Question

我正在使用下面的查询，其中原始转换是为具有不同日期值的另一个表编写的。 [YMDRCVD]值为（INT，null）。我正在使用SQL管理工作室14并接收将表达式转换为数据类型日期时间错误消息的算术溢出错误。

SELECT
   Year(CH.[YMDRCVD]) as [Receipt date Year], 
   Month(CH.[YMDRCVD]) as [Receipt Date Month],
   COUNT(*) as [Count of Claims],
   AVG(convert(numeric(19,2), DATEDIFF(day, CH.[YMDRCVD], CH.[YMDTRANS]))) as [Average Days from Receipt to Initial Finalized] ,
   SUM(
   CASE WHEN  DATEDIFF(day, CH.[YMDRCVD], CH.[YMDTRANS]) > 10 THEN 1
   ELSE 0
   END) as [Count Processed in > 10 Days]
FROM [SERVICE_X] CH WITH (NOLOCK) 
WHERE 
   CH.[YMDRCVD] >= '1/1/2016'
   AND CH.[YMDRCVD] < '6/1/2016'
   AND DATEDIFF(day, CH.[YMDRCVD], CH.[YMDTRANS]) < 20
   AND DATEDIFF(day, CH.[YMDRCVD], CH.[YMDTRANS]) >= 0 
GROUP BY Year(CH.[YMDRCVD]), 
   Month(CH.[YMDRCVD])
ORDER BY 
   Year(CH.[YMDRCVD]), 
   Month(CH.[YMDRCVD])

Answer 1

您收到此错误是因为您尝试在不是数据类型日期/日期时间的值上调用日期函数。

您需要将[YMDRCVD]转换为日期时间。

查看此帖子：Convert INT to DATETIME (SQL)

Answer 2

使用以下内容解码日期：

YYYYMMDD：

 DATEFROMPARTS( FLOOR([YMDRCVD]/10000), FLOOR([YMDRCVD]/100)%100, [YMDRCVD]%100 )

需要将INT转换为值我可以DIFF

2 个答案: