I suppose your right, but I was a little thrown, by the: max(variable_here) That was mentioned was not the solution at all, I kept looking for ways to use max as a function. My issue now, is that the group by doesn't like me getting all the tables I want in the return. BR, Jason Brower
On Thursday, August 25, 2016 at 8:36:58 AM UTC+3, Dave S wrote: > > > > On Wednesday, August 24, 2016 at 10:29:09 PM UTC-7, Encompass solutions > wrote: >> >> Does this seem sensible? It seems to work with my initial tests. >> >> latest_versions = db( (db.item.id == db.item_version.artifact_id) & >> (db.item_version.id > 0) >> >> ).select(db.item.ALL,db.item_version.version_date.max(), groupby= >> db.item.id) >> >> the .max() feature, at least what I found, was totally undocumented. >> > > Totally? > You mean, there isn't > > <URL: > http://web2py.com/books/default/chapter/29/06/the-database-abstraction-layer#sum--avg--min--max-and-len > > > > ? > > We should have an example database as part of the documentation with a >> collection of examples around it so we can all relate better. :/ >> >> > Examples in book, using it on severity of logged events. > > /dps > > >> On Monday, August 22, 2016 at 10:52:55 AM UTC+3, Encompass solutions >> wrote: >>> >>> Consider the following pseudo model. >>> >>> item >>> ->name = "string" >>> >>> version >>> ->item_id = item.id >>> ->version_date = "datetime" >>> >>> >>> While I can easily create a collection of the item with it's versions. >>> all_items = db((db.item.id > 0) & (db.version.item_id == db.item.id >>> )).select(orderby=db.item.name | db.version.version_date) >>> >>> How do get just all items with just the latest version of each item >>> without having to do this.... >>> items = [] >>> current_id = all_items.first().item.id >>> for thing in all_items: >>> if thing.item.id != current_id: >>> current_id = thing.item.id >>> items.append(thing) >>> >>> It seems a bit silly and heavy to be doing this especially since my data >>> could get quite large. I imaging the database has some way to do this, >>> just never learned how. >>> >>> Ideas on how this could be done? >>> >>> BR, >>> Jason Brower >>> >>> -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
