>
> maxID=db(db.node).select(db.node.id.max()).first()[db.node.id.max()]
>
Well, that's what I really recommended:
Instead of this
maxID=db(db.node).select(db.node.id.max()).first()['MAX(node.id)']
I would do
maxID=db(db.node).select(db.node.id.max()).first()[db.node.id.max()]
to make sure it will continue work in the future. The former is
implementation dependent.
On Wednesday, 12 September 2012 11:29:04
Hi Vladimir,
Some time ago Anthony provided me with the following solution:
maxID=db(db.node).select(db.node.id.max()).first()['MAX(node.id)']
rows=db(db.node.id>=maxID-10).select(db.node.id,db.node.computedName,orderby=~db.node.id)
Kind regards,
Annet
--
if you have a SQL db then you can order them by reverse id
db().select(orderby=~db.tablename.id,limitby=(0,10))
otherwise you should have some timestamp field (as created_on in
auth.signature) and use that field.
On Wednesday, 12 September 2012 08:09:07 UTC-5, Vladimir Makarov wrote:
>
> So
10 matches
Mail list logo
