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

2012-03-31 08:39:22
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?

On Sat, Mar 31, 2012 at 8:39 AM, Martin Honnen 
<Martin(_dot_)Honnen(_at_)gmx(_dot_)de> wrote:
David Ryan wrote:

I have an image location name being passed in the xml.

<Photo>/RPT/0000718.jpg</Photo>

The actual image is 8" x 10" and I need to display it on the resulting
.pdf in a size of 3" x 3"

I am using the following to retrieve the file location and display the
image:

  <xsl:element name="fo:external-graphic">
  <xsl:attribute name="src">file:<xsl:value-of
select="/data/Photo"/></xsl:attribute>
  </xsl:element>

I guess I have two questions.


2. Is the way I am retrieving/processing the image filename the
best/most efficient way to do so?


I would simply use a literal result element with e.g.
 <fo:external-graphic src="file:{/data/Photo}"/>
of course making sure the xsl:stylesheet element binds the prefix 'fo' to
the XSL-FO namespace.

--

       Martin Honnen --- MVP Data Platform Development
       http://msmvps.com/blogs/martin_honnen/

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