XSLT dont add XMLNS to elements - java

I have a simple HTML fragment similar to this:
link
I need to transform it to
<abc:href var="123">link</abc:href>
I do it with XSLT, so I had to add the namespace in xsl:stylesheet
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:abc="http://abc.ru">
It works almost fine, unfortunately the XSLT transform keeps on adding a XMLNS to the output, like here:
<abc:href var="123" xmlns:abc="http://abc.ru">link</abc:href>
I don't need the xmlns definition, can I remove it?

Although it really goes against the grain, and I advise strongly against it, if you need to produce this malformed XML, then you can use an instruction like...
<xsl:value-of disable-output-escaping="yes" select="
concat('<abc:href var="',$href,'">',$link,'</abc:href>')
"/>
... where $href and $link are place-markers for the appropriate expression.
Update
In response to the OP's comment, one could use a template like this...
<xsl:template match="a">
<xsl:value-of disable-output-escaping="yes" select="
concat('<abc:href var="',#href,'">',.,'</abc:href>')
"/>
</xsl:template>
This ugly solution should be used only as a last resort. A much better solution would be to use XSLT to produce your WHOLE document, not just an invalid fragment of it. This way you document would be well formed and you could bring to bear the full power and simplicity of XSLT.

It works almost fine, unfortunately the XSLT transform keeps on adding
a XMLNS to the output, like here:
<abc:href var="123" xmlns:abc="http://abc.ru">link</abc:href>
I don't need the xmlns definition, can I remove it?
The wanted removal of the namespace declaration would produce a (namespace-)non-well-formed XML document and for this reason the XSLT processor adds the namespace declaration -- as required by the W3C XSLT specifications.
You can cause these namespace declarations to "disappear" by placing the namespace declaration on a common ancestor (such as the top element of the generated XML document).
Here is a complete example:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/*">
<top xmlns:abc="http://abc.ru">
<xsl:apply-templates/>
</top>
</xsl:template>
<xsl:template match="a[#href]">
<xsl:element name="abc:href" namespace="http://abc.ru">
<xsl:attribute name="var">
<xsl:apply-templates/>
</xsl:attribute>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the following document:
<html>
link1
link2
link3
link4
</html>
the wanted, correct result is produced:
<top xmlns:abc="http://abc.ru">
<abc:href var="link1"/>
<abc:href var="link2"/>
<abc:href var="link3"/>
<abc:href var="link4"/>
</top>

This is sad, but I really need an invalid xml
XSLT is designed to prevent you producing bad XML. If you want to produce bad XML, don't use XSLT.

Try it with exclude-result-prefixes, like this:
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:abc="http://abc.ru"
exclude-result-prefixes="abc">
<xsl:template match="/">
<xsl:apply-templates select="#* | node()"/>
</xsl:template>
<xsl:template match="a">
<href var="{#href}"><xsl:value-of select="."/></href>
</xsl:template>
</xsl:stylesheet>

