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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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