Subject: Re: [xsl] is it possible to resize an image to display smaller than the original From: David Ryan <dgdunk@xxxxxxxxx> Date: Sat, 31 Mar 2012 09:39:10 -0400 |
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.Honnen@xxxxxx> 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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