Question

我在同一列中有两个日期，我需要减去它们之间的天数/小时数。 SELECT [ID],[DATA] FROM [DADOS]

我想像这样显示：

ID Date DiffDate 1 2017-05-12 0 2 2017-05-14 2 3 2017-05-28 14

Answer 1

在使用SQL Server 2008时，需要在同一表上使用LEFT JOIN：

WITH VTE AS (
    SELECT ID,
           CONVERT(date, [date]) AS [date] --That's not confusing
    FROM (VALUES(1,'20170512'),
                (2,'20170514'),
                (3,'20170528')) V(ID, [date]))
SELECT V1.ID,
       V1.[date],
       ISNULL(DATEDIFF(DAY,V2.[date],V1.[date]),0) AS DiffDate
FROM VTE V1
     LEFT JOIN VTE V2 ON V1.ID -1 = V2.ID;

但是，如果您有SQL Server 2012+，则可以使用LAG轻松实现这一点，这意味着您无需扫描表两次：

WITH VTE AS (
    SELECT ID,
           CONVERT(date, [date]) AS [date] --That's not confusing
    FROM (VALUES(1,'20170512'),
                (2,'20170514'),
                (3,'20170528')) V(ID, [date]))
SELECT V.ID,
       V.[date],
       ISNULL(DATEDIFF(DAY,LAG(V.[date]) OVER (ORDER BY V.ID),V.[date]),0) AS DiffDate
FROM VTE V;

Answer 2

对于旧版本的SQL Server 您应该使用如下查询

 SELECT 
    d1.[ID],
    d1.[DATA], 
    DiffDate=DATEDIFF(d,ISNULL(d2.[Data],d1.[Data]),d1.[Data]) 
    FROM [DADOS] d1
     LEFT JOIN [DADOS] d2
     ON d1.[id]=d2.[id]+1

如果您的SQL Server版本支持以下功能，则可以 LEAD/LAG 功能

SELECT 
[ID],
[DATA], 
DiffDate=DATEDIFF(d,ISNULL(LAG([Data]) over( order by [Id]),[Data]),[Data]) 
FROM [DADOS]

请注意，ISNULL用于处理第一行的0值

Answer 3

SQL Server 2008不支持LAG()，而这正是您真正需要的。我建议使用OUTER APPLY：

SELECT d.*,
       COALESCE(DATEDIFF(day, dprev.DATA, d.DATA), 0)
FROM DADOS d OUTER APPLY
     (SELECT TOP (1) d2.*
      FROM DADOS d2
      WHERE d2.DATA < d.DATA
      ORDER BY DATA DESC
     ) dprev;

如何使用显示天数的SQL查询减去同一列中的两行日期？

3 个答案: