Subject: Re: [xsl] Omitting elements from output From: "Joerg Heinicke" <joerg.heinicke@xxxxxx> Date: Sat, 23 Feb 2002 02:54:57 +0100 |
Hello Rick, in XSLT there is a built-in template. This gives you the text-value of all nodes, for which you don't have a specific template. You have two possibilities: Either changing <xsl:apply-templates/> to <xsl:apply-templates select="Element1 | Element2"/>. Or you create another for all other nodes, which does not give you the text-value: <xsl:template match="*"/>. Regards, Joerg > I have the follow structure in my stylesheet. > > <xsl:template match="/"> > <xsl:apply-templates/> > <xsl:template> > > <xsl:template match="Element1"> > ... > </xsl:template> > > <xsl:template match="Element2"> > ... > </xsl:template> > > When I transform the XML file, I only want the content of Element1 and > Element2, but the content of the other elements is output as well. How do I > suppress the output of the other elements. Thanks. > > Rick Quatro XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] Omitting elements from output, Rick Quatro | Thread | [xsl] DocBook XSL - Can't get it to, gary cor |
Re: [xsl] getting a single quote in, Peter Davis | Date | [xsl] Omitting elements from output, Rick Quatro |
Month |