Given the following sample of XML and the select
statement that shreds the xml into a relation, what I need is the second column of the select
to be the ordinal of the category (ie 1 for the directions and 2 for the colours in this case).
Note: The literal value 'rank()' in the select is left a placeholder. I was poking around with using the rank
, but with no success.
declare @x xml
set @x = '
<root>
<category>
<item value="north"/>
<item value="south"/>
<item value="east"/>
<item value="west"/>
</category>
<category>
<item value="red"/>
<item value="green"/>
<item value="blue"/>
</category>
</root>'
select c.value('./@value', 'varchar(10)') as "ItemValue",
'rank()' as "CategoryNumber"
from @x.nodes('//item') as t(c)
Jacob Sebastian also has an interesting solution presented in his blog post:
XQuery Lab 23 - Retrieving values and position of elements
With Jacob's suggestion, I can rewrite your query to be:
SELECT
x.value('@value','VARCHAR(10)') AS 'ItemValue',
p.number as 'CategoryNumber'
FROM
master..spt_values p
CROSS APPLY
@x.nodes('/root/category[position()=sql:column("number")]/item') n(x)
WHERE
p.type = 'p'
and I get the desired output:
ItemValue CategoryNumber
--------- --------------
north 1
south 1
east 1
west 1
red 2
green 2
blue 2
Unfortunately, none of the more obvious solutions like the position()
or fn:id()
functions seem to a) work in SQL Server or b) be supported in SQL Server at all :-(
Hope this helps
Marc