Subject: Re: [xsl] Specifying the src attribute for <fo:externalgraphic> From: Colin Paul Adams <colin@xxxxxxxxxxxxxxxxxx> Date: 22 Aug 2005 06:44:29 +0100 |
>>>>> "Bharathi" == bharathi kongara <bharathikongara@xxxxxxxxx> 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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