Reg: Query group by problem

Question

Hi,
I am trying to write query for report generation.According to my client requirment I need to calculate the difference
between two date columns for "n" number of rows and i need to find the number of rows which are smaller
than 2 and number of rows which are greater than 2 and calculate the percentage for number of rows which are smaller than 2.
It must grouped by a third column
I tried like this, but it shows that date column is not a grouped .....
But I dont want to group that date column... Plz adivce....
SELECT table1.column2,
      table2.column3,
      COUNT(*),
     CASE
             WHEN TO_NUMBER (table1.DTT_column3 - table1.DTT_column4) <=2
             THEN COUNT(*)
             ELSE NULL
        END
         AS sucess_count,
     CASE
             WHEN TO_NUMBER (table1.DTT_column3 - table1.DTT_column4) >2
             THEN COUNT(*)
             ELSE NULL
        END
        AS fail_count,
     FROM
     table1,
     table2
     WHERE table1.column1 =       table2.column2
     AND table1.column4 = value1
     GROUP BY table1.column2,table2.column3;TIA,
Message was edited by: ORCL
ORCLDB

Answer 1

May be something like this.
SQL> select * from v$version;
BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.2.0 - 64bi
PL/SQL Release 10.2.0.2.0 - Production
CORE    10.2.0.2.0      Production
TNS for HPUX: Version 10.2.0.2.0 - Production
NLSRTL Version 10.2.0.2.0 - Production
  1  with t
  2  as
  3  (select 1 id, to_date('01.02.2008','dd.mm.yyyy') date_1,
  4                to_date('03.02.2008','dd.mm.yyyy') date_2 from dual union all
  5   select 1,    to_date('03.02.2008','dd.mm.yyyy'),
  6                to_date('04.02.2008','dd.mm.yyyy') from dual union all
  7   select 2,    to_date('05.02.2008','dd.mm.yyyy'),
  8                to_date('06.02.2008','dd.mm.yyyy') from dual union all
  9   select 2,    to_date('01.02.2008','dd.mm.yyyy'),
10                to_date('05.02.2008','dd.mm.yyyy') from dual
11  )
12  select count(case when to_number(date_2 - date_1) <= 2 then 1 end) diff_2,
13        count(case when to_number(date_2 - date_1) > 2 then 1 end) diff_Greater_2,
14    1 - (  count(case when to_number(date_2 - date_1) > 2 then 1 end)
15         / count(case when to_number(date_2 - date_1) <= 2 then 1 end)
16        ) percent
17  from t
18  group by rollup(id)
19* having grouping_id(id) = 1
SQL> /
    DIFF_2 DIFF_GREATER_2    PERCENT
         3              1 .666666667It will be always useful if you could post your full oracle version and a sample test data.
Regards
Raj