Disable xsl:comment in XSLT transformation - xslt

I have an XSLT file littered with comments such as the following:
<xsl:comment>Entering shipping block</xsl:comment>
Is there a way to explicity disable these comments so that they aren't output at runtime in production? The output of the XSLT file is shown in a public API, so while it is useful for debugging, I would rather be able to switch it off.
The only way I can think of is to have a flag that is set in development mode to turn on the comments:
<xsl:if test="$enableDebug='true'">
<xsl:comment>Entering shipping block</xsl:comment>
</xsl:if>
Is there another way?
(I'm using XSLT 2.0.)

Just include this transformation step in your deployment to production:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="comment()"/>
</xsl:stylesheet>
You my even not specify any indent attribute on <xsl:output> in case readability is not a goal.

You could declare a variable verbose at the very top of your XSL:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output encoding="UTF-8" indent="yes" method="xml"/>
<xsl:variable name="verbose">1</xsl:variable>
<xsl:template match="#*|node()">
<xsl:copy>
...
and wrap your comments in <xsl:if> like this:
<xsl:if test="$verbose">
<xsl:comment>Applying the template...</xsl:comment>
</xsl:if>
If you then want to temporarily deactivate the comments, just remove the 1 from the verbose variable:
<xsl:variable name="verbose"></xsl:variable>

Related

xsl:apply-templates returns nothing − what am I missing?

I have a simple XML response, like
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<searchRetrieveResponse xmlns="http://www.loc.gov/zing/srw/">
<numberOfRecords>1</numberOfRecords>
<records>
<record>
<recordData>
<kitodo xmlns="http://meta.kitodo.org/v1/">
<metadata name="key1">value1</metadata>
<metadata name="key2">value2</metadata>
<metadata name="key3">value3</metadata>
</kitodo>
</recordData>
</record>
</records>
</searchRetrieveResponse>
which I want to transform to this by XSLT
<?xml version="1.0" encoding="utf-8"?>
<mets:mdWrap xmlns:kitodo="http://meta.kitodo.org/v1/"
xmlns:mets="http://www.loc.gov/METS/"
xmlns:srw="http://www.loc.gov/zing/srw/"
MDTYPE="OTHER"
OTHERMDTYPE="Kitodo">
<mets:xmlData>
<kitodo:kitodo>
<kitodo:metadata name="key1">value1</kitodo:metadata>
<kitodo:metadata name="key2">value2</kitodo:metadata>
<kitodo:metadata name="key3">value3</kitodo:metadata>
</kitodo:kitodo>
</mets:xmlData>
</mets:mdWrap>
That is, I want to remove the outside tree searchRetrieveResponse/records/record/recordData, replace it with mdWrap/xmlData and move the contained data node there.
I have a quite short XSLT for it:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:kitodo="http://meta.kitodo.org/v1/" xmlns:mets="http://www.loc.gov/METS/" xmlns:srw="http://www.loc.gov/zing/srw/">
<xsl:output method="xml" indent="yes" encoding="utf-8"/>
<xsl:strip-space elements="*"/>
<xsl:template match="srw:recordData">
<mets:mdWrap MDTYPE="OTHER" OTHERMDTYPE="Kitodo">
<mets:xmlData>
<xsl:apply-templates select="#*|node()"/>
</mets:xmlData>
</mets:mdWrap>
</xsl:template>
<!-- pass-through rule -->
<xsl:template match="#*|node()">
<xsl:apply-templates select="#*|node()"/>
</xsl:template>
</xsl:stylesheet>
However, what I get is:
<?xml version="1.0" encoding="utf-8"?>
<mets:mdWrap xmlns:kitodo="http://meta.kitodo.org/v1/"
xmlns:mets="http://www.loc.gov/METS/"
xmlns:srw="http://www.loc.gov/zing/srw/"
MDTYPE="OTHER"
OTHERMDTYPE="Kitodo">
<mets:xmlData/>
</mets:mdWrap>
Obviously, the template match="srw:recordData" does match, otherwise I would get an empty result. However, the contained apply-templates doesn’t output anything. (I also tried an <xsl:apply-templates/> without a select="" attribute, but it doesn’t output anything either.) What am I missing?
XSLT processor is net.sf.saxon.TransformerFactoryImpl (Java)
I think nothing happens when you are applying templates inside xmlData. There are no templates that would match descendant nodes.
Try using copy-of:
<xsl:template match="srw:recordData">
<mets:mdWrap MDTYPE="OTHER" OTHERMDTYPE="Kitodo">
<mets:xmlData>
<xsl:copy-of select="kitodo:kitodo"/>
</mets:xmlData>
</mets:mdWrap>
</xsl:template>
The problem is not with the xsl:apply-templates instruction. It is with the template being applied. Your "pass-through rule" does not write anything to the output. You probably meant to have the identity transform template in that place - which goes like this:
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>

XML with empty prefix transformation with XSLT [duplicate]

This question already has answers here:
XSLT Transform doesn't work until I remove root node
(2 answers)
Closed 1 year ago.
I'm willing to use XSLT to transform XML files to other XML files by removing (TextLine) elements. However, the elements are not removed as I expect in the output XML files. I imagine that I'll have to modify the XSLT file, but I don't know how. Let me know what should be done.
I suspect that the root cause of the issue is that elements in the XML files have an empty prefix namespace.
The details are the following ones.
An XML test-01.xml file that contains empty prefix namespace elements:
<?xml version="1.0" encoding="UTF-8"?>
<alto xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://www.loc.gov/standards/alto/ns-v4#"
xsi:schemaLocation="http://www.loc.gov/standards/alto/ns-v4# http://www.loc.gov/standards/alto/v4/alto-4-2.xsd">
<TextLine TAGREFS="LT9"/>
<TextLine TAGREFS="LT10"/>
<TextLine TAGREFS="LT9"/>
<TextLine TAGREFS="LT8"/>
</alto>
And I'm using the following date.xslt file:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="TextLine"/>
</xsl:stylesheet>
Note: I'm using python lxml to perform the transformation. However, this shouldn't have any influence on the process as I could use any other XML transformer as xsltproc.
Yes. Your assumption that that the default namespace was the cause of your XSLT not functioning as desired was correct. Try this XSLT-1.0 instead:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:loc="http://www.loc.gov/standards/alto/ns-v4#">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="loc:TextLine"/>
</xsl:stylesheet>

XPath 3.0 Serialize without Namespaces in Scope

While answering this question, it occurred to me that I know how to use the XSLT 3.0 (XPath 3.0) serialize() function, but that I do not know how to avoid serialization of namespaces that are in scope. Here is a minimal example:
XML Input
<?xml version="1.0" encoding="UTF-8" ?>
<ci:cichlids xmlns:ci="http://www.cichlids.com">
<cichlid id="1">
<name>Zeus</name>
<color>gold</color>
<teeth>molariform</teeth>
<breeding-type>lekking</breeding-type>
</cichlid>
</ci:cichlids>
XSLT 3.0 Stylesheet
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:output="http://www.w3.org/2010/xslt-xquery-serialization"
xmlns:ci="http://www.cichlids.com">
<xsl:output method="xml" encoding="UTF-8" indent="yes" />
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/ci:cichlids/cichlid">
<xsl:variable name="serial-params">
<output:serialization-parameters>
<output:omit-xml-declaration value="yes"/>
</output:serialization-parameters>
</xsl:variable>
<xsl:value-of select="serialize(., $serial-params/*)"/>
</xsl:template>
</xsl:stylesheet>
Actual Output
<?xml version="1.0" encoding="UTF-8"?>
<ci:cichlids xmlns:ci="http://www.cichlids.com">
<cichlid xmlns:ci="http://www.cichlids.com" id="1">
<name>Zeus</name>
<color>gold</color>
<teeth>molariform</teeth>
<breeding-type>lekking</breeding-type>
</cichlid>
</ci:cichlids>
The serialization process included the namespace declaration that is in scope for the cichlid element, although it is not used on this element. I would like to remove this declaration and make the output look like
Expected Output
<?xml version="1.0" encoding="UTF-8"?>
<ci:cichlids xmlns:ci="http://www.cichlids.com">
<cichlid id="1">
<name>Zeus</name>
<color>gold</color>
<teeth>molariform</teeth>
<breeding-type>lekking</breeding-type>
</cichlid>
</ci:cichlids>
I know how to modify the cichlid element, removing the namespaces in scope, and serialize this modified element instead. But this seems a rather cumbersome solution. My question is:
What is a canonical way to serialize an XML element using the serialize() function without also serializing unused namespace declarations that are in scope?
Testing with Saxon-EE 9.6.0.7 from within Oxygen.
Serialization will always give you a faithful representation of the data model that you are serializing. If you want to modify the data model, that's called transformation. Run a transformation to remove the unwanted namespaces, then serialize the result.
Michael Kay already gave the correct answer and I have accepted it. This is just to flesh out his comments. By
Run a transformation to remove the unwanted namespaces, then serialize the result.
he means applying a transformation like the following before calling serialize():
XSLT Stylesheet
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:output="http://www.w3.org/2010/xslt-xquery-serialization"
xmlns:ci="http://www.cichlids.com">
<xsl:output method="xml" encoding="UTF-8" indent="yes" />
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:variable name="cichlid-without-namespace">
<xsl:copy-of copy-namespaces="no" select="/ci:cichlids/cichlid"/>
</xsl:variable>
<xsl:template match="/ci:cichlids/cichlid">
<xsl:variable name="serial-params">
<output:serialization-parameters>
<output:omit-xml-declaration value="yes"/>
</output:serialization-parameters>
</xsl:variable>
<xsl:value-of select="serialize($cichlid-without-namespace, $serial-params/*)"/>
</xsl:template>
</xsl:stylesheet>
XML Output
<?xml version="1.0" encoding="UTF-8"?>
<ci:cichlids xmlns:ci="http://www.cichlids.com">
<cichlid id="1">
<name>Zeus</name>
<color>gold</color>
<teeth>molariform</teeth>
<breeding-type>lekking</breeding-type>
</cichlid>
</ci:cichlids>

How to copy a certain node (with children) from a XML with XSLT in Biztalk specifying a custom namespace?

I need to copy a subnode from a XML to a certain node of a new XML in a Biztalk Map using XSLT.
Consider the following input XML:
<?xml version="1.0" encoding="UTF-8"?>
<ns0:root xmlns:ns0="http://not/useful/data/">
<ns0:notuseful>
<ns0:foo></ns0:foo>
<ns0:foo2></ns0:foo2>
<ns0:blabla></ns0:blabla>
</ns0:notuseful>
<ns0:data>
<ns1:usefulDataList xmlns:ns1="http://useful/data/">
<ns1:usefulData>
<ns1:usefulChild1></ns1:usefulChild1>
<ns1:usefulChild2></ns1:usefulChild2>
<ns1:usefulChild3></ns1:usefulChild3>
<ns1:usefulChild4></ns1:usefulChild4>
<ns1:usefulChild5></ns1:usefulChild5>
</ns1:usefulData>
</ns1:usefulDataList>
</ns0:data>
<ns0:root>
What I need is to extract the node called "usefulDataList", so I need to copy it in a new XML like this one:
<?xml version="1.0" encoding="UTF-8"?>
<ns2:root2 xmln:ns2="http://new/xml">
<ns2:blabla>
<ns2:stuff />
</ns2:blabla>
<ns2:data>
<ns2:usefulDataList>
<ns2:usefulData>
<ns2:usefulChild1></ns2:usefulChild1>
<ns2:usefulChild2></ns2:usefulChild2>
<ns2:usefulChild3></ns2:usefulChild3>
<ns2:usefulChild4></ns2:usefulChild4>
<ns2:usefulChild5></ns2:usefulChild5>
</ns2:usefulData>
</ns2:usefulDataList>
</ns2:data>
</ns2:root2>
This should be done inside a Biztalk Functoid, as you see namespaces from source and target are diferent.
I'm an absolute beginner with XSLT, and I've been doing some tests, but I've something wrong with my XSLT expressions:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ns2="http://new/xml">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" />
<xsl:template name="testTemplate" match="//*[local-name() = 'usefulDataList ']">
<xsl:element name="ns0:usefulDataList " namespace="">
<xsl:apply-templates mode="copy-no-ns" select="usefulDataList"/>
</xsl:element>
</xsl:template>
<xsl:template mode="copy-no-ns" match="*">
<xsl:element name="{name(.)}" namespace="{namespace-uri(.)}">
<xsl:copy-of select="#*"/>
<xsl:apply-templates mode="copy-no-ns"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
I'd appreciate any tip, with XSLT or Biztalk mapper. I don't like linking a huge amount of fields one by one if I can solve it with a XSLT expression.
Greetings.
Beware you had a space in *[local-name() = 'usefulDataList ']" so that would never match. this works:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ns2="http://new/xml">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" />
<xsl:strip-space elements="*"/>
<xsl:template match="/">
<ns2:root>
<ns2:blabla>
<ns2:stuff />
</ns2:blabla>
<ns2:data>
<xsl:apply-templates mode="copy-no-ns" select="//*[local-name() = 'usefulDataList']"/>
</ns2:data>
</ns2:root>
</xsl:template>
<xsl:template mode="copy-no-ns" match="*">
<xsl:element name="ns2:{local-name(.)}">
<xsl:copy-of select="#*"/>
<xsl:apply-templates mode="copy-no-ns"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

Removing an XML tag that is named like xfdf:field (with a namespace)

I want to remove an XML element from an XML file. The tag that I want to remove is named as xfdf:field.
How do I specify this in my xslt ? I tried this and I am getting an error saything "org.apache.xpath.domapi.XPathStylesheetDOM3Exception: Prefix must resolve to a namespace: xfdf ".
Here is my xslt.
<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:template match="xfdf:field"></xsl:template>
</xsl:stylesheet>
Here is my xml.
<xfa:datasets
xmlns:xfa="http://www.xfa.org/schema/xfa-data/1.0/"
xmlns:dd="http://ns.adobe.com/data-description/" xmlns:tns="http://hostname"
xmlns:xfdf="http://ns.adobe.com/xfdf/">
<xfa:data>
<tns:form xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<tns:formHeader>
<tns:formId>formid</tns:formId>
<tns:revId>Rev123</tns:revId>
<tns:agencyId>agency</tns:agencyId>
<tns:progId>program</tns:progId>
<tns:serviceId>service</tns:serviceId>
</tns:formHeader>
<tns:formFields>
<tns:date>08-13-1967</tns:date>
<tns:agreementBetween></tns:agreementBetween>
<tns:ncr>xxxx</tns:ncr>
<tns:formConfirmationInfo>
<tns:confNbrLbl>nbrlabel</tns:confNbrLbl>
<tns:confNbrData>1231</tns:confNbrData>
<tns:custNameLbl>3332</tns:custNameLbl>
<tns:custNameData>dasdas</tns:custNameData>
<tns:dateLbl>date</tns:dateLbl>
<tns:dateData>01012001</tns:dateData>
</tns:formConfirmationInfo>
</tns:formFields>
<xfdf:field xmlns:xfdfi="http://ns.adobe.com/xfdf-transition/"
xfdfi:original="FSAPPLICATIONDATA_">
<!--irrelevant data omitted-->
</xfdf:field>
</tns:form>
</xfa:data>
</xfa:datasets>
You need to specify the namespace in your XSLT file:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xfdf="http://ns.adobe.com/xfdf/">