How to extract element-path from XMLType Node? - sql

Question

I would like to have a select statement on an XML document and one column should return me the path of each node.
For example, given the data
SELECT *
FROM TABLE(XMLSequence(
XMLTYPE('<?xml version="1.0"?>
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>').extract('/*//*[text()]'))) t;
Which results in
column_value
--------
<user><name>user1</name></user>
<user><name>user2</name></user>
<user><name>user3</name></user>
<user><name>user4</name></user>
I'd like to have a result like this:
path value
------------------------ --------------
/users/user/name user1
/users/user/name user2
/users/group/user/name user3
/users/user/name user4
I can not see how to get to this. I figure there are two thing that have to work together properly:
Can I extract the path from an XMLType with a single operation or method, or do I have to do this with string-magic?
What is the correct XPath expression so that I do get the whole element path (if thats possible), eg. <users><group><user><name>user3</name></user></group></user> insead of <user><name>user3</name></user>?
Maybe I am not understanding XMLType fully, yet. It could be I need a different approach, but I can not see it.
Sidenotes:
In the final version the XML document will be coming from CLOBs of a table, not a static document.
The path column can of course also use dots or whatever and the initial slash is not the issue, any representation would do.
Also I would not mind if every inner node also gets a result row (possibly with null as value), not only the ones with text() in it (which is what I am really interested in).
In the end I will need the tail element of path separate (always "name" in the example here, but this will vary later), i.e. ('/users/groups/user', 'name', 'user3'), I can deal with that separately.

Answer 1

You can achieve that with help of XMLTable function from Oracle XML DB XQuery function set:
select * from
XMLTable(
'
declare function local:path-to-node( $nodes as node()* ) as xs:string* {
$nodes/string-join(ancestor-or-self::*/name(.), ''/'')
};
for $i in $rdoc//name
return <ret><name_path>{local:path-to-node($i)}</name_path>{$i}</ret>
'
passing
XMLParse(content '
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>'
)
as "rdoc"
columns
name_path varchar2(4000) path '//ret/name_path',
name_value varchar2(4000) path '//ret/name'
)
For me XQuery looks at least more intuitive for XML data manipulation than XSLT.
You can find useful set of XQuery functions here.
Update 1
I suppose that you need totally plain dataset with full data at last stage.
This target can be reached by complicated way, constructed step-by-step below, but this variant is very resource-angry. I propose to review final target (selecting some specific records, count number of elements etc.) and after that simplify this solution or totally change it.
Update 2
All steps deleted from this Update except last because #A.B.Cade proposed more elegant solution in comments.
This solution provided in Update 3 section below.
Step 1 - Constructing dataset of id's with corresponding query results
Step 2 - Aggregating to single XML row
Step 3 - Finally get full plain dataset by querying constracted XML with XMLTable
with xmlsource as (
-- only for purpose to write long string only once
select '
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>' xml_string
from dual
),
xml_table as (
-- model of xmltable
select 10 id, xml_string xml_data from xmlsource union all
select 20 id, xml_string xml_data from xmlsource union all
select 30 id, xml_string xml_data from xmlsource
)
select *
from
XMLTable(
'
for $entry_user in $full_doc/full_list/list_entry/name_info
return <tuple>
<id>{data($entry_user/../#id_value)}</id>
<path>{$entry_user/name_path/text()}</path>
<name>{$entry_user/name_value/text()}</name>
</tuple>
'
passing (
select
XMLElement("full_list",
XMLAgg(
XMLElement("list_entry",
XMLAttributes(id as "id_value"),
XMLQuery(
'
declare function local:path-to-node( $nodes as node()* ) as xs:string* {
$nodes/string-join(ancestor-or-self::*/name(.), ''/'')
};(: function to construct path :)
for $i in $rdoc//name return <name_info><name_path>{local:path-to-node($i)}</name_path><name_value>{$i/text()}</name_value></name_info>
'
passing by value XMLParse(content xml_data) as "rdoc"
returning content
)
)
)
)
from xml_table
)
as "full_doc"
columns
id_val varchar2(4000) path '//tuple/id',
path_val varchar2(4000) path '//tuple/path',
name_val varchar2(4000) path '//tuple/name'
)
Update 3
As mentioned by #A.B.Cade in his comment, there are really simple way to join ID's with XQuery results.
Because I don't like external links in answers, code below represents his SQL fiddle, a little bit adapted to the data source from this answer:
with xmlsource as (
-- only for purpose to write long string only once
select '
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>' xml_string
from dual
),
xml_table as (
-- model of xmltable
select 10 id, xml_string xml_data from xmlsource union all
select 20 id, xml_string xml_data from xmlsource union all
select 30 id, xml_string xml_data from xmlsource
)
select xd.id, x.* from
xml_table xd,
XMLTable(
'declare function local:path-to-node( $nodes as node()* ) as xs:string* {$nodes/string-join(ancestor-or-self::*/name(.), ''/'') }; for $i in $rdoc//name return <ret><name_path>{local:path-to-node($i)}</name_path>{$i}</ret> '
passing
XMLParse(content xd.xml_data
)
as "rdoc"
columns
name_path varchar2(4000) path '//ret/name_path',
name_value varchar2(4000) path '//ret/name'
) x

Answer 2

This is not perfect but can be a start:
Here is a sqlfiddle
with xslt as (
select '<?xml version="1.0" ?><xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<records>
<xsl:apply-templates/>
</records>
</xsl:template>
<xsl:template match="//name">
<columns>
<path>
<xsl:for-each select="ancestor-or-self::*">
<xsl:call-template name="print-step"/>
</xsl:for-each>
</path>
<value>
<xsl:value-of select="."/>
</value>
<xsl:apply-templates select="*"/>
</columns>
</xsl:template>
<xsl:template name="print-step">
<xsl:text>/</xsl:text>
<xsl:value-of select="name()"/>
<xsl:text>[</xsl:text>
<xsl:value-of select="1+count(preceding-sibling::*)"/>
<xsl:text>]</xsl:text>
</xsl:template>
</xsl:stylesheet>'
xsl from dual)
, xmldata as
(select xmltransform(xmltype('<?xml version="1.0"?>
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>'), xmltype(xsl)) xd from xslt)
select XT.*
from xmldata c,
xmltable('$x//columns' passing c.xd
as "x"
columns
path_c VARCHAR2(4000) PATH 'path',
value_c VARCHAR2(4000) PATH 'value'
) as XT
This is what I tried to do:
Since you want the "path" I had to use xslt (credits to this post)
Then I used xmltransform for transforming your original xml with the xsl to the
desired output (path, value)
Then I used xmltable to read it as a table

Answer 3

This improves on above answer by A.B.Cade:
<xsl:template name="print-step">
<xsl:variable name="name" select="name()" />
<xsl:text>/</xsl:text>
<xsl:value-of select="$name"/>
<xsl:text>[</xsl:text>
<xsl:value-of select="1+count(preceding-sibling::*[name()=$name])"/>
<xsl:text>]</xsl:text>
</xsl:template>
With result:
/users[1]/user[1]/name[1] user1
/users[1]/user[2]/name[1] user2
/users[1]/group[1]/user[1]/name[1] user3
/users[1]/user[3]/name[1] user4
Instead of:
/users[1]/user[1]/name[1] user1
/users[1]/user[2]/name[1] user2
/users[1]/group[3]/user[1]/name[1] user3
/users[1]/user[4]/name[1] user4