我有一个示例数据
DECLARE @Table TABLE(
ID INT,
Mon VARCHAR(10),
Dt DateTime
)
INSERT INTO @Table (ID,Mon,Dt) SELECT 1, 'Jan','2016-12-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 2, 'Feb',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 3, 'Mar',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 4, 'Apr',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 5, 'May',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 6, 'Jun',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 7, 'Jul',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 8, 'Aug',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 9, 'Sep',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 10, 'Oct',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 11, 'Nov',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 12, 'Dec',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 1, 'Jan','2017-12-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 2, 'Feb',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 3, 'Mar',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 4, 'Apr',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 5, 'May',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 6, 'Jun',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 7, 'Jul',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 8, 'Aug',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 9, 'Sep',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 10, 'Oct',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 11, 'Nov',NULL
INSERT INTO @Table (ID,Mon,Dt) SELECT 12, 'Dec',NULL
从@Table
中选择*将结果作为
ID Mon Dt
1 Jan 2016-12-23 21:08:22.280
2 Feb NULL
3 Mar NULL
4 Apr NULL
5 May NULL
6 Jun NULL
7 Jul NULL
8 Aug NULL
9 Sep NULL
10 Oct NULL
11 Nov NULL
12 Dec NULL
1 Jan 2017-12-23 21:08:22.280
2 Feb NULL
3 Mar NULL
4 Apr NULL
5 May NULL
6 Jun NULL
7 Jul NULL
8 Aug NULL
9 Sep NULL
10 Oct NULL
11 Nov NULL
12 Dec NULL
我怎样才能得到这样的
id mon dt
1 Jan 2016-01-23 21:08:22.280
2 Feb 2016-02-23 21:08:22.280
3 Mar 2016-03-23 21:08:22.280
4 Apr 2016-04-23 21:08:22.280
5 May 2016-05-23 21:08:22.280
6 Jun 2016-06-23 21:08:22.280
7 Jul 2016-07-23 21:08:22.280
8 Aug 2016-08-23 21:08:22.280
9 Sep 2016-09-23 21:08:22.280
10 Oct 2016-10-23 21:08:22.280
11 Nov 2016-11-23 21:08:22.280
12 Dec 2016-12-23 21:08:22.280
1 Jan 2017-01-23 21:08:22.280
2 Feb 2017-02-23 21:08:22.280
3 Mar 2017-03-23 21:08:22.280
4 Apr 2017-04-23 21:08:22.280
5 May 2017-05-23 21:08:22.280
6 Jun 2017-06-23 21:08:22.280
7 Jul 2017-07-23 21:08:22.280
8 Aug 2017-08-23 21:08:22.280
9 Sep 2017-09-23 21:08:22.280
10 Oct 2017-10-23 21:08:22.280
11 Nov 2017-11-23 21:08:22.280
12 Dec 2017-12-23 21:08:22.280
建议我
答案 0 :(得分：4)
试试这个
SELECT ID,
Mon,
Dateadd(mm, id - 1, Min(Dateadd(mm, -Month(dt) + 1, dt))OVER(partition BY rn)) dt
FROM (SELECT Row_number()OVER(partition BY Mon ORDER BY id) rn,*
FROM @Table) a
ORDER BY dt
结果：
╔════╦═════╦═════════════════════════╗
║ ID ║ Mon ║ dt ║
╠════╬═════╬═════════════════════════╣
║ 1 ║ Jan ║ 2016-01-23 21:08:22.280 ║
║ 2 ║ Feb ║ 2016-02-23 21:08:22.280 ║
║ 3 ║ Mar ║ 2016-03-23 21:08:22.280 ║
║ 4 ║ Apr ║ 2016-04-23 21:08:22.280 ║
║ 5 ║ May ║ 2016-05-23 21:08:22.280 ║
║ 6 ║ Jun ║ 2016-06-23 21:08:22.280 ║
║ 7 ║ Jul ║ 2016-07-23 21:08:22.280 ║
║ 8 ║ Aug ║ 2016-08-23 21:08:22.280 ║
║ 9 ║ Sep ║ 2016-09-23 21:08:22.280 ║
║ 10 ║ Oct ║ 2016-10-23 21:08:22.280 ║
║ 11 ║ Nov ║ 2016-11-23 21:08:22.280 ║
║ 12 ║ Dec ║ 2016-12-23 21:08:22.280 ║
║ 1 ║ Jan ║ 2017-01-23 21:08:22.280 ║
║ 2 ║ Feb ║ 2017-02-23 21:08:22.280 ║
║ 3 ║ Mar ║ 2017-03-23 21:08:22.280 ║
║ 4 ║ Apr ║ 2017-04-23 21:08:22.280 ║
║ 5 ║ May ║ 2017-05-23 21:08:22.280 ║
║ 6 ║ Jun ║ 2017-06-23 21:08:22.280 ║
║ 7 ║ Jul ║ 2017-07-23 21:08:22.280 ║
║ 8 ║ Aug ║ 2017-08-23 21:08:22.280 ║
║ 9 ║ Sep ║ 2017-09-23 21:08:22.280 ║
║ 10 ║ Oct ║ 2017-10-23 21:08:22.280 ║
║ 11 ║ Nov ║ 2017-11-23 21:08:22.280 ║
║ 12 ║ Dec ║ 2017-12-23 21:08:22.280 ║
╚════╩═════╩═════════════════════════╝
答案 1 :(得分：2)
我想这可能是你想要的：
select id, mon,
coalesce(dt, min(dt) over ()) as dt
from @table;
这假设你要做的是用列中的最小日期/时间填充NULL值。
编辑：
修订版增加月份：
select id, mon,
coalesce(dt, dateadd(month, id - 1, min(dt) over ())) as dt
from @table;
答案 2 :(得分：1)
假设id是唯一滞后的，就像这样：
select id, mon,
coalesce(dt, dateadd(month, 1, LAG(dt) over (order by id asc))) as dt
from @table;
因为id在你的例子中并不是唯一的，所以滞后无法工作（当我做出原始评论时我没有注意到）但是，我怀疑id不是唯一的只是懒惰为什么要做例子
答案 3 :(得分：0)
declare @table table (id int, mon varchar(3), Dt datetime)
INSERT INTO @Table (ID,Mon,Dt) SELECT 1, 'Jan','2016-12-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 2, 'Feb','2016-2-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 3, 'Mar','2016-3-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 4, 'Apr','2016-4-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 5, 'May','2016-5-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 6, 'Jun','2016-6-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 7, 'Jul','2016-7-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 8, 'Aug','2016-8-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 9, 'Sep','2016-9-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 10, 'Oct','2016-10-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 11, 'Nov','2016-11-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 12, 'Dec','2016-12-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 1, 'Jan','2017-1-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 2, 'Feb','2016-2-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 3, 'Mar','2016-3-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 4, 'Apr','2016-4-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 5, 'May','2016-5-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 6, 'Jun','2016-6-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 7, 'Jul','2016-7-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 8, 'Aug','2016-8-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 9, 'Sep','2016-9-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 10, 'Oct','2016-10-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 11, 'Nov','2016-11-23 21:08:22.280'
INSERT INTO @Table (ID,Mon,Dt) SELECT 12, 'Dec','2016-12-23 21:08:22.280'
Select * from @table
--<grin />
或实际上，
Select id, mon,
Dateadd(month, mon-1, '23Jan2016 21:08:22.280')
from @table