如何識別連續工作日期？

Question

我正在嘗試編寫一些將假期分組在一起的代碼，併爲每個分組創建一個唯一的ID。我遇到的麻煩是創建團體，以知道他們屬於一起。

默認情況下，如果有人要求一個星期關閉5天將得到一個特定的ID（即223）。因此，會有ID 223的5天假期實例，我可以輕鬆獲得開始和結束日期。但如果有人決定預訂星期五休息（ID 224），然後要求星期三和星期四休息（ID 228），然後要求下一個星期一休息（ID 230），那麼我有3個獨立的實際實際情況一個假期，這裏是我需要將它們組合在一起而忽略週末的地方。

我試過DENSE_RANK OVER PARTION和ROW_NUMBER OVER ORDER的變體，但是我無法得到它返回我需要的數字。

樣本數據和期望的輸出

SELECT '01-March-2017' AS HolidayDate, DATEPART(WK,'01-March-2017') AS Dw, 223 AS ID_Field, 1000 AS HolidayID 
UNION ALL 
SELECT '02-March-2017' AS HolidayDate, DATEPART(WK,'02-March-2017') AS Dw, 223, 1000 
UNION ALL 
SELECT '03-March-2017' AS HolidayDate, DATEPART(WK,'03-March-2017') AS Dw, 223, 1000 
UNION ALL 
SELECT '24-March-2017' AS HolidayDate, DATEPART(WK,'24-March-2017') AS Dw, 230 , 1001 
UNION ALL 
SELECT '27-March-2017' AS HolidayDate, DATEPART(WK,'27-March-2017') AS Dw, 235, 1001 
UNION ALL 
SELECT '20-Sep-2017' AS HolidayDate, DATEPART(WK,'20-Sep-2017') AS Dw, 224, 1002 
UNION ALL 
SELECT '27-Sep-2017' AS HolidayDate, DATEPART(WK,'27-Sep-2017') AS Dw, 228, 1003 

所以上面創建一個例子，節假日ID是所需的列輸出（隨機選擇用於樣品編號）。 ID字段是假期請求提交時生成的編號。

正如你所看到的223請求進來完成，並具有相同的ID，所以我可以選擇MAX（日期），並通過ID_Field得到所需的結果分組。

230和235分別進來但是是同一個假期（作爲一個連續的假期），所以我需要將它們用相同的ID進行分組。

最後225和228是完全獨立的請求，所以需要唯一的號碼。與答案

問題 -

Answer 1

當LAG時支持LAG不支持

with Holidays 
     as 
     (
      select row_number() over (order by HolidayDate) as n 
        ,HolidayDate 

      from mytable 
     ) 

select h.HolidayDate 

     ,count 
     (
      case 
       when not 
         (
          datediff (day,p.HolidayDate,h.holidayDate) = 1 

         or ( datediff (day,p.HolidayDate,h.HolidayDate) = 3 
          and datename (dw,p.HolidayDate) = 'Friday' 
          ) 
         ) 
       then 1 
      end 
     ) over (order by h.HolidayDate) + 1 as Holiday_id 

from    Holidays as h 
     left join Holidays as p 
     on   p.n = h.n - 1 

order by h.HolidayDate 

---

select HolidayDate 
     ,count(is_gap) over (order by HolidayDate) + 1 as holiday_id 

from (select HolidayDate 

       ,case 
        when not 
          (
           datediff (day,prev_HolidayDate,HolidayDate) = 1 

          or ( datediff (day,prev_HolidayDate,HolidayDate) = 3 
           and datename (dw,prev_HolidayDate) = 'Friday' 
           ) 
          ) 
        then 1 
       end  as is_gap 

     from (select HolidayDate 
         ,lag(HolidayDate) over (order by HolidayDate) as prev_HolidayDate 

       from mytable 
       ) t 
     ) t 

order by HolidayDate 

---

+-------------+------------+ 
| HolidayDate | holiday_id | 
+-------------+------------+ 
| 2017-03-01 |   1 | 
| 2017-03-02 |   1 | 
| 2017-03-03 |   1 | 
| 2017-03-24 |   2 | 
| 2017-03-27 |   2 | 
| 2017-09-20 |   3 | 
| 2017-09-27 |   4 | 
+-------------+------------+