Answer by Jon Heller for Nested cos() calculation in Oracle 10

The MODEL clause can solve this:

Test data:

create table test1(n number unique);insert into test1 select * from table(sys.odcinumberlist(1,2,5,10));commit;

Query:

--The last row for each N has the final coscos value.select n, coscosfrom(    --Set each value to the cos() of the previous value.    select * from    (        --Each value of N has N rows, with value rownumber from 1 to N.        select n, rownumber        from        (            --Maximum number of rows needed (the largest number in the table)            select level rownumber            from dual            connect by level <= (select max(n) from test1)        ) max_rows        cross join test1        where max_rows.rownumber <= test1.n        order by n, rownumber    ) n_to_rows    model    partition by (n)    dimension by (rownumber)    measures (0 as coscos)    (        coscos[1] = cos(0),        coscos[rownumber > 1] = cos(coscos[cv(rownumber)-1])    ))where n = rownumberorder by n;

Results:

N   COSCOS1   12   0.540302305868145   0.79348035874256610  0.73140404242251

Let the holy wars begin:

Is this query a good idea? I wouldn't run this query in production, but hopefully it is a useful demonstration that any problem can be solved with SQL.

I've seen literally thousands of hours wasted because people are afraid to use SQL. If you're heavily using a database it is foolish to not use SQL as your primary programming language. It's good to occasionally spend a few hours to test the limits of SQL. A few strange queries is a small price to pay to avoid the disastrous row-by-row processing mindset that infects many database programmers.