我试图计算一件处于REPAIR状态的设备的平均周转时间。
我能够创建一个查询，其中包含每天都有快照状态的设备列表。
+-----------------+--------------+--------+----------------------+------------+------------------+
| equipmentNumber | snapshotDate | status | previousSnapshotDate | prevStatus | statusChangeFlag |
+-----------------+--------------+--------+----------------------+------------+------------------+
| 123456 | 2018-04-29 | ONHIRE | 2018-04-28 | AVAILABLE | 1 |
| 123456 | 2018-04-30 | ONHIRE | 2018-04-29 | ONHIRE | 0 |
| 123456 | 2018-05-01 | ONHIRE | 2018-04-30 | ONHIRE | 0 |
| 123456 | 2018-05-02 | REPAIR | 2018-05-01 | ONHIRE | 1 |
| 123456 | 2018-05-03 | REPAIR | 2018-05-02 | REPAIR | 0 |
| 123456 | 2018-05-04 | ONHIRE | 2018-05-03 | REPAIR | 1 |
| 654321 | 2018-04-30 | REPAIR | 2018-04-29 | AVAILABLE | 1 |
| 654321 | 2018-05-01 | REPAIR | 2018-04-30 | REPAIR | 0 |
| 654321 | 2018-05-02 | REPAIR | 2018-05-01 | REPAIR | 0 |
+-----------------+--------------+--------+----------------------+------------+------------------+
所以，在这个例子中，我们有2台设备，＆＃34; 123456＆＃34;在5/2和5/3处于2天的维修状态，＆＃34; 654321＆＃34;在4月30日，5月1日和5月2日处于REPAIR状态3天。这将是平均修复周转时间（2 + 3）/ 2 = 2.5天。
我尝试了这种算法（Detect consecutive dates ranges using SQL），但它似乎并不能满足我的需求。
答案 0 :(得分：1)
我尝试使用Incrementing ID列回答Gaps和Islands，如果不存在则创建一个，以及CREATE TABLE T1
([equipmentNumber] int, [snapshotDate] datetime, [status] varchar(6), [previousSnapshotDate] datetime, [prevStatus] varchar(9), [statusChangeFlag] int)
;
INSERT INTO T1
([equipmentNumber], [snapshotDate], [status], [previousSnapshotDate], [prevStatus], [statusChangeFlag])
VALUES
(123456, '2018-04-29 00:00:00', 'ONHIRE', '2018-04-28 00:00:00', 'AVAILABLE', 1),
(123456, '2018-04-30 00:00:00', 'ONHIRE', '2018-04-29 00:00:00', 'ONHIRE', 0),
(123456, '2018-05-01 00:00:00', 'ONHIRE', '2018-04-30 00:00:00', 'ONHIRE', 0),
(123456, '2018-05-02 00:00:00', 'REPAIR', '2018-05-01 00:00:00', 'ONHIRE', 1),
(123456, '2018-05-03 00:00:00', 'REPAIR', '2018-05-02 00:00:00', 'REPAIR', 0),
(123456, '2018-05-04 00:00:00', 'ONHIRE', '2018-05-03 00:00:00', 'REPAIR', 1),
(654321, '2018-04-30 00:00:00', 'REPAIR', '2018-04-29 00:00:00', 'AVAILABLE', 1),
(654321, '2018-05-01 00:00:00', 'REPAIR', '2018-04-30 00:00:00', 'REPAIR', 0),
(654321, '2018-05-02 00:00:00', 'REPAIR', '2018-05-01 00:00:00', 'REPAIR', 0)
;
;WITH cteX
AS(
SELECT
Id = ROW_NUMBER()OVER(ORDER BY T.equipmentNumber, T.snapshotDate)
,T.equipmentNumber
,T.snapshotDate
,T.[status]
,T.previousSnapshotDate
,T.prevStatus
,T.statusChangeFlag
FROM dbo.T1 T
),cteIsland
AS(
SELECT
Island = X.Id - ROW_NUMBER()OVER(ORDER BY X.Id)
,*
FROM cteX X
WHERE X.[status] = 'REPAIR'
)
SELECT * FROM cteIsland
窗口函数
Island请注意Island Id equipmentNumber status
3 4 123456 REPAIR
3 5 123456 REPAIR
4 7 654321 REPAIR
4 8 654321 REPAIR
4 9 654321 REPAIR
列
Island使用;WITH cteX
AS(
SELECT
Id = ROW_NUMBER()OVER(ORDER BY T.equipmentNumber, T.snapshotDate)
,T.equipmentNumber
,T.snapshotDate
,T.[status]
,T.previousSnapshotDate
,T.prevStatus
,T.statusChangeFlag
FROM dbo.T1 T
),cteIsland
AS(
SELECT
Island = X.Id - ROW_NUMBER()OVER(ORDER BY X.Id)
,*
FROM cteX X
WHERE X.[status] = 'REPAIR'
)
SELECT
AvgDuration =SUM(Totals.IslandCounts) / (COUNT(Totals.IslandCounts) * 1.0)
FROM
(
SELECT
IslandCounts = COUNT(I.Island)
,I.equipmentNumber
FROM cteIsland I
GROUP BY I.equipmentNumber
) Totals
列，您可以获得此TSQL所需的答案
AvgDuration
2.50000000000000
答案
getChildComponents(children) {
return children.length > 1 ? (
children.map((child, index) => (
<ChildComponent key={`child-component-${index}`}>
{child}
</ChildComponent>
))
) : (
<ChildComponent>{children}</ChildComponent>
);
}
答案 1 :(得分：0)
该方法应该用于识别修复期：
select equipmentNumber, min(snapshotDate), max(snapshotDate)
from (select t.*,
row_number() over (partition by equipmentNumber order by snapshotDate) as seqnum
from t
) t
where status = 'REPAIR'
group by equipmentNumber, dateadd(day, - seqnum, snapshotDate);
您可以使用子查询获得平均值：
select avg(datediff(day, minsd, maxsd) * 1.0)
from (select equipmentNumber, min(snapshotDate) as minsd, max(snapshotDate) as maxsd
from (select t.*,
row_number() over (partition by equipmentNumber order by snapshotDate) as seqnum
from t
) t
where status = 'REPAIR'
group by equipmentNumber, dateadd(day, - seqnum, snapshotDate)
) e;