Related

XSL stylesheet not preserving the encoded values

I have an XML file that i am trying to read & write witin my java program.I am using SAXON 2.0 for this exercise. Below is a sample of the XML file that i have
<?xml version="1.0" encoding="UTF-8"?>
<CustomObject xmlns="http://soap.sforce.com/2006/04/metadata">
<actionOverrides>
<actionName>Accept</actionName>
<type>Default</type>
</actionOverrides>
<fields>
<fullName>Dev__c</fullName>
<externalId>false</externalId>
<formula>Sprint__r.Dev_Lead__r.FirstName + &apos; &apos; + Sprint__r.Dev_Lead__r.LastName</formula>
<formulaTreatBlanksAs>BlankAsZero</formulaTreatBlanksAs>
<label>Dev Lead</label>
<required>false</required>
<trackHistory>false</trackHistory>
<trackTrending>false</trackTrending>
<type>Text</type>
<unique>false</unique>
</fields>
<fields>
<fullName>Manager__c</fullName>
<externalId>false</externalId>
<formula>Sprint__r.Dev_Manager__r.FirstName +&apos; &apos;+
Sprint__r.Dev_Manager__r.LastName</formula>
<formulaTreatBlanksAs>BlankAsZero</formulaTreatBlanksAs>
<label>Dev Manager</label>
<required>false</required>
<trackHistory>false</trackHistory>
<trackTrending>false</trackTrending>
<type>Text</type>
<unique>false</unique>
</fields>
</CustomObject>
What happen is that when i run my data.xsl stylsheet with below code, all of the encoding is replaced with the actual value. In this case :
&apos; &apos;
is replace with ' '. What i want is to preserve the encoded values in the file. How can i make that happen ?. Also, the first line is replaced with . I want to preserve the xmlns value also i the file.
data.xsl
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xpath-default-namespace="http://soap.sforce.com/2006/04/metadata"
exclude-result-prefixes="#all"
version="2.0">
<xsl:output method="xml" indent="yes" />
<xsl:template match="*">
<xsl:element name="{local-name(.)}">
<xsl:apply-templates select="#* | node()" />
</xsl:element>
</xsl:template>
<xsl:template match="#*">
<xsl:attribute name="{local-name(.)}">
<xsl:value-of select="." />
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
Conceptually, an XSLT processor operates on trees of nodes. The input tree of nodes is constructed by an XML parser. It's the job of an XML parser to extract the information content of the XML, and to discard irrelevant distinctions that have no bearing on the information content - such as whitespace between the attributes in a start tag, the difference between single and double quotes around attribute values, the order of attributes, and the distinction between " and ". The XSLT processor doesn't even know how the XML was written, so it can't preserve such details.
As for the namespaces, when you do <xsl:element name="{local-name(.)}"> you are explicitly copying the local name but not the namespace. If you want to copy both, use <xsl:copy>, or <xsl:element name="{local-name(.)}" namespace="{namespace-url(.)">.

How to copy parent node namespace to child element using xslt?

My xml looks like which I created using Java JAXBContext and Marshaller.
I want to format some part of xml only not the whole xml.
<?xml version="1.0" encoding="UTF-8"?>
<ns4:Requests xmlns:ns2="http://www.dummy.com/xsd/tublu/murmur_001" xmlns:ns3="http://www.dummy.com/xsd/CommonObjects_001" xmlns:ns4="http://www.dummy.com/xsd/naku_001">
<ns4:RequestSetId>fhskgvseruigiu</ns4:RequestSetId>
<ns4:RequestStream>CHAPP</ns4:RequestStream>
<ns4:Request>
<ns4:TrackAndTrace>
<ns4:CPAId>003</ns4:CPAId>
<ns4:CorrelationId>ytuty</ns4:CorrelationId>
</ns4:TrackAndTrace>
</ns4:Request>
<ns4:Request>
<ns4:TrackAndTrace>
<ns4:CPAId>003</ns4:CPAId>
<ns4:CorrelationId>cyuri7</ns4:CorrelationId>
</ns4:TrackAndTrace>
</ns4:Request>
</ns4:Requests>
I want to format like
<?xml version="1.0" encoding="UTF-8"?>
<ns4:Requests xmlns:ns2="http://www.dummy.com/xsd/tublu/murmur_001" xmlns:ns4="http://www.dummy.com/xsd/naku_001" xmlns:ns3="http://www.dummy.com/xsd/CommonObjects_001">
<ns4:RequestSetId>fhskgvseruigiu</ns4:RequestSetId>
<ns4:RequestStream>CHAPP</ns4:RequestStream>
<ns4:Request xmlns:ns4="http://www.dummy.com/xsd/naku_001"><ns4:TrackAndTrace><ns4:CPAId>003</ns4:CPAId><ns4:CorrelationId>ytuty</ns4:CorrelationId></ns4:TrackAndTrace></ns4:Request>
<ns4:Request xmlns:ns4="http://www.dummy.com/xsd/naku_001"><ns4:TrackAndTrace><ns4:CPAId>003</ns4:CPAId><ns4:CorrelationId>cyuri7</ns4:CorrelationId></ns4:TrackAndTrace></ns4:Request>
</ns4:Requests>
Here is the solution (by Transforming the XML Data using Java's XSLT APIs),
As you may also have noticed.. JAXB alone cannot meet this requirement, but after marshalling the object to a formatted XML String (as u have shown) you can then post process/transform it accordingly using a suitable XSLT file
So to get a linearized 'Request' element, just make use of the xsl shown below:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:strip-space elements="TrackAndTrace"/>
<xsl:strip-space elements="Request"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Note: Also tested that above method/approach is working properly - used the Stylizer sample code (from https://docs.oracle.com/javase/tutorial/jaxp/xslt/transformingXML.html)
Cheers!
Update: If you want a solution that also preserves the original namespace prefix as shown in your question, follow this variation
Add factory.setNamespaceAware(true); in the Stylizer code
& Use this tweaked XSLT
<xsl:stylesheet version="1.0" xmlns:xsl="w3.org/1999/XSL/Transform" xmlns:ns4="dummy.com/xsd/naku_001">
<xsl:strip-space elements="ns4:TrackAndTrace"/>
<xsl:strip-space elements="ns4:Request"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

add namespace and prefix to xml

I am using camel to route messages to a Webservice. The Messages are like XML but without namespaces/prefixes. The problem now is that the Webservice expects the XML but with the appropriate namespaces for each element. So as an example:
<a>
<b>value_b</b>
<c>value_c</c>
</a>
is what im getting in, but what needs to be sent out should look like this
<a xmlns:n1="http://yadda-ns1.com" xmlns:n2="http://yadda-ns2.com">
<ns1:b>value_b</ns1:b>
<ns2:c>value_c</ns2:c>
</a>
if it was the same namespace on all elements i would have just used an xslt to add it. But its mostly 2 or 3 different namespaces.
Now is it possible to add the namespaces in my camel route? I had the idea to use jaxb to marshal from the "incomplete" XML to the "complete" one (with XML), would this work? I was trying this but did not succeed yet.
Or does someone have a different idea? What i also have in my project is the XSDs and JAXB Annotated Classes so these can also be used and the messages are identical apart from the missing namespace.
Best Regards
Thomas
You could transform the XML with a stylesheet such as the one below to modify the elements to be bound to the appropriate namespaces:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"
xmlns:ns1="http:yadayada-ns1.com"
xmlns:ns2="http:yadayada-ns2.com">
<xsl:output indent="yes"/>
<!-- identity template that copies content(unless more specific templates match) -->
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<!-- Make the "a" elements in the ns1 namespace -->
<xsl:template match="a">
<xsl:element name="ns1:{local-name()}">
<xsl:apply-templates select="#*|node()"/>
</xsl:element>
</xsl:template>
<!-- Make the "b" and "c" elements in the ns2 namespace -->
<xsl:template match="b|c">
<xsl:element name="ns2:{local-name()}">
<xsl:apply-templates select="#*|node()"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

Create xmlns attribute in the XML using XSLT Transformation

I am trying to add the xmlns attribute to the resulting XML with a value passed by parameter during XSLT transformation using JDK Transformer (Oracle XML v2 Parser or JAXP) but it always defaults to http://www.w3.org/2000/xmlns/
My source XML
<test/>
My XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://example.com">
<xsl:param name="myNameSpace" select="'http://neilghosh.com'"/>
<xsl:template match="/">
<process>
<xsl:attribute name="xmlns:neil">
<xsl:value-of select="$myNameSpace"/>
</xsl:attribute>
</process>
</xsl:template>
</xsl:stylesheet>
My Result
<?xml version="1.0"?>
<process xmlns="http://www.w3.org/2000/xmlns/" xmlns:neil="neilghosh.com">
</process>
My Desired Result
<?xml version="1.0"?>
<process xmlns="http://example.com" xmlns:neil="neilghosh.com">
</process>
Firstly, in the XSLT data model, you don't want to create an attribute node, you want to create a namespace node.
Namespace nodes are usually created automatically: if you create an element or attribute in a particular namespace, the requisite namespace node (and hence, when serialized, the namespace declaration) are added automatically by the processor.
If you want to create a namespace node that isn't necessary (because it's not used in the name of any element or attribute) then in XSLT 2.0 you can use xsl:namespace. If you're stuck with XSLT 1.0 then there's a workaround, that involves creating an element in the relevant namespace and then copying its namespace node:
<xsl:variable name="ns">
<xsl:element name="neil:dummy" namespace="{$param}"/>
</xsl:variable>
<process>
<xsl:copy-of select="$ns/*/namespace::neil"/>
</process>
Michael Kay provided you with the correct answer, but based on your comments, you aren't sure how to use it in your transformation.
Here is a complete transformation:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ext="http://exslt.org/common" exclude-result-prefixes="ext">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:param name="pNamespace" select="'neilghosh.com'"/>
<xsl:variable name="vDummy">
<xsl:element name="neil:x" namespace="{$pNamespace}"/>
</xsl:variable>
<xsl:template match="/*">
<xsl:element name="process" namespace="http://example.com">
<xsl:copy-of select="namespace::*"/>
<xsl:copy-of select="ext:node-set($vDummy)/*/namespace::*[.=$pNamespace]"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided XML document:
<test/>
the wanted, correct result is produced:
<process xmlns="http://example.com" xmlns:neil="neilghosh.com" />
Namespace declarations in XML are not attributes even though they look like attributes. In XSLT 2.0 you can use <xsl:namespace name="neil" select="$myNameSpace" /> to add a namespace declaration to the result tree dynamically but that feature is not available in XSLT 1.0.
Don't try to create "xmlns" attributes yourself. Create the namespaces in the XSLT and they will be done automatically.
This XSLT works (tested with Saxon 9.4):
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:neil="neilghosh.com"
xpath-default-namespace="http://example.com"
xmlns="http://example.com" version="2.0">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:param name="myDynamicNamespace" select="'http://neilghosh.com'"/>
<xsl:template match="/">
<xsl:element name="process">
<xsl:namespace name="neil" select="$myDynamicNamespace"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
And gives the following output:
<?xml version="1.0" encoding="UTF-8"?>
<process xmlns="http://example.com" xmlns:neil="http://neilghosh.com"/>
Finally got an workaround which worked with my XSLT Processor (Oracle XML V2 Parser)
I had to transform it to a DOM Document and then persist that DOM to filesystem instead of outputting directly to StreamResult
I used DOMResult in the transform method
Following XSLT fragment worked but there was an extra xmlns:xmlns="http://www.w3.org/2000/xmlns/" which was probably absorbed by Document and did not appear in the final output when I persisted to file system.
<process>
<xsl:attribute name="xmlns">
<xsl:value-of select="'http://example.com'"/>
</xsl:attribute>
<process>
I know this is not the best way to do but given the parse constraint this is the only choice I have now.

Java Style XML one node per line and no whitespace

Just fooling with removing whitespace but keeping each node on its own line from an xml document when adding and removing elements from xml in java and I'm having trouble understanding XML Style Sheets.
Here is what's happening so far.
Firstly I have the following XML,
<jobs>
<job>Job 1</job>
<job>Job 2</job>
<job>Job 3</job>
<job>Job 4</job>
</jobs>
Then I remove one of the elements and it ends up looking like this with the whitespacewhere the element was,
<jobs>
<job>Job 1</job>
<job>Job 3</job>
<job>Job 4</job>
</jobs>
So I tried applying the following style sheet I found,
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" omit-xml-declaration="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Which makes the xml appear on one line because it removes all whitespace. But I'm trying to keep the file readable too.
<jobs><job>Job 1</job><job>Job 2</job><job>Job 3</job><job>Job 4</job></jobs>
I was wondering if anyone has a style sheet to achieve this?
You need to add indent="yes" to <xsl:output:
<xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
(Also, you might want to switch to XSL Version 2.0)
Hope this helps

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