This document doesn't mention your method or using max() http://web2py.com/books/default/chapter/29/06/the-database-abstraction-layer#sum-avg-min-max-and-len
Or I don't understand how you would do it. Could you provide greater detail on how to build that query? BR, Jason 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.
