有没有一种方法可以对值进行分组，只获取验证特定条件的值？
Is there a way to group values getting only the ones that verifies certain condition?
我正在尝试在 Oracle SQL 中编写一个查询，该查询通过一些 ID 聚合值，其中我有以下 table 作为输入：
ID
SOME_DATE
RANK_POSITION
301
20211201
1
301
20211202
2
301
20211203
3
649
20211201
1
649
20211202
2
649
20211206
3
649
20211208
4
649
20211211
5
758
20211212
1
758
20211222
2
你想获得这样的东西：
ID
FIRST_IN_RANK_DATE
SECOND_IN_RANK_DATE
301
01/12/2021
02/12/2021
649
01/12/2021
02/12/2021
758
12/12/2021
22/12/2021
其中 FIRST_IN_RANK_DATE 是 RANK_POSITION 中第一个日期 ID，SECOND_IN_RANK_DATE 是 [=11] 中第二个日期=] 对于特定的 ID.
您可以使用条件聚合：
SELECT id,
MAX(CASE rank_position WHEN 1 THEN some_date END) AS first_in_rank_date,
MAX(CASE rank_position WHEN 2 THEN some_date END) AS second_in_rank_date
FROM table_name
GROUP BY id
或PIVOT:
SELECT *
FROM table_name
PIVOT (
MAX(some_date)
FOR rank_position IN (
1 AS first_in_rank_date,
2 AS second_in_rank_date
)
)
或者，从 Oracle 12，MATCH_RECOGNIZE：
SELECT *
FROM table_name
MATCH_RECOGNIZE (
PARTITION BY id
ORDER BY rank_position
MEASURES
rank1.some_date AS first_in_rank_date,
rank2.some_date AS second_in_rank_date
PATTERN ( ^ rank1 rank2 )
DEFINE
rank1 AS rank_position = 1,
rank2 AS rank_position = 2
)
其中，对于示例数据：
CREATE TABLE table_name (ID, SOME_DATE, RANK_POSITION) AS
SELECT 301, DATE '2021-12-01', 1 FROM DUAL UNION ALL
SELECT 301, DATE '2021-12-02', 2 FROM DUAL UNION ALL
SELECT 301, DATE '2021-12-03', 3 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-01', 1 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-02', 2 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-06', 3 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-08', 4 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-11', 5 FROM DUAL UNION ALL
SELECT 758, DATE '2021-12-12', 1 FROM DUAL UNION ALL
SELECT 758, DATE '2021-12-22', 2 FROM DUAL;
全部输出：
ID
FIRST_IN_RANK_DATE
SECOND_IN_RANK_DATE
301
2021-12-01 00:00:00
2021-12-02 00:00:00
649
2021-12-01 00:00:00
2021-12-02 00:00:00
758
2021-12-12 00:00:00
2021-12-22 00:00:00
db<>fiddle here
我正在尝试在 Oracle SQL 中编写一个查询，该查询通过一些 ID 聚合值，其中我有以下 table 作为输入：
| ID | SOME_DATE | RANK_POSITION |
|---|---|---|
| 301 | 20211201 | 1 |
| 301 | 20211202 | 2 |
| 301 | 20211203 | 3 |
| 649 | 20211201 | 1 |
| 649 | 20211202 | 2 |
| 649 | 20211206 | 3 |
| 649 | 20211208 | 4 |
| 649 | 20211211 | 5 |
| 758 | 20211212 | 1 |
| 758 | 20211222 | 2 |
你想获得这样的东西：
| ID | FIRST_IN_RANK_DATE | SECOND_IN_RANK_DATE |
|---|---|---|
| 301 | 01/12/2021 | 02/12/2021 |
| 649 | 01/12/2021 | 02/12/2021 |
| 758 | 12/12/2021 | 22/12/2021 |
其中 FIRST_IN_RANK_DATE 是 RANK_POSITION 中第一个日期 ID，SECOND_IN_RANK_DATE 是 [=11] 中第二个日期=] 对于特定的 ID.
您可以使用条件聚合：
SELECT id,
MAX(CASE rank_position WHEN 1 THEN some_date END) AS first_in_rank_date,
MAX(CASE rank_position WHEN 2 THEN some_date END) AS second_in_rank_date
FROM table_name
GROUP BY id
或PIVOT:
SELECT *
FROM table_name
PIVOT (
MAX(some_date)
FOR rank_position IN (
1 AS first_in_rank_date,
2 AS second_in_rank_date
)
)
或者，从 Oracle 12，MATCH_RECOGNIZE：
SELECT *
FROM table_name
MATCH_RECOGNIZE (
PARTITION BY id
ORDER BY rank_position
MEASURES
rank1.some_date AS first_in_rank_date,
rank2.some_date AS second_in_rank_date
PATTERN ( ^ rank1 rank2 )
DEFINE
rank1 AS rank_position = 1,
rank2 AS rank_position = 2
)
其中，对于示例数据：
CREATE TABLE table_name (ID, SOME_DATE, RANK_POSITION) AS
SELECT 301, DATE '2021-12-01', 1 FROM DUAL UNION ALL
SELECT 301, DATE '2021-12-02', 2 FROM DUAL UNION ALL
SELECT 301, DATE '2021-12-03', 3 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-01', 1 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-02', 2 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-06', 3 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-08', 4 FROM DUAL UNION ALL
SELECT 649, DATE '2021-12-11', 5 FROM DUAL UNION ALL
SELECT 758, DATE '2021-12-12', 1 FROM DUAL UNION ALL
SELECT 758, DATE '2021-12-22', 2 FROM DUAL;
全部输出：
ID FIRST_IN_RANK_DATE SECOND_IN_RANK_DATE 301 2021-12-01 00:00:00 2021-12-02 00:00:00 649 2021-12-01 00:00:00 2021-12-02 00:00:00 758 2021-12-12 00:00:00 2021-12-22 00:00:00
db<>fiddle here