Subject: Re: [xsl] XSL/XPath to generate a list of ancestors? From: Nathan Potter <ndp@xxxxxxxxxxxxxxxxxxxx> Date: Mon, 12 May 2008 09:54:24 -0700 |
<xsl:template match="*" name="fullNameWorker" mode="fullName"> <xsl:if test=".!=/"> <xsl:apply-templates select=".." mode="fullName"/> <xsl:if test="..!=/">.</xsl:if> <xsl:value-of select="@name"/> </xsl:if> </xsl:template>
Never use != to compare node identity. It can be very expensive and it gives
the wrong answer. For example if your document is
<doc><subdoc> ... </subdoc></doc>
then doc and subdoc both compare equal to "/", and if the document is 100Mb
in size then you will be comparing some very long strings to prove it.
In 2.0, use "is". In 1.0, use generate-id(A)=generate-id(B).
<xsl:template match="*" name="fullNameWorker" mode="fullName"> <xsl:if test="generate-id(.)!=generate-id(/)"> <xsl:apply-templates select=".." mode="fullName"/> <xsl:if test="generate-id(..)!=generate-id(/)">.</xsl:if> <xsl:value-of select="@name"/> </xsl:if> </xsl:template>
<xsl:template name="fullName"> <fullName> <xsl:call-template name="fullNameWorker" /> </fullName > </xsl:template>
<xsl:template match="*" name="fullNameWorker" mode="fullName"> <xsl:if test=".!=/"> <xsl:apply-templates select=".." mode="fullName"/> <xsl:if test="..!=/">.</xsl:if> <xsl:value-of select="@name"/> </xsl:if> </xsl:template>
<xsl:template match="*"> <myNewNode> <xsl:call-template name="fullName"> </myNewNode> </xsl:template>
Michael Kay http://www.saxonica.com/
============================================================ Nathan Potter Oregon State University, COAS ndp at coas.oregonstate.edu 104 Ocean. Admin. Bldg. 541 737 2293 voice Corvallis, OR 97331-5503 541 737 2064 fax
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