Oracle - 值更改的天数

Question

我需要在 Oracle 中编写查询，如下面的屏幕截图所示。任何帮助是极大的赞赏。非常感谢。瓦迪

Table 样本数据：

CREATE TABLE fee_check (
  trans_date DATE,
  fee1 NUMBER(6,3),
  fee2 NUMBER(6,3)
);

INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('18/04/2022','dd/mm/yyyy'), 0.74, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('19/04/2022','dd/mm/yyyy'), 0.75, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('20/04/2022','dd/mm/yyyy'), 0.75, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('21/04/2022','dd/mm/yyyy'), 0.73, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('22/04/2022','dd/mm/yyyy'), 0.73, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('23/04/2022','dd/mm/yyyy'), 0.73, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('24/04/2022','dd/mm/yyyy'), 0.73, 0.87);
INSERT INTO fee_check(trans_date, fee1, fee2) VALUES (to_date('25/04/2022','dd/mm/yyyy'), 0.76, 0.87);
COMMIT;

Answer 1

这就是您的问题的解决方案

SELECT MIN(trans_date) trans_date, COUNT(*) DayCount, fee1, fee2 
FROM fee_check 
GROUP BY fee1,fee2
ORDER BY trans_date

Answer 2

如果您只想计算连续的行（而不是将具有相同费用值的 non-consecutive 组聚合在一起），那么在 Oracle 12 中，您可以使用 MATCH_RECOGNIZE 来执行 row-by-row 处理：

SELECT *
FROM   fee_check
MATCH_RECOGNIZE(
  ORDER BY trans_date
  MEASURES
    FIRST(trans_date) AS trans_date,
    COUNT(trans_date) AS day_count,
    FIRST(fee1) AS fee1,
    FIRST(fee2) AS fee2
  PATTERN (same_fees+)
  DEFINE same_fees AS fee1 = FIRST(fee1) AND fee2 = FIRST(fee2)
)

或者，在早期版本中，您可以使用解析函数：

SELECT MIN(trans_date) AS trans_date,
       COUNT(*) AS day_count,
       MIN(fee1) AS fee1,
       MIN(fee2) AS fee2
FROM   (
  SELECT f.*,
         ROW_NUMBER() OVER (ORDER BY trans_date) -
           ROW_NUMBER() OVER (PARTITION BY fee1, fee2 ORDER BY trans_date) AS grp
  FROM   fee_check f
)
GROUP BY grp

其中，对于示例数据：

CREATE TABLE fee_check (trans_date, fee1, fee2) AS
SELECT DATE '2022-04-18', 0.74, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-19', 0.75, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 0.75, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 0.73, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-22', 0.73, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-23', 0.73, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-24', 0.73, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-25', 0.76, 0.87 FROM DUAL UNION ALL
SELECT DATE '2022-04-26', 0.75, 0.87 FROM DUAL;

注意：在末尾添加了一个额外的行，该行在数据集中前面具有相同的 fee1 和 fee2 值。

双输出：

TRANS_DATE	DAY_COUNT	FEE1	FEE2
2022-04-18 00:00:00	1	.74	.87
2022-04-19 00:00:00	2	.75	.87
2022-04-21 00:00:00	4	.73	.87
2022-04-25 00:00:00	1	.76	.87
2022-04-26 00:00:00	1	.75	.87

db<>fiddle here