SQL时间戳减法以获取天数

在下面的代码中，我需要用开放时间和当前时间进行计算，而不是活动年龄

换句话说，如果当前时间减去开放时间在0到30之间，或者31到60之间，或者超过60，我需要这样做

SELECT COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND CloseTime-OpenTime = 0-30 THEN P_NUMBER END) as crosby_sev4_030,
           COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND Active_Age='0-30' THEN P_NUMBER END) as crosby_sev5_030,
           COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND Active_Age='31-60' THEN P_NUMBER END) as crosby_sev4_3160,
           COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND Active_Age='31-60' THEN P_NUMBER END) as crosby_sev5_3160,
           COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND Active_Age='60+' THEN P_NUMBER END) as crosby_sev4_60,
           COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND Active_Age='60+' THEN P_NUMBER END) as crosby_sev5_60
    FROM dashboard.dbo.SmThings
    WHERE Assignment IN('Crosby')
    AND Severity IN(4,5)

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如果您使用的是MSSQL-请使用

输出将是

DiffDate
61

datediff

SELECT 
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND datediff('day',getdate(), OpenTime) between 0 and 30 THEN P_NUMBER END) as crosby_sev4_030,
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND datediff('day',getdate(), OpenTime between 0 and 30 THEN P_NUMBER END) as crosby_sev5_030,
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND datediff('day',getdate(), OpenTime between 31 and 60 THEN P_NUMBER END) as crosby_sev4_3160,
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND  datediff('day',getdate(), OpenTime between 31 and 30 THEN P_NUMBER END) as crosby_sev5_3160,
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 4 AND  datediff('day',getdate(), OpenTime > 60 THEN P_NUMBER END) as crosby_sev4_60,
COUNT(CASE WHEN Assignment = 'Crosby' AND Severity = 5 AND  datediff('day',getdate(), OpenTime > 60 THEN P_NUMBER END) as crosby_sev5_60
FROM dashboard.dbo.SmThings
WHERE Assignment IN('Crosby')
  AND Severity IN(4,5)

这个怎么样

WITH Assignments AS (
    SELECT
       Grp = Assignment + '_Sev' + Convert(varchar(11), Severity) + '_' + Label
    FROM
       dashboard.dbo.SmThings
       CROSS APPLY (
          SELECT Age = DateDiff(OpenTime, GetDate())
       ) A
       INNER JOIN (
          SELECT '030', 0, 30, '030'
          UNION ALL SELECT '3160', 31, 59 -- notice 59 here!
          UNION ALL SELECT '60', 60, 2147483647
       ) X (Label, Low, High)
          ON A.Age BETWEEN X.Low AND X.High
    WHERE Assignment IN ('Crosby')
    AND Severity IN (4, 5)
)
SELECT
   Grp,
   Value = Count(*)
INTO #Data
FROM
   Assignments
GROUP BY
   Grp
;

然后，您可以自动使用这些生成的值来透视（需要SQL Server 2005或更高版本）：

---

看着魔法发生！（假设我没有犯任何错误，这不太可能。）

我如何使用当前日期而不是2008-08-05？我还需要它像“和DATEDIFF（day，'open_time，'current_date'）=0-31或31-60或60+”，虽然这有一个逻辑问题，因为你可能有超过30和<31的日期，这不会显示在计数中，但你原来的帖子也有这个问题。这可能可以通过舍入datediff来解决，但我不知道您是要舍入还是截断。请稍候，不是关闭时间，而是当前时间。

WITH Assignments AS (
    SELECT
       Grp = Assignment + '_Sev' + Convert(varchar(11), Severity) + '_' + Label
    FROM
       dashboard.dbo.SmThings
       CROSS APPLY (
          SELECT Age = DateDiff(OpenTime, GetDate())
       ) A
       INNER JOIN (
          SELECT '030', 0, 30, '030'
          UNION ALL SELECT '3160', 31, 59 -- notice 59 here!
          UNION ALL SELECT '60', 60, 2147483647
       ) X (Label, Low, High)
          ON A.Age BETWEEN X.Low AND X.High
    WHERE Assignment IN ('Crosby')
    AND Severity IN (4, 5)
)
SELECT
   Grp,
   Value = Count(*)
INTO #Data
FROM
   Assignments
GROUP BY
   Grp
;

DECLARE @SQL nvarchar(max);
SET @SQL = '
SELECT *
FROM
   #Data 
   PIVOT (Max(Value) FOR Grp IN (' 
+ Stuff(
     (SELECT ', ' + Convert(nvarchar(max), QuoteName(Grp)) 
     FROM #Data FOR XML PATH(''), TYPE).value('.[1]', 'nvarchar(max)'),
     1, 2, 0
  ) + ')) P
';

EXEC (@SQL);