xslt 1.0 - translate stirng backslash to double backslash - json

Question

My xml:
<workorder>
<specifications>
<hpath>94 \ 72</hpath>
<classdesc></classdesc>
</specifications>
</workorder>
my xslt:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" version="1.0">
<xsl:output method="text" encoding="UTF-8" />
<xsl:template match="/">
<xsl:apply-templates />
</xsl:template>
<xsl:template match="workorder">
<xsl:apply-templates select="specifications" />
</xsl:template>
<xsl:template match="specifications">
<xsl:text>{
"Hierarchy Path"="</xsl:text>
<xsl:value-of select="translate(//workorder/specifications/hpath,'\','\\')" />
<xsl:text>"
}</xsl:text>
</xsl:template>
</xsl:stylesheet>
Expecting output is:
{
"Hierarchy Path"="94 \\ 72"
}
Current output is :
{
"Hierarchy Path"="94 \ 72"
}
Problem is: When I send json format as "Current output", not a valid json format.
If I send as "Expected output", it is a valid json format.
Please help. Thanks in advance.

Answer 1

With XSLT 1.0 you need a recursive approach.
You may try something like this named template.
<xsl:template name="jsonescape">
<xsl:param name="str" select="."/>
<xsl:choose>
<xsl:when test="contains($str, '\')">
<xsl:value-of select="concat(substring-before($str, '\'), '\\' )"/>
<xsl:call-template name="jsonescape">
<xsl:with-param name="str" select="substring-after($str, '\')"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$str"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
And call it with your strings(for example).
<xsl:call-template name="jsonescape">
<xsl:with-param name="str" select="//workorder/specifications/hpath"/>
</xsl:call-template>