[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Re: [xsl] Problem in xsl:for-each
Subject: Re: [xsl] Problem in xsl:for-each From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Thu, 10 Jan 2002 16:58:40 +0000 |
Hi Jam, > I'm trying to access to all items in node 'Parrafo'. > Variable $Nombre_Fichero contains a valid filename . > This XSL does not output all elements. (This is the real problem ) > > <xsl:for-each > select="$Nombre_Fichero//Documento/Noticia/Cuerpo/Parrafo"> > <xsl:copy-of select="$Nombre_Fichero//."/> > <br> > </br> > </xsl:for-each> Perhaps you want: <xsl:for-each select="$Nombre_Fichero//Documento/Noticia/Cuerpo/Parrafo"> <xsl:copy-of select="." /> <br /> </xsl:for-each> This will give you a copy of each Parrafo element, separated by br elements. But you said that $Nombre_Fichero contained a filename (and not a node set, which is what it would have to hold for the above file to work), so possibly you're actually after: <xsl:for-each select="document($Nombre_Fichero)//Documento/Noticia/Cuerpo/Parrafo"> <xsl:copy-of select="." /> <br /> </xsl:for-each> Feel free to post more details about your source document and the output that you want to generate if the above doesn't work. Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] Problem in xsl:for-each, jam | Thread | RE: [xsl] Problem in xsl:for-each, Zwetselaar M. van (M |
Re: [xsl] Sorting a variable that c, rafael vazquez | Date | RE: [xsl] Problem with IE5.5 and MS, Andrew Welch |
Month |
Keywords