For this xml (in a SQL 2005 XML column):
1
Declare @xml Xml = '<doc><a>1</a><b ba="1" bb="2" bc="3" /><c bd="3"/></doc>'
Select n.value('local-name(.)', 'varchar(max)') from @xml.nodes('/doc/*/@*') a(n)
Returns ba bb bc bd
DECLARE @xml as xml
SET @xml =
'<doc>
<a>1</a>
<b ba="1" bb="2" bc="3" />
<c bd="3"/>
</doc>'
SELECT DISTINCT
CAST(Attribute.Name.query('local-name(.)') AS VARCHAR(100)) Attribute,
Attribute.Name.value('.','VARCHAR(100)') Value
FROM @xml.nodes('//@*') Attribute(Name)
Returns:
Attribute Value
ba 1
bb 2
bc 3
bd 3
DECLARE @xml as xml
DECLARE @path as varchar(max)
DECLARE @index int, @count int
SET @xml =
'<doc>
<a>1</a>
<b ba="1" bb="2" bc="3" />
<c bd="3"/>
</doc>'
SELECT @index = 1
SET @count = @xml.query('count(/doc/b/@*)').value('.','int')
WHILE @index <= @count
BEGIN
SELECT @xml.value('local-name((/doc/b/@*[sql:variable("@index")])[1])', 'varchar(max)')
SET @index = @index + 1
END
for element 'b'
it returns
You can build a loop to get attributes for each element in the xml.
BTW The XML in your sample should be closed at closing doc tag.
this:
declare @xml as xml
set @xml =
'<doc>
<a>1</a>
<b ba="1" bb="2" bc="3" />
<c bd="3"/>
</doc>'
select @xml.query('
for $attr in /doc/b/@*
return local-name($attr)')
returns:
ba bb bc