Вы можете использовать RANK
(или DENSE_RANK
) аналитические функции, чтобы найти максимальное значение (ы) в пределах группы:
SQL Fiddle
Oracle 11g R2 Схема установки:
CREATE TABLE table_name (id, col1, col2, col3, col4, col5, date_col) AS
SELECT 1, 1, 1, 1, 1, 1, DATE '2015-11-13' FROM DUAL
UNION ALL SELECT 2, 1, 1, 1, 1, 2, DATE '2015-11-12' FROM DUAL
UNION ALL SELECT 3, 1, 1, 1, 1, 3, DATE '2015-11-11' FROM DUAL
UNION ALL SELECT 4, 1, 1, 1, 1, 4, DATE '2015-11-13' FROM DUAL
UNION ALL SELECT 5, 1, 1, 1, 1, 5, DATE '2015-11-12' FROM DUAL
UNION ALL SELECT 5, 1, 1, 1, 1, 5, DATE '2015-11-12' FROM DUAL
UNION ALL SELECT 6, 1, 1, 1, 2, 1, DATE '2015-11-12' FROM DUAL
UNION ALL SELECT 7, 1, 1, 1, 2, 2, DATE '2015-11-13' FROM DUAL
UNION ALL SELECT 8, 1, 1, 1, 2, 3, DATE '2015-11-11' FROM DUAL
UNION ALL SELECT 9, 1, 1, 1, 2, 4, DATE '2015-11-12' FROM DUAL
UNION ALL SELECT 10, 1, 1, 1, 2, 5, DATE '2015-11-13' FROM DUAL
Запрос 1:
SELECT *
FROM (
SELECT t.*,
RANK() OVER (PARTITION BY col1, col2, col3, col4 ORDER BY date_col DESC) AS rnk
FROM table_name t
)
WHERE rnk = 1
Results:
| ID | COL1 | COL2 | COL3 | COL4 | COL5 | DATE_COL | RNK |
|----|------|------|------|------|------|----------------------------|-----|
| 1 | 1 | 1 | 1 | 1 | 1 | November, 13 2015 00:00:00 | 1 |
| 4 | 1 | 1 | 1 | 1 | 4 | November, 13 2015 00:00:00 | 1 |
| 7 | 1 | 1 | 1 | 2 | 2 | November, 13 2015 00:00:00 | 1 |
| 10 | 1 | 1 | 1 | 2 | 5 | November, 13 2015 00:00:00 | 1 |
очень подробный ответ, спасибо –