Sorry for my delay in answer.
Anthony:
Your solution works well:
db(db.articulos.id.belongs(list_of_ids)).select()
Creating a list of ids and using belongs was the correct for me.
Thanks guys for the help.
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- http://web2py.com/book (Documentation)
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> Antony: Your solution sounds well, but, How I can build the query
> dinamically as you propose? Can you write an example?
>
I don't think you want to simply take the 11 consecutive integers preceding
the max id, as there may be gaps due to deletions. Anyway, if you did have
a list of ids, r
Here's a possibility
from operator import or_
query_list = []
for i in range(limite_bajo,maxID):
cuenta+=1
if cuenta%2==0: # only the even records are saved
query_list.append(db.articulos.id== i)
even_query = reduce(or_, query_list)
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uhm, not quite. Leonel has the "mathematically right" solution...or you
didn't explain your requirements well.
Given ids ranging from 1 to 31, i.e.
1,2,.,30,31
your requirement, i.e. "I need to get the last 11 records from my database,
but only the even" needs to fetch
11,13,15,17,19,21,2
Thanks Leonel and Anthony for your answers.
Leonel: Your solution works well if the id are consecutive (this fail if
user delete a record).
Antony: Your solution sounds well, but, How I can build the query
dinamically as you propose? Can you write an example?
Regards.
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Resources:
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If you are writing the SQL manually, you cannot have "db." in the table
names, as that is Python code, not SQL. Anyway, this is probably not the
best approach because IDs may not remain consecutive integers (e.g., if you
delete a record). Instead, maybe it would be simpler to just select that
l
This one is simpler than it looks. Heres how you could get the even numbers
from the last 11 rows in just one line:
db((db.articulos.id % 2) == 0).select(orderby=~db.articulos.id, limitby=(0,
11))
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Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web
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