COUNT the same column multiple time with 2 different condition

Question

Hi everyone,
I have a small query but I don't find the answer. Can anyone help me with this one ? Look at this example:
WITH T AS (
  SELECT 'a' as Name, 1 as v FROM DUAL
  UNION ALL
  SELECT 'a' as Name, 2 as v FROM DUAL
  UNION ALL
  SELECT 'a' as Name, 2 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 1 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 1 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 2 as v FROM DUAL
) SELECT DISTINCT name,
         COUNT(DECODE(v, 1, 1, null)) OVER (PARTITION BY name ORDER BY v) as v1,
         COUNT(DECODE(v, 2, 1, null)) OVER (PARTITION BY name ORDER BY v) as v2
    FROM T
  ORDER BY 1;This returns :
NAME V1                     V2                    
a    1                      0                     
a    1                      2                     
b    2                      0                     
b    2                      1                      But what I need is only one row with the count of 1 and 2 by name. How can I achieve that without using subqueries like (the following query returns the correct output)
WITH T AS (
  SELECT 'a' as Name, 1 as v FROM DUAL
  UNION ALL
  SELECT 'a' as Name, 2 as v FROM DUAL
  UNION ALL
  SELECT 'a' as Name, 2 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 1 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 1 as v FROM DUAL
    UNION ALL
  SELECT 'b' as Name, 2 as v FROM DUAL
) SELECT DISTINCT t.name,
         (SELECT COUNT(*) FROM T z WHERE z.name = t.name and z.v = 1),
         (SELECT COUNT(*) FROM T z WHERE z.name = t.name and z.v = 2)
    FROM T t
  ORDER BY 1;The table I'm working on has millions of records and with subqueries it takes ages to executes. The one with analytical function takes less than 3 sec.
Thanks for your help!

Answer 1

Hi,
user13117585 wrote:
Hi everyone,
I have a small query but I don't find the answer. Can anyone help me with this one ? Look at this example:
WITH T AS (
SELECT 'a' as Name, 1 as v FROM DUAL
UNION ALL
SELECT 'a' as Name, 2 as v FROM DUAL
UNION ALL
SELECT 'a' as Name, 2 as v FROM DUAL
UNION ALL
SELECT 'b' as Name, 1 as v FROM DUAL
UNION ALL
SELECT 'b' as Name, 1 as v FROM DUAL
UNION ALL
SELECT 'b' as Name, 2 as v FROM DUAL
) SELECT DISTINCT name,
COUNT(DECODE(v, 1, 1, null)) OVER (PARTITION BY name ORDER BY v) as v1,
COUNT(DECODE(v, 2, 1, null)) OVER (PARTITION BY name ORDER BY v) as v2
FROM T
ORDER BY 1;But what I need is only one row with the count of 1 and 2 by name. How can I achieve that without using subqueries Use the aggregate COUNT function instead of the analytic COUNT:
SELECT   name,
          COUNT(DECODE(v, 1, 1, null))  as v1,
          COUNT(DECODE(v, 2, 1, null))  as v2
     FROM T
GROUP BY  name
ORDER BY  name;