Question

10/03/2011 11/03/2011 12/03/2011 13/03/2011 14/03/2011  15/03/2011  QUERYRESULT
a             a          a                      a            a       14/03/2011
                                    b           b            b       13/03/2011
c                        c          c           c                    12/03/2011

因此，对于最近的连续日期块，请返回该特定块的最早日期，如“queryresult”列中所示。

我现在意识到我在我给出的例子中无意中转换了数据对于源可以假设模式是：[ID] 1,2,3，...; [type] a，b，c ..; [日期]

我可以通过递归的CTE /窗口/排名功能看到一些东西，但无法弄明白。

Answer 1

比我想象的更容易，不需要CTE或窗口/排名功能：

假设数据是

ID  mydate  pfo
1   12/03/2011  a
4   13/03/2011  a
6   14/03/2011  a
10  16/03/2011  a
2   12/03/2011  b
5   13/03/2011  b
8   15/03/2011  b
3   12/03/2011  c
7   14/03/2011  c
9   15/03/2011  c
11  16/03/2011  c

然后

SELECT 

    MAX( n.mydate)
   ,n.pfo

 FROM testtable n

 LEFT OUTER JOIN
 testtable n2 ON  n.pfo                  = n2.pfo 
              AND DATEADD(D,-1,n.mydate) = n2.mydate

 WHERE n2.mydate IS NULL

 GROUP BY n.pfo

Answer 2

;WITH data (type, date) AS (
  SELECT 'a', CAST('20110310' AS datetime) UNION ALL
  SELECT 'a', CAST('20110311' AS datetime) UNION ALL
  SELECT 'a', CAST('20110312' AS datetime) UNION ALL
  SELECT 'a', CAST('20110314' AS datetime) UNION ALL
  SELECT 'a', CAST('20110315' AS datetime) UNION ALL
  SELECT 'b', CAST('20110313' AS datetime) UNION ALL
  SELECT 'b', CAST('20110314' AS datetime) UNION ALL
  SELECT 'b', CAST('20110315' AS datetime) UNION ALL
  SELECT 'c', CAST('20110310' AS datetime) UNION ALL
  SELECT 'c', CAST('20110312' AS datetime) UNION ALL
  SELECT 'c', CAST('20110313' AS datetime) UNION ALL
  SELECT 'c', CAST('20110314' AS datetime)
),
grouped AS (
  SELECT
    type,
    date,
    groupID = DATEADD(day, -ROW_NUMBER() OVER (PARTITION BY type ORDER BY date), date)
  FROM data
),
startdates AS (
  SELECT
    type,
    groupStartDate = MIN(date)
  FROM grouped
  GROUP BY type, groupID
)
SELECT
  type,
  LastGroupStartDate = MAX(groupStartDate)
FROM startdates
GROUP BY type

间歇时间序列。返回T-SQL中最新数据的最后一个连续块的第一个日期

2 个答案: