SQL SERVER - OPENROWSET与文件路径参数

Question

我有这样的SQL脚本THT正常工作：

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


SELECT @XML = XMLData FROM #XMLwithOpenXML 

现在我需要通过xml文件路径有变数。

如何更改脚本？

谢谢支持

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