这篇文章与我的问题非常相似: SQL Server : how many days each item was in each state
但是我没有列修订来查看以前的状态,而且我想获得状态的全部时间,我b
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我想知道一件物品一般处于一种状态多长时间,我的表格如下:
ID DATE STATUS
3D56B7B1-FCB3-4897-BAEB-004796E0DC8D 2016-04-05 11:30:00.000 1
3D56B7B1-FCB3-4897-BAEB-004796E0DC8D 2016-04-08 11:30:00.000 13
274C5DA9-9C38-4A54-A697-009933BB7B7F 2016-04-29 08:00:00.000 5
274C5DA9-9C38-4A54-A697-009933BB7B7F 2016-05-04 08:00:00.000 4
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 2016-04-14 07:50:00.000 1
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 2016-04-21 14:00:00.000 2
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 2016-04-23 12:15:00.000 3
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 2016-04-23 16:15:00.000 1
BF122AE1-CB39-4967-8F37-012DC55E92A7 2016-04-05 10:30:00.000 1
BF122AE1-CB39-4967-8F37-012DC55E92A7 2016-04-20 17:00:00.000 5
我想得到这个
Column 1 : ID Column 2 : Status Column 3 : Time with the status
第3列:状态的时间 = NextDate - PreviosDate + 1
我应该是这样的:
ID STATUS TIME
3D56B7B1-FCB3-4897-BAEB-004796E0DC8D 1 3
3D56B7B1-FCB3-4897-BAEB-004796E0DC8D 13 1
274C5DA9-9C38-4A54-A697-009933BB7B7F 5 5
274C5DA9-9C38-4A54-A697-009933BB7B7F 4 1
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 1 8
A70A66DC-9D9E-49BE-93CF-00F9E3E06CE2 2 2
BF122AE1-CB39-4967-8F37-012DC55E92A7 1 15
BF122AE1-CB39-4967-8F37-012DC55E92A 5 1
答案 0 :(得分:1)
感谢@ConradFrix评论,这是有效的方法..
WITH CTE
AS
(
SELECT
ID,
STATUS,
DATE,
LEAD(DATE, 1) over (partition by ID order by DATE) LEAD,
ISNULL(DATEDIFF(DAYOFYEAR, DATE,
LEAD(DATE, 1) over (partition by ID order by DATE)), 1) DIF_BY_LEAD
FROM TABLE_NAME
)
SELECT ID, STATUS, SUM(DIF_BY_LEAD) AS TIME_STATUS
FROM CTE GROUP BY ID, STATUS
ORDER BY ID, STATUS