Subject: RE: [xsl] Ignoring Child Nodes using XPath From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Mon, 26 Mar 2001 18:08:22 +0100 |
> I should have mentioned the fact that I'm not using the > XPath Query in an XSL stylesheet but rather to select the > node in MSXML i.e. dom.selectsinglenode("XPath Query"). yes you should! But I'm now completely mystfied, the usual form of XPath used in selectSingleNode() is a downwards selection from the root, e.g. "/html/body/p" - so where do the children come into it? What are you actually doing that doesn't work? Mike Kay > > I know I could concat the element together by getting the element > name and attributes but I thought there might have been a > more efficient solution using an XPath Query. > > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Michael Kay > Sent: Monday, March 26, 2001 1:58 AM > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: RE: [xsl] Ignoring Child Nodes using XPath > > > > Is there a way in which I can > > select a node ignoring the child nodes > > as in the following example: > > > > What I have... > > > > <p align="left"> > > Here <b>is</b> some content<br/> > > </p> > > > > What I want... > > <p alignt="left"></p> > > There are many ways of ignoring the child nodes, in fact they > will always be > ignored unless you specifically do something to access them. > > Try > <xsl:template match="p"> > <xsl:copy/> > <xsl:copy-of select="@*"/> > </xsl:template> > > Mike Kay > Software AG > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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