SQL SERVER - OPENROWSET with filepath parameter

Question

>我有这个SQL脚本正常工作：

INSERT INTO #XMLwithOpenXML(XMLData, LoadedDateTime)
SELECT CONVERT(XML, BulkColumn) AS BulkColumn, GETDATE() 
FROM OPENROWSET(BULK 'C:temptest.wordpress.2017-05-22.xml', SINGLE_BLOB) AS x;

SELECT @XML = XMLData FROM #XMLwithOpenXML

现在我需要用变量传递 xml 文件路径。

如何更改脚本？

感谢支持

Answer 1

您可以使用动态 SQL：

create table #XMLwithOpenXML(XMLData xml, LoadedDateTime DateTime)
declare @xml xml
declare @filename nvarchar(100)
declare @sql nvarchar(max)
set  @filename ='F:a.xml'
set  @sql = 'INSERT INTO #XMLwithOpenXML(XMLData, LoadedDateTime) '
set  @sql = @sql +' SELECT CONVERT(XML, BulkColumn) AS BulkColumn, GETDATE() '
set  @sql = @sql +' FROM OPENROWSET(BULK ''' + @filename +''', SINGLE_BLOB) AS x;'
EXEC (@Sql)
SELECT @XML = XMLData FROM #XMLwithOpenXML
SELECT @XML
drop table #XMLwithOpenXML