Subject: Re: [xsl] XSL comparing nodesets by name only From: Lars Huttar <lars_huttar@xxxxxxx> Date: Mon, 15 Nov 2010 06:38:45 -0600 |
On 11/15/2010 6:24 AM, Piet van Oostrum wrote: > Markus Ohlenroth wrote: > > > I use XSLT 1.0 > > > > Given the following nodesets: > > > > <data:me1> <a></a> <b/> </data:me1> > > > > <data:me2> <a>value</a> <dd></dd> </data:me2> > > > > > > <xsl:variable name="me1" select="//data:me1/*"/> <xsl:variable > > name="me2" select="//data:me2/*"/> > > > > I want to find out if the two nodesets share one or more elements. I > > only want a comparison regarding their nodenames not the values of > > the nodes. In the above example $me1 and $me2 share the name of one > > element: and that is the element "<a/>". So my nodeset comparison > > should return "true". > > In XSLT 1.0: > > <xsl:value-of select="name($me1) = name($me2)"/> > Piet, it seems to me that will work like <xsl:value-of select="name($me1[1]) = name($me2[1])"/> In other words, it will only evaluate to true if the *first* element of $me1 and the *first* element of $me2 have the same name. "The *name <http://www.w3.org/TR/xpath/#function-name>* function returns a string containing a QName <http://www.w3.org/TR/REC-xml-names#NT-QName> representing the expanded-name <http://www.w3.org/TR/xpath/#dt-expanded-name> of the node in the argument node-set that is first in document order <http://www.w3.org/TR/xpath/#dt-document-order>." Lars
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