enumerating paths to leaf nodes in XML Schema

Discussion in 'XML' started by dkacher, Dec 4, 2006.

  1. dkacher

    dkacher Guest

    Hi -
    I'm looking for a way to generate a list of the fully-qualified paths
    to all of the leaf nodes in an XML Schema. The reason: I have a large
    schema for which I'm building a transform stylesheet; I need to be sure
    I've covered everything. With a list of the paths to all the leaves, I
    can check off my progress.

    Have you encountered a program that can generate such a list? Or any
    pointers about how to approach it? It seems that there should be a
    generic stylesheet to do this, but I don't know how to tackle it. The
    schema that I want to analyze is complex -- it has many Type
    definitions and references, and uses includes. Any pointers greatly
    appreciated.
    - Don
    dkacher, Dec 4, 2006
    #1
    1. Advertising

  2. There may not be a complete enumeration, if any of the data structures
    are recursive..

    --
    Joe Kesselman / Beware the fury of a patient man. -- John Dryden
    Joseph Kesselman, Dec 4, 2006
    #2
    1. Advertising

  3. For that matter, as soon as you allow axes other than child the concept
    of "all possible paths" breaks down, since you then have
    multiple/redundant ways to express the same request.

    Folks have analysed schemas to express them as data-structure trees --
    IBM has published papers on schema-driven specialization of parsing and
    XSLT processing -- but that's a matter of deciding in advance exactly
    which paths are going to be considered significant/useful (and,
    sometimes, rewriting the application so it only uses those forms)

    --
    Joe Kesselman / Beware the fury of a patient man. -- John Dryden
    Joseph Kesselman, Dec 4, 2006
    #3
  4. dkacher

    dkacher Guest

    Joseph Kesselman wrote:
    > For that matter, as soon as you allow axes other than child the concept
    > of "all possible paths" breaks down, since you then have
    > multiple/redundant ways to express the same request.
    >
    > Folks have analysed schemas to express them as data-structure trees --
    > IBM has published papers on schema-driven specialization of parsing and
    > XSLT processing -- but that's a matter of deciding in advance exactly
    > which paths are going to be considered significant/useful (and,
    > sometimes, rewriting the application so it only uses those forms)
    >
    > --
    > Joe Kesselman / Beware the fury of a patient man. -- John Dryden


    Thanks. You're right that there might be an unlimited number of paths,
    and that it's necessary to limit the axes. Taking what you say into
    account, I can restate my goal: I want to build a tree structure that
    represents the minimal set of nodes an xml document instance would have
    if it had one occurence of each node allowed by the schema. The result
    should look like the output of the tree /f /a command in windows, or
    the find . -print command in unix.

    Thinking about it that way, I realized that there was a way to get what
    I was looking for. IDEs like XML Spy allow you to visualize a schema as
    such a tree. But, as best I could tell, XML Spy at least doesn't offer
    a text listing of the tree. But IDEs do offer to generate a sample
    instance, and you can tune whether to include optional nodes. So I
    generated a sample instance for my schema. Then I wrote a small
    transform to re-present the nodes in the generated xml in a more
    compact form:

    <xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:fo="http://www.w3.org/1999/XSL/Format">
    <!-- enumerate_paths: lists every path (to internal nodes and leaves)
    in the input xml -->
    <xsl:eek:utput method="text"/>
    <xsl:template match="/">
    <xsl:for-each select="child::*">
    <xsl:call-template name="listNodes">
    <xsl:with-param name="prefix" select="''"/>
    </xsl:call-template>
    </xsl:for-each>
    </xsl:template>
    <xsl:template name="listNodes">
    <xsl:param name="prefix"/>
    <xsl:variable name="myName" select="local-name()"/>
    <xsl:value-of select="concat($prefix, '/', $myName, '
    ')"/>
    <xsl:for-each select="attribute::*">
    <xsl:value-of select="concat($prefix, '/', $myName, '/@',
    local-name(), '
    ')"/>
    </xsl:for-each>
    <xsl:for-each select="child::*">
    <xsl:call-template name="listNodes">
    <xsl:with-param name="prefix" select="concat($prefix, '/',
    $myName)"/>
    </xsl:call-template>
    </xsl:for-each>
    </xsl:template>
    </xsl:stylesheet>

    I fed my generated document instance in to this transform, and got back
    a listing like the one I want, except that it had duplicates. This was
    easy to fix, with sort -u in unix. The end result was the listing I
    was looking for. So I'm all set. Thanks for your help.
    - Don
    dkacher, Dec 5, 2006
    #4
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. Clive Moore

    Hide Leaf nodes in a jtree

    Clive Moore, Oct 29, 2003, in forum: Java
    Replies:
    1
    Views:
    789
    Harald Hein
    Oct 30, 2003
  2. Markus
    Replies:
    1
    Views:
    1,497
    Markus
    Nov 23, 2005
  3. gavnosis
    Replies:
    0
    Views:
    500
    gavnosis
    Aug 2, 2003
  4. Stanimir Stamenkov
    Replies:
    3
    Views:
    1,173
    Stanimir Stamenkov
    Apr 25, 2005
  5. Replies:
    7
    Views:
    865
Loading...

Share This Page