Question

好的，所以我一直收到这个错误：

Msg 245，Level 16，State 1，Line 1 转换varchar值时转换失败＆＃39;天＆＃39;数据类型int。

下面是我的代码.....我在这里做错了什么....我试图显示日，小时和分钟差异...我在我的约会中使用了CASTS因为那些是不同的列所以我使用演员来结合他们，以便更容易做我的约会.......

DATEDIFF(DD, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME)) + ' Days ' 
        + DATEDIFF(HH, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME)) + ' Hours '
        + DATEDIFF(MINUTE, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME)) + ' Minutes'
AS Time_diff2,

Answer 1

DateDiff函数将以整数形式返回结果。在这里，您将结合Integer值和字符串值。因此，您将面临类型差异错误。

您必须将datediff输出转换为varchar，如下所述

cast(DATEDIFF(DD, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME))as varchar(50)) + ' Days ' 
        + cast(DATEDIFF(HH, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME))as varchar(50)) + ' Hours '
        + cast(DATEDIFF(MINUTE, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME) + CAST(txt_signoff_ml_time AS DATETIME)) as varchar(50)) + ' Minutes'

Answer 2

此：

DATEDIFF(DD, checkin_datetime, CAST(txt_signoff_ml_date AS DATETIME)

需要一个结束括号。这应该让你开始。

我坚持这个DATEDIFF

2 个答案: