Changing the value of an attribute of the document element in an xml file using XSLT

Question

I am trying to change the value of an attribute of the document element in an XML file using XSLT transformation. For example,

<?xml version="1.0" encoding="UTF-8"?>
<ns1:xmlgMsc xmlns:ns1="org.example" formatVersion="1.0" name="BlaBlah" pathName="/system/abc.xml" writtenBy="Me me me">
   <ns1:blockRoot someAtt="0" anotherAtt="1" />
</ns1:xmlgMsc>

Here I would like to change the "pathName" to some another path (say "/local/xyz.xml"). Can somebody please provide the syntax or point me into the right direction for doing this in XSLT?

Thanks in advance!

Tomalak · Accepted Answer · 2013-11-05 13:35:56Z

3

You will need a stylesheet that consists of two templates. The identity template (look it up) and this one:

<xsl:template match="/*/@pathName">
  <xsl:attribute name="{name()}">
    <xsl:value-of select="'/local/xyz.xml'" />
  </xsl:attribute>
</xsl:template>

You could use an <xsl:param> to pass in the new path dynamically, if you don't want to hard-code the new value.

Minor correction: The root node (/) of an XML document does not have attributes. You mean the document element (/ns1:xmlgMsc), that's one level down the hierarchy.

edited Nov 5, 2013 at 13:35

answered Nov 4, 2013 at 15:36

Tomalak

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1 Comment

schwarz Over a year ago

I am afraid it didn't work me ... any idea if I am doing anything wrong? This is a small xslt I made; <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="w3.org/1999/XSL/Transform"> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()" /> </xsl:copy> </xsl:template> <xsl:template match="/*/@pathName"> <xsl:copy> <xsl:value-of select="'/local/xyz.xml'" /> </xsl:copy> </xsl:template> </xsl:stylesheet>

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