Crosstab query using pure SQL only

Question

Hi all,
Found a lot of threads on crosstab, but none seems to address what I need. I need to perform crosstab query using pure SQL only & the number of columns are dynamic. From a query, I obtained the below table:
Name Date Amount
Alex 2005-06-10 1000
Alex 2005-06-20 1000
Alex 2005-07-10 1000
Alex 2005-07-20 1000
Alex 2005-08-10 1000
Alex 2005-08-20 1000
John 2005-06-10 2000
John 2005-06-20 2000
John 2005-07-10 2000
John 2005-07-20 2000
John 2005-08-10 2000
John 2005-08-20 2000
And I need to transform it into:
Name 06-2005 07-2005 08-2005
Alex 2000 2000 2000
John 4000 4000 4000
Reason for the columns being dynamic is because they'll be a limit on the date ranges to select the data from. I'd have a lower & upper bound date say June-2005 to August-2005, which explains how I got the data from the above table.
Please advise.
Thanks!

Answer 1

Hi,
I couldn't resist the intellectual challenge of a pure SQL solution for a pivot table with a dynamic number of columns. As Laurent pointed out, a SQL query can only have a fixed number of columns. You can fake a dynamic number of columns, though, by selecting a single column containing data at fixed positions.
<br>
<br>
If it were me, I'd use a PL/SQL solution, but if you must have a pure SQL solution, here is an admittedly gruesome one. It shows the sum of all EMP salaries per department over a date range defined by start and end date parameters (which I've hardcoded for simplicity). Perhaps some of the techniques demonstrated may help you in your situation.
<br>
<br>
set echo off
set heading on
set linesize 100
<br>
select version from v$instance ;
<br>
set heading off
<br>
column sort_order noprint
column sal_sums format a80
<br>
select -- header row
  1        as sort_order,
  'DEPTNO' as DEPTNO ,
  sys_connect_by_path
    ( rpad
        ( to_char(month_column),
          10
      ' | '
    ) as sal_sums
from
    select
      add_months( first_month, level - 1 ) as month_column
    from
      ( select
          date '1981-01-01' as first_month,
          date '1981-03-01' as last_month,
          months_between( date '1981-03-01', date '1981-01-01' ) + 1 total_months
        from dual
    connect by level < total_months + 1
  ) months
where
  connect_by_isleaf = 1
connect by
  month_column = add_months( prior month_column, 1 )
start with
  month_column = date '1981-01-01'
union all
select -- data rows
  2 as sort_order,
  deptno,
  sys_connect_by_path( sum_sal, ' | ' ) sal_sums
from
  select
    dept_months.deptno,
    dept_months.month_column,
    rpad( to_char( nvl( sum( emp.sal ), 0 ) ), 10 ) sum_sal
  from
      select
        dept.deptno,
        reporting_months.month_column
      from
        dept,
        ( select
            add_months( first_month, level - 1 ) as month_column
          from
            ( select
                date '1981-01-01' as first_month,
                date '1981-03-01' as last_month,
                months_between( date '1981-03-01', date '1981-01-01' ) + 1 total_months
              from
                dual
          connect by level < total_months + 1
        ) reporting_months
    ) dept_months,
    emp
  where
    dept_months.deptno = emp.deptno (+) and
    dept_months.month_column = trunc( emp.hiredate (+), 'MONTH' )
  group by
    dept_months.deptno,
    dept_months.month_column
) dept_months_sal
where
  month_column = date '1981-03-01'
connect by
  deptno = prior deptno and
  month_column = add_months( prior month_column, 1 )
start with
  month_column = date '1981-01-01'
order by
  1, 2
<br>
VERSION
10.1.0.3.0
<br>
DEPTNO      | 81-01-01   | 81-02-01   | 81-03-01
10          | 0          | 0          | 0
20          | 0          | 0          | 0
30          | 0          | 2850       | 0
40          | 0          | 0          | 0
<br>
Now, if we substitute '1981-03-01' with '1981-06-01', we see 7 columns instead of 4
<br>
DEPTNO      | 81-01-01   | 81-02-01   | 81-03-01   | 81-04-01   | 81-05-01   | 81-06-01
10          | 0          | 0          | 0          | 0          | 0          | 2450
20          | 0          | 0          | 0          | 2975       | 0          | 0
30          | 0          | 2850       | 0          | 0          | 2850       | 0
40          | 0          | 0          | 0          | 0          | 0          | 0
<br>To understand the solution, start by running the innermost subquery by itself and then work your way outward.