SQL Server 2012中的编号岛

Question

我需要在SQL Server
2012中对孤岛进行编号。孤岛定义为一组行，其中同一行之间DateFrom和DateTo之内没有日间空白ItemId。

以下数据集：

CREATE TABLE #Ranges (ItemId INT, DateFrom DATETIME, DateTo DATETIME)

INSERT INTO #Ranges VALUES (1,'2015-01-31','2015-02-17')
INSERT INTO #Ranges VALUES (1,'2015-02-18','2015-03-31')
INSERT INTO #Ranges VALUES (1,'2015-04-14','2015-05-21')
INSERT INTO #Ranges VALUES (2,'2015-07-12','2015-07-19')
INSERT INTO #Ranges VALUES (2,'2015-07-20','2015-07-24')
INSERT INTO #Ranges VALUES (2,'2015-07-26','2015-08-02')
INSERT INTO #Ranges VALUES (2,'2015-08-03','2015-08-07')

应该编号如下：

ItemId;  DateFrom;    DateTo;      Number
1;       2015-01-31;  2015-02-17;  1
1;       2015-02-18;  2015-03-31;  1
1;       2015-04-14;  2015-05-21;  2
2;       2015-07-12;  2015-07-19;  3
2;       2015-07-20;  2015-07-24;  3
2;       2015-07-26;  2015-08-02;  4
2;       2015-08-03;  2015-08-07;  4

任何帮助，不胜感激。

此致Przemek

Answer 1

如果您只想对它们进行编号，那么我建议lag()使用一个累加的总和：

select t.*,
       sum(case when datefrom = dateadd(day, 1, prev_dateto
                then 0 else 1
           end) over (order by itemId, datefrom)
from (select t.*,
             lag(dateto) over (partition by itemid order by datefrom) as prev_dateto
      from table t
     ) t;

在case确定新的岛屿开始的地方。累计总和只是将此标志求和。