Re: [xsl] is it possible to resize an image to display smaller than the original

[David Ryan <dgdunk@xxxxxxxxx>]

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