XQUERY SQL Server中节点XML文件中的解析器属性如何

Question

I have this XML:

i want to get the value on Property name = "ParticipTypeName" i am using something like that:

;WITH XMLNAMESPACES(DEFAULT     'http://xml.common.asset.aoma.sonymusic.com/ProductMetadata.xsd')
SELECT   
    x.u.value('(/BusinessUnitProperties/Property[@name = "ParticipTypeName"])[1]', 'varchar(100)') as ParticipTypeName
from 
    @XML.nodes('/ProductMetadata/Tracks/Track/Participants/Participant') x(u)

it doesn't work. How I should get the value in this property?

Answer 1

Try this:

SELECT x.u.value('(//*:Property[@*:name="ParticipTypeName"])[1]','nvarchar(max)')

The // will search for any element <Property> . The XQuery -filter will choose the one with the name you are looking for. The *: will allow you to ignore the namespace.