Thanks Anthony, i understand clearly now, getcourseids was indeed a Rows object and it works now with just this line: s = db(db.item.id.belongs([c.item_id for c in getcourseids])).select(db.item .course_title)
Thanks a lot really! On Wednesday, September 21, 2016 at 8:59:05 PM UTC+2, Anthony wrote: > > On Wednesday, September 21, 2016 at 3:30:49 AM UTC-4, Meinolf wrote: >> >> OMG, Is this normal????? >> i got chills telling me it had something to do with the *For Loop*, >> played around with the LBs and UBs and the following worked: >> >> * for i in range(-1, (len(getcourseids)-1)):* >> > > First, if you want the subscripts of a list, just do range(len(thelist)). > Do not subtract 1 from the length of the list -- range already handles that > for you (it returns integers that are strictly less than the second > argument). Also, no need to start with 0, as that is the default. And don't > start with -1 -- when used as a subscript, that will simply retrieve the > *last* item in the list, not the first. Also, in Python 2, use xrange() > in for loops, which is more memory efficient (in Python 3, range has been > replaced by xrange). > > >> * c = db.item.id <http://db.item.id>==getcourseids[i].item_id* >> * s.append(db(c).select(db.item.course_title)[i])* >> > > Here it is not clear why you are subscripting the result of the select > with [i]. Presumably each query returns a single matching row, which means > you want the subscript to always be [0] -- otherwise, you will get a list > index out of range error. > > Actually, given that you don't really need the index of each element in > the list, you can instead just iterate over the list itself: > > for item in getcourseids: > > And if you did need the index values in addition to the elements, the more > typical approach in Python would be: > > for i, item in enumerate(getcourseids): > > Anyway, in this case you should not be using the above method at all, as > it is very inefficient (you are doing a separate database query for every > single item in getcourseids, when you could instead use a single query). > Villas has the right idea, but getcourseids.values() assumes getcourseids > is a dictionary -- presumably it is actually a Rows object. So, you can do: > > s = db(db.item.id.belongs([c.item_id for c in > getcourseids])).select(db.item.course_title) > > Note, the above is a Rows object. You can convert it to a list if you > really need a list, but most likely you can work with the Rows object (it > can be iterated and indexed just like a list). > > Assuming getcourseids is a Rows object and the only reason it was created > was to get the item_id values to be used in this subsequent query, you can > instead skip the creation of getcourseids and just use a nested select > <http://web2py.com/books/default/chapter/29/06/the-database-abstraction-layer#belongs> > : > > id_query = db([your query to retrieve course > ids])._select(db.your_course_table.item_id) > s = db(db.item.id.belongs(id_query)).select(db.item.course_title) > > > Anthony > -- 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 web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
