Question

我需要将最大日期减去上一个上一个状态日期，但无法弄明白。我将使用FindingID和UpdatedEstimatedRemediationDate。

例如：

FindingID'FND-5645'已更新3次：

UpdatedEstimatedRemediationDate
--------------------------------
NULL
2015-06-15
2015-12-30
2016-06-30

我需要从2016年6月30日到2015年12月12日获得日差。我正在使用SQL Server 2008 R2。提前谢谢。

Answer 1

如果我理解正确，这基本上是一个datediff()的聚合查询：

select findingid, datediff(day, min(UpdatedEstimatedRemediationDate), max(UpdatedEstimatedRemediationDate)
from t
group by findingid;

Answer 2

您可以使用ROW_NUMBER()按FindingId进行分区，然后按UpdateDate desc排序，选择第一个和最后一个日期，并将日期差异化为天：

<强>设定：

-- drop table UpdatedEstimatedRemediationDate
create table UpdatedEstimatedRemediationDate
(
    FindingId INT,
    UpdateDate DATE
)

insert into UpdatedEstimatedRemediationDate values 
(1, '2015-06-15'), (1, '2015-12-30'), (1, '2016-06-30'), (2, '2015-07-13'), (2, '2016-05-01')
GO

<强>查询：

;WITH Cte AS (
    SELECT FindingId, UpdateDate, ROW_NUMBER() OVER (PARTITION BY FindingId ORDER BY UpdateDate DESC) AS RowNo
    FROM UpdatedEstimatedRemediationDate
)
SELECT LU1.FindingId, DATEDIFF(day, LU1.UpdateDate, LU2.UpdateDate) AS DaysDiff
 FROM Cte LU1
    JOIN Cte LU2 ON LU2.FindingId = LU1.FindingId AND LU1.RowNo = 2 AND LU2.RowNo = 1

[没有自我加入版本]

对于SQL Server 2012，使用SELF JOIN函数可以避免LAG/LEAD：

WITH CTE AS (
    SELECT FindingId, DATEDIFF(day, UpdateDate, LEAD(UpdateDate, 1, NULL) OVER (PARTITION BY FindingId ORDER BY UpdateDate)) DayDiff,
        ROW_NUMBER() OVER (PARTITION BY FindingId ORDER BY UpdateDate DESC) RowNo 
FROM UpdatedEstimatedRemediationDate)
SELECT CTE.FindingId, CTE.DayDiff
FROM CTE 
WHERE RowNo = 2

SQL-尝试从同一列中减去日期值

2 个答案: