...
Using this technique will allow you to compare the content of <FieldWithData> as XML, not as a one long value of a single field.
In case this XML is in addition stored in the CDATA, you can try to reuse the following XSLT mapping:
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<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:element name="FieldWithData">
<xsl:value-of select="//FieldWithData" disable-output-escaping="yes"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet> |
XSLT for other encodings than UTF-8
In case Suppose your XML is encoded not in UTF-8, and you are missing some characters during output comparison in IFTT. In that case, you can use below the simple XSLT that provided below, which will remove encoding and allow the XSLT engine to convert the message to UTF-8. This is implicit XSLT engine conversion.
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<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes"/> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> </xsl:stylesheet> |
Different order of XML nodes in a new solution
Different order of XML nodes in a new solution
Sorting XML
If your reference and current XML structures are in a different order you can run below XSLT to sort them before comparision:
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<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="<PARENT>">
<xsl:copy>
<xsl:apply-templates select="@* | node()[not(self::<CHILD>)]"/>
<xsl:apply-templates select="<CHILD>">
<xsl:sort select="<SORT_BY>" order="ascending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet> |
<PARENT> - node name of the parent node of the node to be sorted
<CHILD> - node to be sorted
<SORT_BY> - by which field in <CHILD> it should be sorted