How to remove some content from the elements using XSLT - json

Question

For the image tag, I need to keep only the image file with "file/" content using XSLT in the JSON output:
My input XML file is:
<image>binary/alias/my.jpg</image>
XSL which is used as:
<xsl:template match="image">
image: <xsl:apply-templates/>,
</xsl:template>
JSON output which I get is:
image: binary/alias/my.jpg
I need the output as:
image: files/my.jpg
Please help me on this. Thanks in advance.

Answer 1

In XSLT 2.0 you can do:
<xsl:template match="image">
<xsl:text>image: files/</xsl:text>
<xsl:value-of select="tokenize(., '/')[last()]"/>
</xsl:template>

Answer 2

To get the string after the last occurrence of a char (in this case '/') you need (in XSLT-1.0) a recursive template. Applying this template, the solution is straightforward: Output the desired prefix text 'image: files/' and append the result of the recursive template:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="image">
image: files/<xsl:call-template name="LastOccurrence">
<xsl:with-param name="value" select="text()" />
<xsl:with-param name="separator" select="'/'" />
</xsl:call-template>
</xsl:template>
<xsl:template name="LastOccurrence">
<xsl:param name="value" />
<xsl:param name="separator" select="'/'" />
<xsl:choose>
<xsl:when test="contains($value, $separator)">
<xsl:call-template name="LastOccurrence">
<xsl:with-param name="value" select="substring-after($value, $separator)" />
<xsl:with-param name="separator" select="$separator" />
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$value" />
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
The LastOccurrence template was inspired by this SO post.