Finding node order in XML document in SQL Server

Question

How can I find the order of nodes in an XML document?

What I have is a document like this:

Answer 1

According to this document and this connect entry it is not possible, but the Connect entry contains two workarounds.

I do it like this:

WITH n(i) AS (SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9),
     o(i) AS (SELECT n3.i * 100 + n2.i * 10 + n1.i FROM n n1, n n2, n n3)
SELECT v.value('@code', 'varchar(20)') AS code,
       v.value('../@code', 'varchar(20)') AS parent,
       o.i AS ord
  FROM o
 CROSS APPLY @xml.nodes('/root//value[sql:column("o.i")]') x(v)
 ORDER BY o.i

Answer 2

The answer by erikkallen is absolutely correct.

However, if the original document/schema may be modified, an alternative is to store the position/index in an attribute. I use a mix of both approaches, depending who the "originator" of the XML is and the type of queries that need to be performed upon it. At the end of the day I rue most use of XML except possibly "dumb storage" in SQL Server and am usually happy when I can dump it (XML) for normalized tables.

Happy dealing with the unmentioned limitations of "enterprise-grade" products -- the wonders never end.

Answer 3

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-like current() 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.

Answer 4

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)

Answer 5

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.

Answer 6

I see answer by @Ben and... get new sollution

 row_number() over (order by (select null))

as

  SELECT value.value('@code', 'varchar(20)') code, 
  value.value('../@code', 'varchar(20)') parent, 
  row_number() over (order by (select null))
  FROM @xml.nodes('/root//value') n(value)