Question

我的问题是，有一个历史记录表，它每天提取一张表并为其提供时间戳。不幸的是，过去每天都多次加载数据，而实际上并非如此。

就像：

时间戳/ id
13.07.2020 15:01 ... / 123
13.07.2020 15:02 ... / 123
13.07.2020 15:03 ... / 123
14.07.2020 15:01 ... / 123
14.07.2020 15:02 ... / 123
14.07.2020 15:03 ... / 123

并且应该像这样：

13.07.2020 15:01 ... / 123
14.07.2020 15:01 ... / 123

我正在寻找一种基于每天的第一个时间戳删除重复项的方法。

您有任何想法以这种方式删除重复项吗？

提前谢谢！

Answer 1

我建议使用CTE删除

：

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY id, CONVERT(date, ts_col) ORDER BY ts_col) rn
    FROM yourTable
)

DELETE
FROM cte
WHERE rn > 1;     -- targets all records per day except for the first one

Answer 2

如果只有两列，请使用聚合：

select id, cmin(timestamp) as timestamp
from t
group by id, convert(date, timestamp);

如果您有很多列并且想要完整的行，那么row_number()可能是最佳选择：

select t.*
from (select t.*,
             row_number() over (partition by id, convert(date, timestamp) order by timestamp) as seqnum
      from t
     ) t
where seqnum = 1;

Answer 3

您可以使用此选择来控制：

select  a.* from yourtable a
inner join
(
    select id,convert(date,[datetime]) [date], MIN([datetime]) [datetime]
    from yourtable
    group by id,convert(date,[datetime])
) b on a.id = b.id and convert(date,a.[datetime]) = b.[date] and a.[datetime] <> b.[datetime]

删除：

delete  a from yourtable a
inner join
(
    select id,convert(date,[datetime]) [date], MIN([datetime]) [datetime]
    from yourtable
    group by id,convert(date,[datetime])
) b on a.id = b.id and convert(date,a.[datetime]) = b.[date] and a.[datetime] <> b.[datetime]

SQL-根据时间戳删除重复项

3 个答案: