CODEWITHSUNDEEP
xsltxslt-1.0xslt-2.0
tomato
tomato

Reputation: 5694

XSLT: How to change an attribute value during <xsl:copy>?

I have an XML document, and I want to change the values for one of the attributes.

First I copied everything from input to output using:

<xsl:template match="@*|node()">
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>

And now I want to change the value of the attribute "type" in any element named "property".

Upvotes: 68

Views: 107234

Answers (8)

Ashish Lahoti
Ashish Lahoti

Reputation: 1008

I also came across same issue and i solved it as follows:

<!-- identity transform -->
<xsl:template match="@*|node()">
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>

<!-- copy property element while only changing its type attribute -->
<xsl:template match="property">
  <xsl:copy>
    <xsl:attribute name="type">
      <xsl:value-of select="'your value here'"/>
    </xsl:attribute>
    <xsl:apply-templates select="@*[not(local-name()='type')]|node()"/>
  </xsl:copy>
</xsl:template>

Upvotes: 2

michael.hor257k
michael.hor257k

Reputation: 116992

If your source XML document has its own namespace, you need to declare the namespace in your stylesheet, assign it a prefix, and use that prefix when referring to the elements of the source XML - for example:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xhtml="http://www.w3.org/1999/xhtml">

<xsl:output method="xml" encoding="utf-8" indent="yes" omit-xml-declaration="yes" />

<!-- identity transform -->
<xsl:template match="node()|@*">
    <xsl:copy>
        <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
</xsl:template>

<!-- exception-->    
<xsl:template match="xhtml:property/@type">
    <xsl:attribute name="type">
        <xsl:text>some new value</xsl:text> 
    </xsl:attribute>
</xsl:template>

</xsl:stylesheet>

Or, if you prefer:

...
<!-- exception-->    
<xsl:template match="@type[parent::xhtml:property]">
  <xsl:attribute name="type">
        <xsl:text>some new value</xsl:text> 
  </xsl:attribute>
</xsl:template>
...

ADDENDUM: In the highly unlikely case where the XML namespace is not known beforehand, you could do:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output method="xml" encoding="utf-8" indent="yes" omit-xml-declaration="yes" />

<!-- identity transform -->
<xsl:template match="node()|@*">
    <xsl:copy>
        <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
</xsl:template>

<!-- exception -->
<xsl:template match="*[local-name()='property']/@type">
    <xsl:attribute name="type">
        <xsl:text>some new value</xsl:text> 
    </xsl:attribute>
</xsl:template>

Of course, it's very difficult to imagine a scenario where you would know in advance that the source XML document contains an element named "property", with an attribute named "type" that needs replacing - but still not know the namespace of the document. I have added this mainly to show how your own solution could be streamlined.

Upvotes: 1

astonia
astonia

Reputation: 548

The top two answers will not work if there is a xmlns definition in the root element:

<?xml version="1.0"?>
<html xmlns="http://www.w3.org/1999/xhtml">
    <property type="old"/>
</html>

All of the solutions will not work for the above xml.

The possible solution is like:

<?xml version="1.0"?> 

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

  <xsl:output omit-xml-declaration="yes" indent="yes"/>
  <xsl:template match="node()[local-name()='property']/@*[local-name()='type']">
      <xsl:attribute name="{name()}" namespace="{namespace-uri()}">
                some new value here
          </xsl:attribute>
  </xsl:template>

  <xsl:template match="@*|node()|comment()|processing-instruction()|text()">
      <xsl:copy>
          <xsl:apply-templates select="@*|node()|comment()|processing-instruction()|text()"/>
      </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

Upvotes: 12

Dimitre Novatchev
Dimitre Novatchev

Reputation: 243449

This problem has a classical solution: Using and overriding the identity template is one of the most fundamental and powerful XSLT design patterns:

<xsl:stylesheet version="1.0" 
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output omit-xml-declaration="yes" indent="yes"/>

    <xsl:param name="pNewType" select="'myNewType'"/>

    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="property/@type">
        <xsl:attribute name="type">
            <xsl:value-of select="$pNewType"/>
        </xsl:attribute>
    </xsl:template>
</xsl:stylesheet>

When applied on this XML document:

<t>
  <property>value1</property>
  <property type="old">value2</property>
</t>

the wanted result is produced:

<t>
  <property>value1</property>
  <property type="myNewType">value2</property>
</t>

Upvotes: 66

rwrobson
rwrobson

Reputation: 21

I had a similar case where I wanted to delete one attribute from a simple node, and couldn't figure out what axis would let me read the attribute name. In the end, all I had to do was use

@*[name(.)!='AttributeNameToDelete']

Upvotes: 2

Welbog
Welbog

Reputation: 60418

Tested on a simple example, works fine:

<xsl:template match="@*|node()">
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>
<xsl:template match="@type[parent::property]">
  <xsl:attribute name="type">
    <xsl:value-of select="'your value here'"/>
  </xsl:attribute>
</xsl:template>

Edited to include Tomalak's suggestion.

Upvotes: 41

Andrew Hare
Andrew Hare

Reputation: 351516

For the following XML:

<?xml version="1.0" encoding="utf-8"?>
<root>
    <property type="foo"/>
    <node id="1"/>
    <property type="bar">
        <sub-property/>
    </property>
</root>

I was able to get it to work with the following XSLT:

<?xml version="1.0" encoding="utf-8"?>
<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="//property">
        <xsl:copy>
            <xsl:attribute name="type">
                <xsl:value-of select="@type"/>
                <xsl:text>-added</xsl:text>
            </xsl:attribute>
            <xsl:copy-of select="child::*"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

Upvotes: 1

Richard
Richard

Reputation: 109005

You need a template that will match your target attribute, and nothing else.

<xsl:template match='XPath/@myAttr'>
  <xsl:attribute name='myAttr'>This is the value</xsl:attribute>
</xsl:template>

This is in addition to the "copy all" you already have (and is actually always present by default in XSLT). Having a more specific match it will be used in preference.

Upvotes: 5

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