SQL DATEDIFF结果与Excel在小时数上的差异

我们一直在努力寻找解决这个问题的方法

我们正在远离excel，通过减去两个日期，在几小时内将计算结果添加到SQL中

问题是我们在excel中的结果与SQL不同

这是excel中的内容：

Change Date: 30/10/2018      
Change Time: 10:53

Service Date: 29/10/2018    
Service Time: 9:37      

The Formula in excel is: =((Change Date+Change Time)-(Service Date+Service Time))*24

以小时为单位的结果为：25.266667

使用DATEDIFF将相同的数据导入SQL（语法如下）：

给出结果：26.350000

有人对如何解决这一矛盾有想法吗

提前感谢你

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朱莉安娜检查你的输入。你的逻辑是正确的，尽管可以简化。为清晰起见，请在SSMS中运行下面的代码示例

declare @Service_Date varchar(max) = '30/10/2018'
declare @Service_Time varchar(max) = '10:53'

declare @Change_Date varchar(max) = '29/10/2018'
declare @Change_Time varchar(max) = '9:37'

--
-- Original 
--
select datediff(second,CAST(CAST(CONVERT(DATETIME,@Service_Date,103) as smalldatetime) 
            + CAST(CONVERT(DATETIME,@Service_Time,103) as smalldatetime) as smalldatetime),
            CAST(CAST(CONVERT(DATETIME,@Change_Date,103) as smalldatetime) 
            + CAST(CONVERT(DATETIME,@Change_Time,103) as smalldatetime) as smalldatetime)) /3600.0

--
-- Simplified - same result
--
select datediff(second,
    (CONVERT(smalldatetime,@Service_Date,103) + CONVERT(smalldatetime,@Service_Time,103)),
    (CONVERT(smalldatetime,@Change_Date,103) + CONVERT(smalldatetime,@Change_Time,103))) /3600.0

--
-- Simplfied - shows error is with inpuyts, not logic
--
set @Service_Time = '11:57'
set @Change_Time = '9:36'
select datediff(second,
    (CONVERT(smalldatetime,@Service_Date,103) + CONVERT(smalldatetime,@Service_Time,103)),
    (CONVERT(smalldatetime,@Change_Date,103) + CONVERT(smalldatetime,@Change_Time,103))) /3600.0

---

我确实得到了同样的价值，但我无法重现这个问题。我也没有得到不同。此外，在您的情况下，不需要将值强制转换为SMALLDATETIME。不强制转换就检查它：DATEDIFF（秒，CONVERT（DATETIME，Service_-Date，103）+CONVERT（DATETIME，Service_-Time，103），CONVERT（DATETIME，Change_-Date，103）+CONVERT（DATETIME，Change_-Time，103））/3600.0请检查

declare @Service_Date varchar(max) = '30/10/2018'
declare @Service_Time varchar(max) = '10:53'

declare @Change_Date varchar(max) = '29/10/2018'
declare @Change_Time varchar(max) = '9:37'

--
-- Original 
--
select datediff(second,CAST(CAST(CONVERT(DATETIME,@Service_Date,103) as smalldatetime) 
            + CAST(CONVERT(DATETIME,@Service_Time,103) as smalldatetime) as smalldatetime),
            CAST(CAST(CONVERT(DATETIME,@Change_Date,103) as smalldatetime) 
            + CAST(CONVERT(DATETIME,@Change_Time,103) as smalldatetime) as smalldatetime)) /3600.0

--
-- Simplified - same result
--
select datediff(second,
    (CONVERT(smalldatetime,@Service_Date,103) + CONVERT(smalldatetime,@Service_Time,103)),
    (CONVERT(smalldatetime,@Change_Date,103) + CONVERT(smalldatetime,@Change_Time,103))) /3600.0

--
-- Simplfied - shows error is with inpuyts, not logic
--
set @Service_Time = '11:57'
set @Change_Time = '9:36'
select datediff(second,
    (CONVERT(smalldatetime,@Service_Date,103) + CONVERT(smalldatetime,@Service_Time,103)),
    (CONVERT(smalldatetime,@Change_Date,103) + CONVERT(smalldatetime,@Change_Time,103))) /3600.0