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Failed to compile stylesheet: Focus for child axis is abs...
» include-wrapper-for-json-input.xsl
Martin Honnen
, 2020-06-28 16:04
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
xmlns:xsl=
"http://www.w3.org/1999/XSL/Transform"
xmlns:xs=
"http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes=
"#all"
version=
"3.0"
>
<xsl:param
name=
"method"
as=
"xs:string"
static=
"yes"
select=
"'json'"
/>
<xsl:param
name=
"xslt-uri"
static=
"yes"
as=
"xs:string?"
select=
"'group-sales-sum-product.xsl'"
/>
<xsl:include
_href=
"{$xslt-uri}"
/>
<xsl:param
name=
"json-uri"
as=
"xs:string?"
select=
"'sales.json'"
/>
<xsl:param
name=
"json-content"
as=
"xs:string?"
select=
"()"
/>
<xsl:variable
name=
"input"
as=
"item()"
select=
"if (not(empty($json-uri))) then json-doc($json-uri) else parse-json($json-content)"
/>
<xsl:output
_method=
"{$method}"
indent=
"yes"
/>
<xsl:template
name=
"xsl:initial-template"
>
<xsl:apply-templates
select=
"$input"
/>
</xsl:template>
</xsl:stylesheet>
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