How to get tag name of root element in an XML docu

问题:

I'm interested in assigning the tag name of the root element in an xml document to an xslt variable. For instance, if the document looked like (minus the DTD):

<foo xmlns="http://.....">
    <bar>1</bar>
</foo>

and I wanted to assign the string 'foo' to an xslt variable. Is there a way to reference that?

Thanks, Matt

回答1:

I think you want to retrieve the name of the outermost XML element. This can be done like in the following XSL sample:

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

  <xsl:variable name="outermostElementName" select="name(/*)" />

  <xsl:template match="/">
    <xsl:value-of select="$outermostElementName"/>
  </xsl:template>
</xsl:stylesheet>

Please note that there is a slight difference in XPath terminology:

The top of the tree is a root node (1.0 terminology) or document node (2.0). This is what "/" refers to. It's not an element: it's the parent of the outermost element (and any comments and processing instructions that precede or follow the outermost element). The root node has no name.

See http://www.dpawson.co.uk/xsl/sect2/root.html#d9799e301

回答2:

Use the XPath name() function.

One XPath expression to obtain the name of the top (not root!) element is:

       name(/*)

The name() function returns the fully-qualified name of the node, so for an element <bar:foo/> the string "bar:foo" will be returned.

In case only the local part of the name is wanted (no prefix and ":"), then the XPath local-name() function should be used.

回答3:

Figured it out. The function name() given the parameter * will return foo.

回答4:

you want local-name()