CODEWITHSUNDEEP
xsltconditional-statements
The Nightmare
The Nightmare

Reputation: 701

Condition in xsl:stylesheet

I have a few the same files with one little difference. Only difference between this files is declaration in xsl:stylesheet. Example:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions" xmlns:sf="http://www.company.pl/sf"
        xmlns:fo="http://www.w3.org/1999/XSL/Format" exclude-result-prefixes="fo" xmlns:iit="http://www.abc.qwerty.com/scheme/AFQ/DataTypeStructureDef/2018/07/09/InsuranceInThousands" xmlns:is="http://www.abc.qwerty.com/scheme/AFQ/DataTypeStructureDef/2018/07/09/InsuranceStruct">

and in other file I have only difference in iit param

xmlns:iit="http://www.abc.qwerty.com/scheme/AFQ/DataTypeStructureDef/2018/07/09/InsuranceInHundreds"

In effect I have many the same files with this one difference... In other cases I resolve this problem by pass param to my xslt, example:

<xsl:param name="tagCount"/>

but In xsl:stylesheet I don't know If I can pass params and add if condition? How I can reach this effect?

Upvotes: 0

Views: 52

Answers (1)

Michael Kay
Michael Kay

Reputation: 163342

When you're dealing with multiple source documents that have very similar structure but different namespace URIs, I think the best strategy is often to preprocess the source documents so they all use the same namespace URI. That can be done with a simple transformation like this:

<xsl:param name="input-namespace"/>
<xsl:param name="output-namespace"/>
<xsl:template match="*[namespace-uri() = $input-namespace]">
  <xsl:element name="name()" namespace-uri="{$output-namespace}">
    <xsl:copy-of select="@*"/>
    <xsl:apply-templates/>
  </xsl:element>
</xsl:template>

Alternatively, write a SAX filter.

Once you've eliminated the namespace differences, the main stylesheet becomes much simpler and less cluttered.

Upvotes: 1

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