Question

我有这样的表

我从连续日期中得到了不同，但我想计算每次从1开始的日期差异，比如（1-2和1-3再次1-4）

ChildID   ID      Date
1         4    2005-12-07 00:00:00.000
1         3    2005-11-07 00:00:00.000
1         2    2005-10-09 00:00:00.000
1         1    2005-04-09 00:00:00.000

预期

ChildID   ID      Date                         Difference
1         4    2005-12-07 00:00:00.000            243 days (1-4)
1         3    2005-11-07 00:00:00.000            207 days (1-3)
1         2    2005-10-09 00:00:00.000            183 days (1-2)
1         1    2005-04-09 00:00:00.000            NULL

Answer 1

您可以通过简单查询

来完成

declare @temp table(childId int, id int, date datetime)

insert into @temp values(1,4,'2005-12-07 00:00:00.000')
insert into @temp values(1,3,'2005-11-07 00:00:00.000')
insert into @temp values(1,2,' 2005-10-09 00:00:00.000')
insert into @temp values(1,1,'2005-04-09 00:00:00.000')


SELECT t1.*, datediff(day, t2.date, t1.date) AS DateDifference
FROM @temp AS t1
INNER JOIN @temp AS t2 ON t1.childId = t2.id

更新：添加完整示例

1   4   2005-12-07 00:00:00.000 242
1   3   2005-11-07 00:00:00.000 212
1   2   2005-10-09 00:00:00.000 183
1   1   2005-04-09 00:00:00.000 0

结果

await result

Answer 2

使用窗口函数获取此记录的child_id的最早日期：

 SELECT datediff(day, min(date_field) OVER (PARTITION BY child_id) - date_field), child_id, id FROM your_table

这没有考虑到child_id的min（date）可能不是id=1的记录。因此，如果这是一个问题，那么子查询解决方案：

SELECT datediff(day, t2.datefield, t1.date_field), child_id, id
FROM your_table t1
    JOIN (SELECT child_id, date_field FROM your_table where id = 1) t2
         ON t1.child_id = t2.child_id;

Answer 3

1）SQL2012：

SELECT DATEDIFF(DAY, FIRST_VALUE(DATECOL) OVER(ORDER BY ID),  DATECOL) AS DIFF ...
FROM ...

2) If ID column has unique values:

SELECT DATEDIFF(DAY, (SELECT A.DATECOL FROM TABLE1 A WHERE A.ID=1), B.DATECOL) AS DIFF ...
FROM TABLE1 B ...

Answer 4

包含其他样本数据

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        return nil
    }

    if #available(iOS 11.0, *) {
        let annoView = MKMarkerAnnotationView(annotation: annotation, reuseIdentifier: "HunterAnno")
        annoView.canShowCallout = true
        return annoView
    }

    let annoView = MKPinAnnotationView(annotation: annotation, reuseIdentifier: "HunterAnnoLow")
    annoView.canShowCallout = true
    return annoView
}

输出

如果您希望最终列看起来像这样

DECLARE @tab TABLE (ChildId INT, Id INT, Date DATETIME) INSERT INTO @tab (ChildId, Id, Date) SELECT 1,4,'2005-12-07 00:00:00.000' UNION SELECT 1,3,'2005-11-07 00:00:00.000' UNION SELECT 1,2,'2005-10-09 00:00:00.000' UNION SELECT 1,1,'2005-04-09 00:00:00.000' UNION SELECT 2,1,'2005-12-09 00:00:00.000' UNION SELECT 2,2,'2005-12-15 00:00:00.000' UNION SELECT 2,3,'2005-12-31 00:00:00.000' ;WITH cteX AS( SELECT T.ChildId , T.Id , T.Date , Diff = DATEDIFF(DAY, MIN(T.Date) OVER ( PARTITION BY T.ChildId), T.Date) FROM @tab T ) SELECT X.ChildId , X.Id , X.Date , Diff = CASE WHEN X.Diff = 0 THEN NULL ELSE X.Diff END FROM cteX X ORDER BY X.ChildId , id DESC

然后像这样修改242 days (1-4)

Diff

每个第一行日期

4 个答案: