Finding node order in XML document in SQL Server
You can emulate the position()
function by counting the number of sibling nodes preceding each node:
SELECT
code = value.value('@code', 'int'),
parent_code = value.value('../@code', 'int'),
ord = value.value('for $i in . return count(../*[. << $i]) + 1', 'int')
FROM @Xml.nodes('//value') AS T(value)
Here is the result set:
code parent_code ord
---- ----------- ---
1 NULL 1
11 1 1
111 11 1
12 1 2
121 12 1
1211 121 1
1212 121 2
How it works:
- The
for $i in .
clause defines a variable named$i
that contains the current node (.
). This is basically a hack to work around XQuery's lack of an XSLT-likecurrent()
function. - The
../*
expression selects all siblings (children of the parent) of the current node. - The
[. << $i]
predicate filters the list of siblings to those that precede (<<
) the current node ($i
). - We
count()
the number of preceding siblings and then add 1 to get the position. That way the first node (which has no preceding siblings) is assigned a position of 1.
You can get the position of the xml returned by a x.nodes()
function like so:
row_number() over (order by (select 0))
For example:
DECLARE @x XML
SET @x = '<a><b><c>abc1</c><c>def1</c></b><b><c>abc2</c><c>def2</c></b></a>'
SELECT
b.query('.'),
row_number() over (partition by 0 order by (select 0))
FROM
@x.nodes('/a/b') x(b)
SQL Server's row_number()
actually accepts an xml-nodes column to order by. Combined with a recursive CTE you can do this:
declare @Xml xml =
'<value code="1">
<value code="11">
<value code="111"/>
</value>
<value code="12">
<value code="121">
<value code="1211"/>
<value code="1212"/>
</value>
</value>
</value>'
;with recur as (
select
ordr = row_number() over(order by x.ml),
parent_code = cast('' as varchar(255)),
code = x.ml.value('@code', 'varchar(255)'),
children = x.ml.query('./value')
from @Xml.nodes('value') x(ml)
union all
select
ordr = row_number() over(order by x.ml),
parent_code = recur.code,
code = x.ml.value('@code', 'varchar(255)'),
children = x.ml.query('./value')
from recur
cross apply recur.children.nodes('value') x(ml)
)
select *
from recur
where parent_code = '121'
order by ordr
As an aside, you can do this and it'll do what do you expect:
select x.ml.query('.')
from @Xml.nodes('value/value')x(ml)
order by row_number() over (order by x.ml)
Why, if this works, you can't just order by x.ml
directly without row_number() over
is beyond me.