That's great. There will not be so many records, so first() should work in this case but i really appreciate the added insight, that is really useful to know.
Thank you. On Saturday, January 2, 2016 at 2:43:59 PM UTC-8, Niphlod wrote: > > I second this BUT I still want to point out that .first() is a great tool > but it works at python level. In a "people" table holding 8M results, > you'll wait a lot. Use - WHENEVER and WHEREVER possible - orderby. > > db(db.people.city == 'London').select(orderby=db.people.age, > limitby=(0,1)).first() > > will land you the youngest of London. If several people are the same > "youth" there's an "having" method (but it requires a "groupby", which > seems to counteract the "complete record in the first query" requirement ). > The DBA in me would suggest a crafted executesql query involving WINDOWING > functions (that are the only ones giving the resultset in a single query) > but I guess you're not that into database internals. > > I'd go for 2 queries or a subselect then > > db( > (db.people.city == 'London') & > (db.people.age.belongs( > db(db.people.city == 'London')._select(db.people.age.min()) > )).select() > ).select(orderby=db.people.age, limitby=(0,1)) > > On Saturday, January 2, 2016 at 3:41:49 PM UTC+1, DenesL wrote: >> >> You could do >> >> db(db.people.city == "London").select(orderby=db.people.age).first() >> >> >> >> >> >> On Saturday, January 2, 2016 at 8:12:31 AM UTC-5, UG wrote: >>> >>> Hi, >>> >>> If i have the following table >>> ID >>> fist_name >>> last_name >>> city >>> age >>> >>> >>> I use the MIN to find the lowest age >>> >>> youngest = db(db.people.city == >>> "London").select(db.people.age.min()).first() >>> >>> This will only give me the lowest age in the city. Is there a way to get >>> the complete record in the first query without having to resort to a second >>> query like >>> >>> result = db(db.people.age == youngest & db.people.city == >>> "London").select() >>> >>> Thanks >>> >> -- 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 [email protected]. For more options, visit https://groups.google.com/d/optout.
