max(version_date) ..... group by item.id On Monday, August 22, 2016 at 9:52:55 AM UTC+2, 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 > >
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