比较并从同一个表中获取列详细信息
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【中文标题】比较并从同一个表中获取列详细信息【英文标题】：compare and fetch column detail from same table 【发布时间】：2017-05-09 09:54:08 【问题描述】：我需要一些帮助来解决以下问题。
表格数据格式
Value | MATERIAL DATE
2949.00 | 000000000430450051 | 03/11/2013 00:00:00
2949.00 | 000000000430450051 | 04/15/2013 00:00:00
2949.00 | 000000000430450051 | 05/13/2013 00:00:00
2949.00 | 000000000430450051 | 06/24/2013 00:00:00
2949.00 | 000000000430450051 | 09/03/2013 00:00:00
2949.00 | 000000000430450051 | 10/14/2013 00:00:00
2949.00 | 000000000430450051 | 11/11/2013 00:00:00
2949.00 | 000000000430450051 | 12/16/2013 00:00:00
2949.00 | 000000000430450051 | 02/24/2014 00:00:00
2425.26 | 000000000430450051 | 04/21/2014 00:00:00 <===
2425.26 | 000000000430450051 | 05/05/2014 00:00:00
2425.26 | 000000000430450051 | 08/11/2014 00:00:00
2425.26 | 000000000430450051 | 09/04/2014 00:00:00
2425.26 | 000000000430450051 | 10/13/2014 00:00:00
2425.26 | 000000000430450051 | 12/10/2014 00:00:00
2425.26 | 000000000430450051 | 02/02/2015 00:00:00
2425.26 | 000000000430450051 | 03/18/2015 00:00:00
3032.00 | 000000000430450051 | 04/13/2015 00:00:00 <===
3032.00 | 000000000430450051 | 05/26/2015 00:00:00
3032.00 | 000000000430450051 | 06/15/2015 00:00:00
3032.00 | 000000000430450051 | 07/20/2015 00:00:00
我怎样才能实现如下输出，我需要选择与值列更改对应的日期列，如下所示
2425.26 | 04/21/2014 00:00:00
3032.00 | 04/13/2015 00:00:00
【问题讨论】：
到目前为止你有什么尝试，你有没有做过任何研究，你有什么想法？我们不是来为您工作，而是帮助您自己找到答案。 This question 可能会帮助您删除中心列，而 this 可能会帮助您检查值是否已更改。 【参考方案1】：您可以使用滞后并获取下一个值并找到如下变化
select * from (
select *, NextValue = lag(value,1,null) over (order by date) from #yourMaterial
) a
where value <> nextvalue
你的输出：
+---------+--------------------+-------------------------+
| value | material | date |
+---------+--------------------+-------------------------+
| 2425.26 | 000000000430450051 | 2014-04-21 00:00:00.000 |
| 3032 | 000000000430450051 | 2015-04-13 00:00:00.000 |
+---------+--------------------+-------------------------+
【讨论】：
【参考方案2】：您可以对值和材料列进行分组，并在选择语句中获取 MIN 日期
SELECT Value,
Material,
MIN(Date)
FROM [YourTable]
GROUP BY Value,
Material
【讨论】：
【参考方案3】：这是完整的场景，使用 WITH 子句以提高可读性：
WITH
-- your input
input(value,material,dt) AS (
SELECT 2949.00,'000000000430450051',DATE '03/11/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '04/15/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '05/13/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '06/24/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '09/03/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '10/14/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '11/11/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '12/16/2013'
UNION ALL SELECT 2949.00,'000000000430450051',DATE '02/24/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '04/21/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '05/05/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '08/11/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '09/04/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '10/13/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '12/10/2014'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '02/02/2015'
UNION ALL SELECT 2425.26,'000000000430450051',DATE '03/18/2015'
UNION ALL SELECT 3032.00,'000000000430450051',DATE '04/13/2015'
UNION ALL SELECT 3032.00,'000000000430450051',DATE '05/26/2015'
UNION ALL SELECT 3032.00,'000000000430450051',DATE '06/15/2015'
UNION ALL SELECT 3032.00,'000000000430450051',DATE '07/20/2015'
)
-- end of your input, start "real" WITH clause - and your query - here
,
input_w_change_marker AS (
SELECT
*
, value <> LAG(value) OVER(PARTITION BY material ORDER BY dt) AS val_has_changed
FROM input
)
SELECT
value
, dt
FROM input_w_change_marker
WHERE val_has_changed
;
value |dt
2,425.26|2014-04-21
3,032.00|2015-04-13
【讨论】：
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