我有以下数据结构:
表 A:
ID | RequestNumber | Date
----+-----------------+-----------
1 | 1 | 2017/09/27
2 | 1 | 2018/06/02
表 B:
RequestNumber | Serial | Date
---------------+----------+-----------
1 | 1 | 2017/09/27
1 | 2 | 2017/09/27
1 | 6 | 2018/06/03
1 | 7 | 2018/06/03
1 | 8 | 2018/06/03
正如我们所看到的,Table A
第一行最近的日期是Table B
中的 2017/09/27,距离第二行最近的日期是 2018/06/03 在Table B
所以。。。
我需要一个查询来让 Table A
的每一行都与最接近Table A
记录的Table B
的所有行(这意味着第一条记录应返回 2 条记录,第二条记录应返回 3 条记录)
预期结果将是:
ID | RequestNumber | Serial | Date
----+-----------------+----------+------------
1 | 1 | 1 | 2017/09/27
1 | 1 | 2 | 2017/09/27
2 | 1 | 6 | 2018/06/03
2 | 1 | 7 | 2018/06/03
2 | 1 | 8 | 2018/06/03
提前致谢
此查询将执行您想要的操作。它将TableA
连接到RequestNumber
上的TableB
,然后连接到 TableB
和 TableA
之间的最小DATEDIFF
值表,确保我们只得到结果中最接近的日期:
SELECT a.ID, a.RequestNumber, b.Serial, b.Date
FROM TableA a
JOIN TableB b ON b.RequestNumber = a.RequestNumber
JOIN (SELECT a.ID AS ID, MIN(ABS(DATEDIFF(day, b.Date, a.Date))) AS days
FROM TableA a
JOIN TableB b ON b.RequestNumber = a.RequestNumber
GROUP BY a.ID) c ON c.ID = a.ID AND c.days = ABS(DATEDIFF(day, b.Date, a.Date))
输出:
ID RequestNumber Serial Date
1 1 1 27/09/2017 09:30:00
1 1 2 27/09/2017 09:30:00
2 1 6 03/06/2018 09:30:00
2 1 7 03/06/2018 09:30:00
2 1 8 03/06/2018 09:30:00
在 dbfiddle 上演示
另一种可能的方法是使用 LEFT JOIN 和 DENSE_RANK()。我假设,你想要最近和大于日期:
CREATE TABLE #TableA (
ID int,
RequestNumber int,
[Date] date
)
CREATE TABLE #TableB (
RequestNumber int,
Serial int,
[Date] date
)
INSERT INTO #TableA (ID, RequestNumber, [Date])
VALUES
(1, 1, '2017-09-27'),
(2, 1, '2018-06-02')
INSERT INTO #TableB (RequestNumber, Serial, [Date])
VALUES
(1, 1, '2017-09-27'),
(1, 2, '2017-09-27'),
(1, 6, '2018-06-03'),
(1, 7, '2018-06-03'),
(1, 8, '2018-06-03'),
(1, 9, '2018-06-05'),
(1, 10, '2018-06-07')
; WITH cte AS (
SELECT
a.ID,
a.RequestNumber,
b.Serial,
b.[Date],
DENSE_RANK() OVER (PARTITION BY a.ID, a.RequestNumber ORDER BY a.ID, a.RequestNumber, b.[Date]) AS rn
FROM #TableA a
LEFT JOIN #TableB b ON (a.RequestNumber = b.RequestNumber) AND (a.[Date] <= b.[Date])
)
SELECT
ID,
RequestNumber,
Serial,
[Date]
FROM cte
WHERE rn = 1
ORDER BY ID, RequestNumber
输出:
ID RequestNumber Serial Date
1 1 1 27/09/2017 00:00:00
1 1 2 27/09/2017 00:00:00
2 1 6 03/06/2018 00:00:00
2 1 7 03/06/2018 00:00:00
2 1 8 03/06/2018 00:00:00
这将为您提供所需的内容,尽管我认为您实际上并没有为从所需输出中看到的日期正确指定较小的日期项。
Select * from table B
left join table A
on
B.requestNumber=A.requestNumber
and B.date >=A.Date;
这是使用横向连接(apply
关键字)的好地方:
select a.*, b.*
from tablea a cross apply
(select top (1) with ties b.*
from tableb b
order by abs(datediff(day, a.date, b.date))
) b;
这是一个数据库<>小提琴。