XSLT: How to change an attribute value during <xsl:copy>?
XsltXslt 1.0Xslt 2.0Xslt Problem Overview
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".
Xslt Solutions
Solution 1 - Xslt
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>
Solution 2 - Xslt
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.
Solution 3 - Xslt
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>
Solution 4 - Xslt
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.
Solution 5 - Xslt
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']
Solution 6 - Xslt
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>
Solution 7 - Xslt
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>
Solution 8 - Xslt
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.