Getting DISTINCT count from two different columns

Question

Hi all,
I have following query which gives currency code from two different tables. I would like to get the distinct count of currency codes from these two different columns.
SELECT eb.person_seq_id, eb.bonus_amount, eb.currency_cd, ed.currency_cd_host
FROM fr_emp_bonuses eb, fr_emp_details ed, fr_periods p
WHERE eb.person_seq_id = ed.person_seq_id AND ed.period_seq_id = eb.period_seq_id
AND ed.period_seq_id = p.period_seq_id AND p.period_status = 'CURRENT'
AND eb.bonus_amount >= 0 AND eb.person_seq_id = 3525125;
This query gives following result
3525125     240000     USD     INR
3525125     0      USD     INR
3525125     60000      USD     INR
3525125     50000      USD     INR
There are two distinct currency codes (USD, INR) and total amount is 350000. So I am looking for a query to give me the following result
3525125     350000 2
Thanks in advance

Answer 1

Hi,
Here's one way:
WITH     original_query     AS
     SELECT  eb.person_seq_id
     ,     eb.bonus_amount
     ,     eb.currency_cd
     ,     ed.currency_cd_host
     FROM     fr_emp_bonuses         eb
     ,     fr_emp_details          ed
     ,     fr_periods          p
     WHERE      eb.person_seq_id    = ed.person_seq_id
     AND      ed.period_seq_id    = eb.period_seq_id
     AND      ed.period_seq_id    = p.period_seq_id
     AND      p.period_status         = 'CURRENT'
     AND      eb.bonus_amount     >= 0
     AND     eb.person_seq_id    = 3525125
,     unpivoted_data     AS
     SELECT     person_seq_id
     ,     bonus_amount
     ,     currency_cd
     FROM     original_query
    UNION ALL
        SELECT  person_seq_id
     ,     0               AS bonus_amount
     ,     currency_cd_host     AS currency_cd
     FROM     original_query
SELECT       person_seq_id
,       SUM (bonus_amount)          AS total_bonus_amount
,       COUNT (DISTINCT currency_cd)     AS distinct_currency_cds
FROM       unpivoted_data
GROUP BY  person_seq_id
;There may be a shorter, more efficient way to get the same results, but without knowing more about your tables, I can't tell.
The tricky thing is getting two columns (currency_cd and currencuy_cd_host in this case) counted together. You can't simply say
COUNT (DISTINCT eb.currency_cd) +
COUNT (DISTINCT ed.currency_code_host)That happens to get the correct result with the sample data you posted, but what if you had data like thEe following?
currency_cd     currency_cd_host
INR          USD
USD          INRHere, the count of distinct currency_cds is 2, and the count of distinct currency_cd_hsots is also 2. Does that mean the grand total is 2 + 2 = 4? No, the 2 codes in one column arte the same 2 codes as in the other column. We need to get both currency_cd and currency_cd_hsot into the same column, and then do COUNT (DISTINCT ...) on that combined column. A UNION, as shown above, will certainly do that, starting with your query as you posted it. The query you posted isn't necessarily the best frist step towards this result, however, so there may be a much better approach, depending on your tables.
Edited by: Frank Kulash on Feb 1, 2012 6:21 PM
Here's a slightly shorter, and probably more efficient way to get the same results:
WITH     cntr     AS
     SELECT     LEVEL     AS n
     FROM     dual
     CONNECT BY     LEVEL     <= 2
SELECT       eb.person_seq_id
,       SUM (eb.bonus_amount)          AS total-amount
,       COUNT ( DISTINCT CASE
                    WHEN  c.n = 1
                    THEN  eb.currency_cd
                    ELSE  ed.currency_cd_host
                  END
          )               AS distinct_currency_cds
FROM       fr_emp_bonuses    eb
,       fr_emp_details    ed
,       fr_periods          p
,       cntr              c
WHERE        eb.person_seq_id  = ed.person_seq_id
AND        ed.period_seq_id  = eb.period_seq_id
AND        ed.period_seq_id  = p.period_seq_id
AND        p.period_status   = 'CURRENT'
AND        eb.bonus_amount   >= 0
AND       eb.person_seq_i   = 3525125
--       NOTE: no join condition involving c; we really do want a cross-join
GROUP BY  eb.person_seq_id
;