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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Determining whether no single instance of a specif
Well, it should be in the same namespace as count (that is, in the default fn: namespace). " The default prefix for the function namespace is fn:, and the URI is: http://www.w3.org/2005/02/xpath-functions. " However, I see you already have a working solution... michael -----Original Message----- From: Mark Peters [mailto:markpeters.work@xxxxxxxxx] Sent: 4. jzna 2009 21:07 To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: Determining whether no single instance of a specific node exists Hi Mtekel, I'm encountering the following error: XSLT Error (javax.xml.transform.TransformerException): Error in expression not(exists( (b/c) ) ): Unknown system function: exists Do I need to specify the URI for the XSLT namespace somewhere? Thanks, Mark On Thu, Jun 4, 2009 at 2:51 PM, Mtekel <thx@xxxxxxxxx> wrote: > Hello, > > <xsl: if test="not(exists( (b/c) ) )" > ... </xsl:if> > > Check other functions here: > http://www.w3schools.com/xpath/xpath_functions.asp#sequence > > You can use all xpath functions in the if/when test condition. > > m. > > -----Original Message----- > From: Mark Peters [mailto:markpeters.work@xxxxxxxxx] > Sent: 4. jzna 2009 20:21 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: Determining whether no single instance of a specific node > exists > > Hi Everyone, > > I have the following XML document: > > <a> > <b/> > <b/> > <b/> > </a> > <a> > <b/> > <b/> > <b> > <c/> > </b> > > </a> > <a> > <b/> > <b/> > <b/> > </a> > > For each <b/> set, I want to apply a transformation only if no single > instance of <c/> exists. I know how to check whether a node does > exist, but not how to check whether no single instance of the node > exists. > > I've tried the following conditional statements without success: > > 1, > <xsl:when test="not(a/b/c)"> > > 2. > <xsl:when test="count(not(a/b/c)) >= 1"> > > > What conditional statement should I use? > > Thanks in advance for any help. > > Mark
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