XSL:FO How to get the img src from xml and then use it as src for fo:external-graphic

Question

Question: How to get the src attribute from the img element, then copy the src's value it into a variable, and then set the variable as the src for the fo:external-graphic element?

Assume that I have a XML document with an image that looks like this:

<diffreport>
<css />
<diff>
    <p>
        <span class="diff-html-removed" id="removed-diff-0" previous="first-diff"
            changeId="removed-diff-0" next="added-diff-0">
            <img alt="" id="Rxed6OQAKXfCYA"
                src="D:\udu\img1.jpg" changeType="diff-removed-image" />
        </span>         
    </p>
</diff>

Note: Assume the src paths for every img element will be dynamic.

I confirmed that this piece of code bellow works ok, but it is not good because it is hard coded. I would really like to know how to replace the "url('D:\udu\img1.jpg')" with a variable so that the code is dynamic.

<!-- Image -->
<xsl:template match="img">
    <fo:external-graphic
        src="url('D:\udu\img1.jpg')"></fo:external-graphic>
</xsl:template>

Is there any way to doing this? Thank you. :)

Answer 1

Use an attribute value template that will be evaluated to get the value of the @src from your XML. See https://www.w3.org/TR/xslt#attribute-value-templates

<!-- Image -->
<xsl:template match="img">
  <fo:external-graphic src="url('{@src}')" />
</xsl:template>

Answer 2

you can try

<xsl:template match="img">
    <fo:external-graphic
        src="{concat('file:///', translate(@src, '\', '/'))}" />
</xsl:template>

which will get D:\udu\img1.jpg and have it called as file:///D:/udu/img1.jpg