(not) evaluating a document name and a path in xslt

[markoniinimaki]

Hi,

suppose I get a document name a.xml and a path //foo/bar (which nodes
eventually to read from it) from another doc.

Problem: how to combine them so that the whole expression gets
evaluated. I've tried:

<xsl:variable name="imported" select="concat( document('$doc' ),$path)"
/>

But this makes the value of imported [a sequence of linefeeds]
//foo/bar.

Another possibility:
<xsl:for-each select="xalan:e valuate(concat( document('$doc' ),$path))">

...does not seem to work: the $path is ok but probably the
document('$doc' ) part is still a sequence of linefeeds so as a whole
the expression is meaningless.

Is there any way of defining "set the value of the variable to be
document('$doc' ) but do not evaluate it yet"?

Marko

[Andy Dingley]

markoniinimaki wrote:

suppose I get a document name a.xml and a path //foo/bar (which nodes
eventually to read from it) from another doc.

Not sure I've read your question quite carefully enough, but what's
wrong with this?
<xsl:variable name="imported" select="documen t ($doc)//foo/bar" />

[George Bina]

> Is there any way of defining "set the value of the variable to be

document('$doc' ) but do not evaluate it yet"?

<xsl:variable name="test">doc ument('$doc')</xsl:variable>

Best Regards,
George
---------------------------------------------------------------------
George Cristian Bina
<oXygen/> XML Editor, Schema Editor and XSLT Editor/Debugger
http://www.oxygenxml.com

[markoniinimaki]

Hi,

thanks for your answer. However, the following code

<xsl:variable name="nfile">y1 980.xml</xsl:variable>
<xsl:variable name="ndoc">doc ument('$nfile') </xsl:variable>
<xsl:value-of select="$nfile"/>
<xsl:value-of select="$ndoc"/>

results:
y1980.xml
document('$nfil e')

Marko

[Martin Honnen]

markoniinimaki wrote:

<xsl:variable name="nfile">y1 980.xml</xsl:variable>
<xsl:variable name="ndoc">doc ument('$nfile') </xsl:variable>
<xsl:value-of select="$nfile"/>
<xsl:value-of select="$ndoc"/>

results:
y1980.xml
document('$nfil e')

Perhaps you want e.g.
<xsl:variable name="nfile" select="'y1980. xml'" />
<xsl:variable name="ndoc" select="concat( 'document(&quot ;', $nfile,
'&quot;)')" />

--

Martin Honnen
http://JavaScript.FAQTs.com/

[markoniinimaki]

Hi,

<xsl:variable name="imported" select="documen t ($doc)//foo/bar" />

could work but "//foo/bar" is a variable. Thus concat($docvar, $pathvar)
does not work.
xalan:evaluate( concat($docvar, $pathvar)) could work but seemingly does
not..

It might be possible to refer to "docvar" by using a namespace. Is
there a way of forcing a new namespace for a file that streams trough
the processor ("-IN" file in xalan)?

Marko

[Andy Dingley]

markoniinimaki wrote:

could work but "//foo/bar" is a variable.

OK, that's hard (more than I can think of offhand)

But you can use a cascade of individual "name()=$fo o" which will work
for dynamic selections