Hi
Maybe you could try something like
db().select(db.myTable.name=='Alex', groupby=db.myTable.name)
Le mercredi 24 avril 2013 13:24:01 UTC+2, Ramos a écrit :
hello,
i have a table like
Name date
John 2013-01-01
John2013-01-01
Alex2013-02-02
Alex2013-02-03
Alex
Not sure if this can be done completely with the DAL, but doing some
post-processing in Python:
count = db.mytable.Name.count()
rows = db(db.mytable).select(db.mytable.Name, count, groupby=db.mytable.Name
)
odd_count_names = [r.mytable.Name for r in rows if r[count] % 1 == 0]
Anthony
On
Thanks
That is how i did it reading the book.
2013/4/24 Anthony abasta...@gmail.com
Not sure if this can be done completely with the DAL, but doing some
post-processing in Python:
count = db.mytable.Name.count()
rows = db(db.mytable).select(db.mytable.Name, count, groupby=db.mytable.
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