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.
We should have an example database as part of the documentation with a
collection of examples around it so we can all relate better. :/
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
>
>
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