SQL：計算不同行中一列日期之間的天數

Question

根據我的數據，我讓個人在不同的日期多次進行評估。 它看起來像這樣：

╔════════╦═══════════╦═══════════╦═══════╗
║ Person ║ ID Number ║ Date      ║ Score ║
║ John   ║ 134       ║ 7/11/2013 ║ 18    ║
║ John   ║ 134       ║ 8/23/2013 ║ 16    ║
║ John   ║ 134       ║ 9/30/2013 ║ 16    ║
║ Kate   ║ 887       ║ 2/28/2013 ║ 21    ║
║ Kate   ║ 887       ║ 3/16/2013 ║ 19    ║
║ Bill   ║ 990       ║ 4/18/2013 ║ 15    ║
║ Ken    ║ 265       ║ 2/12/2013 ║ 23    ║
║ Ken    ║ 265       ║ 4/25/2013 ║ 20    ║
║ Ken    ║ 265       ║ 6/20/2013 ║ 19    ║
║ Ken    ║ 265       ║ 7/15/2013 ║ 19    ║
╚════════╩═══════════╩═══════════╩═══════╝

我希望它在末尾有另一列，用於計算自該人第一次評估以來的天數。 如果這更容易的話，我也會確定自上次對該人進行評估以來的天數。

理想情況下，它看起來像這樣：

╔════════╦═══════════╦═══════════╦═══════╦══════════════════╗
║ Person ║ ID Number ║ Date      ║ Score ║ Days Since First ║
║ John   ║ 134       ║ 7/11/2013 ║ 18    ║ 0                ║
║ John   ║ 134       ║ 8/23/2013 ║ 16    ║ 43               ║
║ John   ║ 134       ║ 9/30/2013 ║ 16    ║ 81               ║
║ Kate   ║ 887       ║ 2/28/2013 ║ 21    ║ 0                ║
║ Kate   ║ 887       ║ 3/16/2013 ║ 19    ║ 16               ║
║ Bill   ║ 990       ║ 4/18/2013 ║ 15    ║ 0                ║
║ Ken    ║ 265       ║ 2/12/2013 ║ 23    ║ 0                ║
║ Ken    ║ 265       ║ 4/25/2013 ║ 20    ║ 72               ║
║ Ken    ║ 265       ║ 6/20/2013 ║ 19    ║ 128              ║
║ Ken    ║ 265       ║ 7/15/2013 ║ 19    ║ 153              ║
╚════════╩═══════════╩═══════════╩═══════╩══════════════════╝

Answer 1

select  *
,       datediff(day, min(Date) over (partition by [ID Number]), Date)
from    YourTable

SQL Fiddle 上的實時示例。

Answer 2

我喜歡 Andomar 的回答，但如果你想找到從第一次開始的天數和總天數，你可以這樣做：

SELECT a.*
        ,ISNULL(DATEDIFF(day,b.Date,a.Date),0)'Since Previous'
        ,datediff(day, min(a.Date) over (partition by a.[ID Number]), a.Date)'Since First'
FROM (select  *,ROW_NUMBER() OVER(PARTITION BY [ID Number] ORDER BY DATE)RowRank
      from    YourTable
      )a
LEFT JOIN (select  *,ROW_NUMBER() OVER(PARTITION BY [ID Number] ORDER BY DATE)RowRank
      from    YourTable
      )b
ON a.[ID Number] = b.[ID Number]
 AND a.RowRank = b.RowRank + 1

演示： SQL 小提琴

Answer 3

您可以將選項與APPLY運算符一起使用

1.當前行日期與上一個日期的差異

SELECT t1.*, 
       DATEDIFF(dd, ISNULL(o.[Date], t1.[Date]), t1.[Date]) AS [Days Since First]
FROM YourTable t1 OUTER APPLY (
                               SELECT TOP 1 [Date]
                               FROM YourTable t2
                               WHERE t1.[ID Number] = t2.[ID Number]
                                 AND t1.[Date] > t2.[Date]
                               ORDER BY t2.[Date] DESC
                               ) o

請參閱SQLFiddle上的示例

2.自第一次評估以來的天數

SELECT t1.*, 
       DATEDIFF(dd, ISNULL(o.[Date], t1.[Date]), t1.[Date]) AS [Days Since First]
FROM YourTable t1 OUTER APPLY (
                               SELECT MIN(t2.[Date]) AS [Date]
                               FROM YourTable t2
                               WHERE t1.[ID Number] = t2.[ID Number]                                   
                               ) o

請參閱SQLFiddle上的示例