you can do this:
results = list()
for row in big_long_hairy_select_with_joins_and_stuff
vals=list()
for f in rows.colnames:
vals.append(row[f.split('.')[1])
results.append(dict(id=row[rows.colnames[0].split('.')
[1]],cell=vals))
On Oct 23, 11:08 pm, BigBaaadBob <bigbaaad...@gmail.com> wrote:
> I must be expressing myself poorly. I'm saying that if the dal (and
> sql) Row class had a method like this:
>
> def column(self, colname):
> (table, field) = colname.split('.')
> return self[table][field]
>
> I could do this (which is what jqGrid wants):
>
> results = list()
> for row in big_long_hairy_select_with_joins_and_stuff
> vals=list()
> for f in rows.colnames:
> vals.append(row.column(f))
> results.append(dict(id=row.column(rows.colnames[0]),cell=vals))
>
> return dict(results=results)
>
> Am I alone in thinking this is something the Row class should know how
> to do, or is there some trivial way of doing this already?
>
> It seems weird that no one else has run into this problem...
