"Bharathi" == bharathi kongara <bharathikongara(_at_)yahoo(_dot_)com>
writes:
Bharathi> Sorry the last mail was sent by mistake my xml is:
Bharathi> <CoverPage> <Graphics>C:\baby.bmp</Graphics>
Bharathi> <Title>Cover Page</Title> </CoverPage>
Bharathi> my xsl is:
Bharathi> <xsl:template match="CoverPage"> <fo:block
Bharathi> break-after="page"> <xsl:apply-templates
Bharathi> select="Graphics"/> <xsl:apply-templates
Bharathi> select="Title"/> <xsl:apply-templates
Bharathi> select="CoverPageText"/> </fo:block> </xsl:template>
Bharathi> <xsl:template match="Graphics"> <fo:block>
Bharathi> <fo:external-graphic src="" height="40mm" width="40mm"/>
Bharathi> </fo:block> </xsl:template>
Bharathi> My problem is how to refer to the graphic location
Bharathi> specified in the xml, in the src attribute of
Bharathi> fo:externalgraphic of xsl stylesheet(I left it blank).
Well, the first thing to note is that the src attribute must take a
URI, not a file name.
So, assuming you can correct your xml file to specify a url rather
than a file name, you can then use xsl:element plus xsl:attribute
(instead of coding fo:external-graphic as a literal result element).
Within the xsl:attribute for the src attribute, you can specify
select="CoverPage/Graphics/text()".
--
Colin Adams
Preston Lancashire
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