Question

In my XML document, I am pulling the content of a <TextBlock> that contains images. The XML shows:

<img src="/templates_soft/images/facebook.png" alt="twitter" />

When I view the page, the image doesn’t show up because it is not at the same path as the original page.

I need to add the rest of the URL for the images to display. Something like http://www.mypage.com/ so that the image displays from http://www.mypage.com/templates_soft/images/facebook.png

Is there a way to do this?

Answer 1

Use:

<img src="{$imageBase/}templates_soft/images/facebook.png" alt="twitter" />

where the xsl:variable named $imageBase is defined to contain the necessary prefix (in your case "http://www.mypage.com").

Here is a complete XSLT solution:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:param name="pimageBase" select="'http://www.mypage.com'"/>

    <xsl:template match="img">
   <img src="{concat($pimageBase, @src)}" alt="{@alt}"/>
  </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the following XML document:

<img src="/templates_soft/images/facebook.png" alt="twitter" />

the wanted, correct result is produced:

<img src="http://www.mypage.com/templates_soft/images/facebook.png" alt="twitter"/>