Import XML into SQL server using OPENXML command with XMLNS

4
votes

I have the following code to import an xml into SQL

DECLARE @XML AS XML, @hDoc AS INT, @SQL NVARCHAR (MAX)

SELECT @XML = XMLData FROM XMLwithOpenXML

EXEC sp_xml_preparedocument @hDoc OUTPUT, @XML

SELECT rid, uid
FROM OPENXML(@hDoc, '/PportTimetable/Journey')
WITH 
(
rid [varchar](50) '@rid',
uid [varchar](100) '@uid'
)

EXEC sp_xml_removedocument @hDoc
GO

I can get the code to work but only when it does not contain the xmlns information as seen below why is this?

xmlns:xsd="http://www.w3.org/2001/XMLSchema"

xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"

xmlns="http://www.thalesgroup.com/rtti/XmlTimetable/v8"

XML header

<PportTimetable xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" timetableID="20161018020822" xmlns="http://www.thalesgroup.com/rtti/XmlTimetable/v8">
  <Journey rid="201610188012733" uid="P12733" trainId="2J27" ssd="2016-10-18" toc="AW">
</Journey>
</PportTimetable>
2
sqlsql-serverxml

2 Answers

3
votes

I would recommend to skip the OPENXML stuff altogether and use the built-in, native XQuery support in SQL Server:

declare @input XML = '<PportTimetable xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" timetableID="20161018020822" xmlns="http://www.thalesgroup.com/rtti/XmlTimetable/v8">
    <Journey rid="201610188012733" uid="P12733" trainId="2J27" ssd="2016-10-18" toc="AW">
    </Journey>
</PportTimetable>'

-- define your XML namespaces - here, there's only a single "default" namespace    
;WITH XMLNAMESPACES(DEFAULT 'http://www.thalesgroup.com/rtti/XmlTimetable/v8')
SELECT 
    RID = XC.value('@rid', 'varchar(50)'),
    UID = XC.value('@uid', 'varchar(20)'),
    TrainId = XC.value('@trainId', 'varchar(25)'),
    SSD = XC.value('@ssd', 'varchar(25)'),
    TOC = XC.value('@toc', 'varchar(20)')
FROM 
    @input.nodes('/PportTimetable/Journey') AS XT(XC)

Use the XQuery .nodes() function to "shred" your XML into an "inline" table of XML fragments (one for each <Journey> node, in this example), and then use the .value() function to grab the individual elements and attributes from those XML fragments one by one.

3
votes

You need to specify namespace as a third argument of sp_xml_preparedocument

DECLARE @XML AS XML='<PportTimetable xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" timetableID="20161018020822" xmlns="http://www.thalesgroup.com/rtti/XmlTimetable/v8">
  <Journey rid="201610188012733" uid="P12733" trainId="2J27" ssd="2016-10-18" toc="AW">
</Journey>
</PportTimetable>';
DECLARE @hDoc AS INT, @SQL NVARCHAR (MAX);

EXEC sp_xml_preparedocument @hDoc OUTPUT, @XML,'<PportTimetable xmlns:b="http://www.thalesgroup.com/rtti/XmlTimetable/v8"/>';

SELECT rid, uid
FROM OPENXML(@hDoc, '/b:PportTimetable/b:Journey')
WITH 
(
rid [varchar](50) '@rid',
uid [varchar](100) '@uid'
)

EXEC sp_xml_removedocument @hDoc
GO