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 11:13:17 -0400 |
Thank you. I've not tried the scaling yet as I can't get the image to display. I keep getting: Apr 2, 2012 10:05:14 AM org.apache.fop.fo.flow.ExternalGraphic bind SEVERE: Image not available: No ImagePreloader found for Here is the line in the xml: (which the image changes w/ every new xml). <photo>/RPT/0000718.jpg</photo> and here is the xsl: <fo:external-graphic src="'url({photo})'"/> Many thanks for the replys and my apologies as I am trying to learn this "on the fly" unfortunately. 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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