Hi,
You mean the original xml would have markers in place of actual location
and later you would replace the markers with the actual location like
below:
main.xml:
<foo>
<img id="img1"/>
</foo>
and in imgpath.xml you would have
<images>
<img id="img1" path="c:\images\ball.gif"/>
</images>
In the above case you could use the document function to load the second
xml into your xslt as follows:
<xsl:variable name="imgpaths" select="document('imgpath.xml')"/>
<xsl:template match="img">
<xsl:variable name="imgid" select="@id"/>
<fo:external-graphic>
<xsl:attribute name="src">
<xsl:value-of select="concat('URI(', $imgpaths/images/img/@path[(_at_)id =
$imgid], ')')"/>
</xsl:attribute>
</fo:external-graphic>
</xsl:template>
Hope this helps.
cheers,
prakash
bharathi kongara
<bharathikongara To:
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
@yahoo.com> cc: (bcc:
omprakash.v/Polaris)
Subject: [xsl] Specifying the
src attribute for <fo:externalgraphic>
08/22/2005 09:25
AM
Please respond
to xsl-list
Hi
I want to get the src attribute for
<fo:externalgraphic> in xsl stylesheet from an xml
document. I mean the location of the graphic would be
specified in the xml document, and my style sheet
should be able to get that location while it is
converting xml to xsl-fo.(This is because the
locations are generated dynamically by my Java
program, I cannot hardcode them into the stylesheet).
Do anyone know how to specify the src attribute in
this case? Any help would be greatly appreciated
Bharathi
Bharathi Kongara,
Graduate Student,
Department of Computer Science,
Utah State University,
Logan,UT 84321
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