Can you embed XML in XSL and access it?

[johkar]

I would like to embed some XML within my XSL as either a param or a
global variable. Is this possible? If so, what is the syntax to
access it?

<xsl:call-template name="securetemplate">
<xsl:with-param name="leftnav">
<links>
<link>Link 1</link>
<link>Link 2</link>
</links>
</xsl:with-param>
</xsl:call-template>

Thanks, John

 

[Andy Dingley]

I would like to embed some XML within my XSL as either a param or a
global variable. Is this possible?

Yes. Stick it into your XSL as XML elements within a separate
namespace (and an easily findable root element). The use the XPath
document() function to access it.

 

[Martin Honnen]

I would like to embed some XML within my XSL as either a param or a
global variable. Is this possible? If so, what is the syntax to
access it?

You can embed XML data as a top level element (child of the
xsl:stylesheet element) if the element is in a different namespace e.g.

<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:data="http://example.com/2006/some-data"
version="1.0">

<data:data>
<links xmlns="">
<link>Link 1</link>
<link>Link 2</link>
</links>
</data:data>

The XSLT stylesheet itself can be accessed with calling the document
function with '' e.g.
document('')
so you would get at that data with e.g.
document('')/xsl:stylesheet/data:data/links/link
to access all the link elements.


If you use xslaram or xsl:variable and have contents in these elements
then you have a result tree fragment as the parameter or variable value
that you can later copy to the result tree with xsl:copy-of e.g.

<xsl:call-template name="securetemplate">
<xsl:with-param name="leftnav">
<links>
<link>Link 1</link>
<link>Link 2</link>
</links>
</xsl:with-param>
</xsl:call-template>

<xsl:template name="securetemplate">
<xslaram name="leftnav"/>
<xsl:copy-of select="$leftnav"/>
</xsl:template>

 

[johkar]

I would like to embed some XML within my XSL as either a param or a
global variable. Is this possible? If so, what is the syntax to
access it?

Excellent. Thank you both. That was record response time to boot.

John

 

[johkar]

You can embed XML data as a top level element (child of the
xsl:stylesheet element) if the element is in a different namespace e.g.

<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:data="http://example.com/2006/some-data"
version="1.0">

<data:data>
<links xmlns="">
<link>Link 1</link>
<link>Link 2</link>
</links>
</data:data>

The XSLT stylesheet itself can be accessed with calling the document
function with '' e.g.
document('')
so you would get at that data with e.g.
document('')/xsl:stylesheet/data:data/links/link
to access all the link elements.


If you use xslaram or xsl:variable and have contents in these elements
then you have a result tree fragment as the parameter or variable value
that you can later copy to the result tree with xsl:copy-of e.g.

<xsl:call-template name="securetemplate">
<xsl:with-param name="leftnav">
<links>
<link>Link 1</link>
<link>Link 2</link>
</links>
</xsl:with-param>
</xsl:call-template>

<xsl:template name="securetemplate">
<xslaram name="leftnav"/>
<xsl:copy-of select="$leftnav"/>
</xsl:template>

One more question. If data:data is on the main page, how can you
access it in an included xsl?

John

 

[Joe Kesselman]

One more question. If data:data is on the main page, how can you
access it in an included xsl?

I'm not sure what you mean by "main page" in this context. The main
stylesheet? If you know the URI for that stylesheet, you can ask the
document function to retrieve from that specific URI. If you don't know
the URI, you may have to pass it down as a parameter ... or just pass a
reference to that data:data element down as a parameter.

 

[johkar]

I'm not sure what you mean by "main page" in this context. The main
stylesheet? If you know the URI for that stylesheet, you can ask the
document function to retrieve from that specific URI. If you don't know
the URI, you may have to pass it down as a parameter ... or just pass a
reference to that data:data element down as a parameter.

I have only been able to get one scenario to work, if I have the
data:data namespace and the document function on the same stylesheet.
If I want to output various nodes from the data:data XML from and
included stylesheet, I cannot get it to work. I have tried passing
references and other things suggested, but my syntax is apparently
incorrect. We use Xalan if that makes a difference. Below is an
example.

Thanks, John

Main template
<?xml version='1.0'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:data="http://example.com/2006/some-data">
<xsl:include href="common/leftcol.xsl" />

<data:data>
<links xmlns="">
<link name="selected">Link 1</link>
<link>Link 2</link>
</links>
</data:data>

<xsl:template match="/">
<xsl:call-template name="leftcol" />
</xsl:template>
</xsl:stylesheet>

Left Column Template

<?xml version='1.0'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:data="http://example.com/2006/some-data">
<xsl:template name="leftcol">
<xsl:value-of
select="document(common/mainPage.xsl)/data:data/links/link[@name='selected']"
/>
</xsl:template>
</xsl:stylesheet>

 

[Martin Honnen]

johkar wrote:

<xsl:value-of
select="document(common/mainPage.xsl)/data:data/links/link[@name='selected']"
/>

I think you want/need
document('common/mainPage.xsl')/xsl:stylesheet/data:data/links/link[@name='selected']

 

[johkar]

I think you want/need
document('common/mainPage.xsl')/xsl:stylesheet/data:data/links/link[@name='selected']

Sorry, this is my directory structure:

common/leftcol.xsl
main/mainPage.xsl (this is the main stylesheet)

I tried both of these in leftcol.xsl, but nothing has worked so far:

document('main/mainPage.xsl')/xsl:stylesheet/data:data/links/link[@name='selected']

and

document('../main/mainPage.xsl')/xsl:stylesheet/data:data/links/link[@name='selected']

 

[johkar]

johkar wrote:

<xsl:value-of
select="document(common/mainPage.xsl)/data:data/links/link[@name='selected']"
/>

I think you want/need
document('common/mainPage.xsl')/xsl:stylesheet/data:data/links/link[@name='selected']

Ok, I don't know what happened, but since I couldn't get the path
correct, I tried passing a reference to the document() and it worked
this time...go figure.

So within my call-template, I included a reference. I know I tried
this before.
<xsl:with-param name="leftcol" select="document('')" />

Thanks everyone.