Subject: Re: [xsl] is it possible to resize an image to display smaller than the original From: David Ryan <dgdunk@xxxxxxxxx> Date: Mon, 2 Apr 2012 17:54:34 -0400 |
ok. the resize finally works using content-height="50pt" content-width="50pt" the final issue is getting the fo:external-graphic src to pull the file location/name that is being passed in the xml. I have tried every combination I could find and trying to assign a variable equal to the location. <xsl:variable name="photo" select="data/photoimg"/> <fo:block-container height="28.68mm"> <fo:block-container overflow="hidden" display-align="before"> <fo:block text-align="start"> <fo:external-graphic src="'url({$photo})'" content-height="50pt" content-width="50pt"/> </fo:block> </fo:block-container> </fo:block-container> if I hardcode the file location in the url - it will work but I need to pass that in the xml as this routine gets called for many different xml's. Does anyone know what I am doing wrong or have a solution to this issue? I can't begin to grateful enough if someone can help solve this. David On Sun, Apr 1, 2012 at 3:09 AM, davep <davep@xxxxxxxxxxxxx> wrote: > On 31/03/12 14:39, David Ryan wrote: >> >> I changed the xsl to use your suggestion: >> >> <fo:external-graphic src="file:{/data/Photo}"/> >> >> but when I run it, it doesn't find the image file saying Image not >> available. I confirmed the image is in the referenced location. any >> ideas? > > > http://www.w3.org/TR/xsl/#fo_external-graphic > > try > <fo:external-graphic src="url(file.jpg)" /> > If you can, make the file relative rather than absolute and > don't have spaces in path or filenames. > > > Two attributes useful for scaling are > content-height="3" scaling="uniform" > or content-width > > HTH > > > > regards > > -- > Dave Pawson > XSLT XSL-FO FAQ. > http://www.dpawson.co.uk
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