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Message Posted: Wed, 06 Jun 2001 @ 21:45:36 GMT

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Thanks to all that offered ideas on how to solve today's puzzle. The solution that seemed to offer the best performance and flexibility in our case involved using derived tables along the lines of the following:

SELECT A.col1,
       sum(A.col2),
       sum(A.col2),
       sum(B.col3),
       sum(A.col4),
       sum(A.col5),
       sum(A.col6)
  FROM
   (SELECT column_name_1 col1,
           count(DISTINCT column_name_2) col2,
           sum(column_name_4) col4,
           sum(column_name_5) col5,
           sum(column_name_6) col6
      FROM table_name GROUP BY column_name_1) A,
   (SELECT column_name_1 col1,
           count(DISTINCT column_name_3) col3
      FROM table_name GROUP BY column_name_1) B
WHERE A.col1 = B.col1
GROUP BY A.col1 ;
;

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