SQL服務器：我們會根據持續時間繼續我有一個列

-1

ID   VehicleID    Time    EngineStatus 
--   ----------  ------------------  ---------------- 
0    0001   1/11/2016 00:00    off 
0    0002   1/11/2016 00:10    off 
0    0003   1/11/2016 00:10    off 
0    0001   1/11/2016 00:10    ON 
0    0001   1/11/2016 00:11    ON 
0    0003   1/11/2016 00:20    off 
0    0002   1/11/2016 00:15    off 
0    0001   1/11/2016 00:11    off 
0    0001   1/11/2016 00:12    off 
0    0001   1/11/2016 00:20    ON 
0    0001   1/11/2016 00:25    ON

我想每一輛汽車，它的持續時間EngineStatus。我需要一個查詢導致這樣

 Start    End     Status  VehileID 
--------------- --------------   --------  ---------- 
1/11/2016 00:00 1/11/2016 00:10   OFF   0001 
1/11/2016 00:10 1/11/2016 00:11   ON    0001 
1/11/2016 00:11 1/11/2016 00:12   Off   0001 
1/11/2016 00:20 1/11/2016 00:25   ON    0001 

1/11/2016 00:10 1/11/2016 00:15   OFF   0002 

1/11/2016 00:10 1/11/2016 00:20   OFF   0003

我想，不知何故，我必須組基於數據繼續列的值，然後從各組獲得第一和最後一排。

來源

2016-01-11 Sepehr Davarnia

您能否包含您正在使用的SQL Server版本？ 2012及更高版本，包括其他選項在早期版本中不可用。 –

我*認爲*你的問題已被低估，因爲它沒有包括你到目前爲止嘗試過的例子（但很難說，因爲downvoters沒有包括評論）。 –

這是SQL服務器2008R2 謝謝你的時間提前 –

像這樣的東西應該工作：

SELECT MIN([Time]) AS [Start], MAX([Time]) AS [End], 
     EngineStatus, VehicleID 
FROM (
    SELECT VehicleID, [Time], EngineStatus, 
     ROW_NUMBER() OVER (PARTITION BY VehicleID ORDER BY [Time]) - 
     ROW_NUMBER() OVER (PARTITION BY VehicleID, EngineStatus 
          ORDER BY [Time]) AS grp 
    FROM mytable) AS t 
GROUP BY VehicleID, EngineStatus, grp

grp標識連續記錄具有相同VehicleID, EngineStatus值切片。

來源

2016-01-11 12:57:46

嘗試;

;with data_tbl as (
    select 
     [VehicleID], 
     [Time], 
     [EngineStatus], 
     dense_rank() over(partition by [VehicleID] order by [VehicleID], [Time]) rnk 
    from tbl  
) 
select 
    t1.[Time] Start, 
    t2.[Time] End, 
    t1.[EngineStatus] Status, 
    t1.[VehicleID] VehileID 
from data_tbl t1 
join data_tbl t2 
on t1.[VehicleID] = t2.VehicleID 
and t1.rnk = t2.rnk - 1 
and t1.[EngineStatus] = t2.[EngineStatus] 
order by t1.[VehicleID], t1.[Time]

來源

2016-01-11 13:26:47 Praveen

SQL服務器：我們會根據持續時間繼續我有一個列

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