SQL 按日期和顺序分区

Question

下面是我的数据。

Date	Item	Location	Seqn
2/25/2022 17:00	A1234	L1	2
3/4/2022 18:00	A1234	L1	1
4/7/2022 3:00	A1234	L1	2
4/7/2022 3:00	A1234	L1	4
4/7/2022 3:00	A1234	L1	6
4/8/2022 11:00	A1234	L1	1
4/13/2022 8:00	A1234	L1	2
4/16/2022 17:00	A1234	L1	2
4/29/2022 15:00	A1234	L1	1
5/10/2022 5:00	A1234	L1	2
5/10/2022 5:00	A1234	L1	4
5/10/2022 5:00	A1234	L1	6
5/10/2022 5:00	A1234	L1	8
5/10/2022 5:00	A1234	L1	10
5/19/2022 4:00	A1234	L1	1

我需要根据日期和顺序使用分区依据和 row_number 进行分隔。因为我想获得每个日期的第一条记录。 table 下方将是上述示例的结果：

Date	Item	Location	Seqn	Row_number
2/25/2022 17:00	A1234	L1	2	1
3/4/2022 18:00	A1234	L1	1	1
4/7/2022 3:00	A1234	L1	2	1
4/7/2022 3:00	A1234	L1	4	2
4/7/2022 3:00	A1234	L1	6	3
4/8/2022 11:00	A1234	L1	1	1

这是我的查询：

select * from (
WITH test AS (
SELECT 
    row_number() OVER(
        PARTITION BY  Item, Location, Date, Seqn
        ORDER by date, Seqn
    ) row_num,
    date,
    item,
    location,
    seqn
FROM 
    table1
)
SELECT * FROM test WHERE row_num = 1
)T1
where
item='A1234'
and location='L1'
Order by 
trdt,
seqn

我已经弄明白这两天了。请帮助

Answer 1

我认为你缺少的是 date 列包含时间部分，你需要在 PARTITION BY.

中使用它之前删除它

WITH
  sorted AS
(
  SELECT 
    row_number() OVER (
      PARTITION BY item, location, seqn, TRUNC(date)
          ORDER BY date
    )
      AS row_num,
    date,
    item,
    location,
    seqn
  FROM 
    table1
)
SELECT
  *
FROM
  sorted
WHERE
      row_num  = 1
  AND item     = 'A1234'
  AND location = 'L1'
ORDER BY 
  location,
  item,
  seqn,
  date

Answer 2

WITH
  sorted AS
(
  SELECT 
    row_number() OVER (
      PARTITION BY item, location, TRUNC(date)
          ORDER BY date,seqn
    )
      AS row_num,
    item,
    location,
    seqn
  FROM 
    table1
)
SELECT
  *
FROM
  sorted
WHERE row_num  = 1

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