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
>>
>>

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