Question

我继承了大量的存储过程，这些存储过程将XML块作为唯一参数，然后作者将其破解（使用OPENXML），然后更新一行。

我在更新包含datetime列的表时遇到问题，以下是当前失败的数据片段：

declare @stuff nvarchar (max);

set @stuff =
'<Booking xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">  
  <UpdatedDateTime>2009-08-14T15:32:09.7240556+01:00</UpdatedDateTime>
</Booking>';

declare @iDoc int;

exec sp_xml_preparedocument @idoc OUTPUT, @stuff;

SELECT UpdatedDateTime
    FROM 
        OPENXML(@idoc, '/*')
    WITH
    (
        UpdatedDateTime datetime 'UpdatedDateTime'
    )

运行它会导致以下错误：

Msg 241，Level 16，State 1，Line 12

转换时转换失败   来自字符串的日期时间。

如何将日期时间从XML片段转换为基于SQL的日期时间？

由于

Answer 1

尝试：

declare @stuff nvarchar (max);

    set @stuff =
    '<Booking xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">  
      <UpdatedDateTime>2009-08-14T15:32:09.7240556+01:00</UpdatedDateTime>
    </Booking>';

    declare @iDoc int;

    exec sp_xml_preparedocument @idoc OUTPUT, @stuff;

    SELECT CONVERT(datetime,REPLACE(LEFT(UpdatedDateTime,23),'T',' ')),RIGHT(UpdatedDateTime,6)
        ,CASE
             WHEN LEFT(RIGHT(UpdatedDateTime,6),1)='+' THEN DATEADD(hh,CONVERT(int,LEFT(RIGHT(UpdatedDateTime,5),2)),CONVERT(datetime,REPLACE(LEFT(UpdatedDateTime,23),'T',' ')))
             ELSE DATEADD(hh,-1*CONVERT(int,LEFT(RIGHT(UpdatedDateTime,5),2)),CONVERT(datetime,REPLACE(LEFT(UpdatedDateTime,23),'T',' ')))
         END

        FROM 
            OPENXML(@idoc, '/*')
        WITH
        (
            UpdatedDateTime varchar(100) 'UpdatedDateTime'
        )

结果集：

----------------------- ------ -----------------------
2009-08-14 15:32:09.723 +01:00 2009-08-14 16:32:09.723

(1 row(s) affected)

SQL，OPENXML和DateTime对话

1 个答案: