XML XSL Namespace

Question

I am creating a report from an auto-generated XML file using XSL. In the file my root node contains the time stamp as a namespace. When I generate my HTML report with time stamp included in the XML file, I am not able to pull the attribute in the node. When I delete it, it works fine. Can someone shed a little light on this topic. Thanks.

with the timestamp:

<AdminReports xmlns="30/11/2011 09:25:58">

<?xml version="1.0" encoding="ISO-8859-1"?><!DOCTYPE AdminReports []>
<?xml-stylesheet type="text/xsl" href="AdminReport.xsl"?>

<AdminReports xmlns="30/11/2011 09:25:58">
  <AdminReport ID="1">
  </AdminReport >
</AdminReports>

Daniel Haley · Answer 1 · 2011-12-05T20:54:18.067

0

This value shouldn't be stored in xmlns, but you can access it in XPath 2.0 with namespace-uri():

 namespace-uri(/*:AdminReports)

edited Dec 05 '11 at 20:54

answered Dec 05 '11 at 20:49

Daniel Haley

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I really don't need to access it. I am grabbing the information from a different attribute and when this is in the xml file, it prevents me from accessing it. – NewUser101 Dec 05 '11 at 20:51
@NewUser101 - Ahh...what you're dealing with is a default namespace. There are lots of questions/answers on this topic. I will try to locate some. – Daniel Haley Dec 05 '11 at 20:57
Take a look at this: http://stackoverflow.com/questions/7920450/xhtml-to-xml-xslt-conversion/7920612#7920612 It has options for XSLT 1.0 and 2.0. The problem that you will run into if you're using XSLT 1.0 is that your `xmlns` is going to vary if it's based on a timestamp. You'll end up modifying your XSLT every time the `xmlns` changes. – Daniel Haley Dec 05 '11 at 21:00
Also take a look at Mads' answer here http://stackoverflow.com/a/6710125/317052 This is a good example of using `local-name()` – Daniel Haley Dec 05 '11 at 21:08

score 0 · Answer 2 · answered Dec 05 '11 at 21:07

xmlns="..." is not an attribute declaration but a namespace declaration. It means that the element AdminReports is in a namespace.

I presume that in your XSL, there's no namespace declaration corresponding to the following URI "30/11/2011 09:25:58". Therefore, the XSL doen't give you the expected result. If you have the good result after deleting the namespace declaration, then it means that your XSL works for an XML without namespace.

Before transformation, you can add a first XSL to delete the namespaces. This one for example :

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">
    <xsl:template match="*">
        <xsl:element name="{local-name()}">
            <xsl:for-each select="@*">
                <xsl:attribute name="{local-name()}" >
                    <xsl:value-of select="."/>
                </xsl:attribute>
            </xsl:for-each>
            <xsl:apply-templates/>
        </xsl:element>
    </xsl:template>    
</xsl:stylesheet>

Another alternative could be to change your XSL so that it becomes "namespace aware".

If your namespace is fixed, you can add the `xsl:xpath-default-namespace` attribute with the fixed namespace in it (but it's only for xslt 2.0). If your namespace changes often, maybe it's not that easy, and the possiblity is to change your xsl so that it ignores namespaces, but the code is not easy to maintain after (for example instead of `match="elt"`, write `match="*[local-name()='elt']"`) — Vincent Biragnet, Dec 05 '11 at 21:33

score 0 · Accepted Answer · answered Dec 06 '11 at 05:13

Because the namespace-uri for the default namespace is a timestamp -- which means that it is potentially different for every auto-generated document, there is no way to define "this" namespace in the XSLT code.

There exists another way -- not using unprefixed names directly, but comparing them to the name() of the element:

/*/*[name()='AdminReport']/@ID

correctly selects the ID attribute of any element named 'AdminReport' (regardles to which namespace it belongs), that is a child of the top element of the XML document.

Here is a complete XSLT example:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>

 <xsl:template match="/">
  <xsl:value-of select=
    "/*/*[name()='AdminReport']/@ID"/>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the provided XML document:

<AdminReports xmlns="30/11/2011 09:25:58">
    <AdminReport ID="1">
    </AdminReport >
</AdminReports>

the wanted, correst result is produced:

Thanks Dimitre!!! That worked perfect! Can you refer me to further reading on this topic and other challenges that I may run into to. — NewUser101, Dec 06 '11 at 13:36
@NewUser101: You are welcome. Both the books of Michael Kay and Jeni Tennison are a good read. See my recommendations here: http://stackoverflow.com/questions/339930/any-good-xslt-tutorial-book-blog-site-online/341589#341589 — Dimitre Novatchev, Dec 06 '11 at 13:44

XML XSL Namespace

3 Answers3