Sql 获取datediff结果的和

我有一个SQL语句MS SQL Server 2005，它对一些记录上的日期差异进行简单计算。我还想返回DateDiff的总和

SELECT     (DATEDIFF(day, StartDate, EndDate)+1) AS myTotal
FROM         myTable
WHERE     (Reason = '77000005471247')

我如何从我的总数中得到总数？这就是我想要回报的一切

提前感谢

使用总和：

SELECT     SUM(DATEDIFF(day, StartDate, EndDate)+1) AS myTotal
FROM         myTable
WHERE     (Reason = '77000005471247')

使用总和：

SELECT     SUM(DATEDIFF(day, StartDate, EndDate)+1) AS myTotal
FROM         myTable
WHERE     (Reason = '77000005471247')

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如果包含任何其他列，则还需要包含GROUPBY子句

SELECT     AnotherColumn, SUM(DATEDIFF(day, StartDate, EndDate)+1) AS myTotal
FROM         myTable
WHERE      (Reason = '77000005471247')
GROUP BY   AnotherColumn

---

如果包含任何其他列，则还需要包含GROUPBY子句

SELECT     AnotherColumn, SUM(DATEDIFF(day, StartDate, EndDate)+1) AS myTotal
FROM         myTable
WHERE      (Reason = '77000005471247')
GROUP BY   AnotherColumn

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谢谢你，也许你需要它！这将为另一列中的每个值提供单独的总和。谢谢，可能需要这样做！这将为另一列中的每个值提供单独的总和。