[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: How to declare a param which name is in a XML file?
To clear this up - The XSL in the original message was this: 1) <xsl:template match="parameters"> 2) <xsl:for-each select="param"> 3) <xsl:param name="@ident"/> 4) </xsl:for-each> 5) </xsl:template> The original XML was this (which is not well-formed): <parameters> <param ident="param1"> <param ident="param2"> </parameters> The xsl:param on line 3 is what I was referring to. I fixed the xml and compiled and got the following: MSXML3 says: "Keyword xsl:param may not be used here." (Instant) Saxon says: "Failed to compile style sheet. At xsl:param on line 6 of file:/c:\test.xml: Name @ident contains invalid characters" (line 6 being its position in my XSL file which includes the <?xml header, etc.) Don >-----Original Message----- >From: Gary L Peskin [mailto:garyp@xxxxxxxxxxxx] >Sent: Thursday, October 26, 2000 4:16 PM >To: xsl-list@xxxxxxxxxxxxxxxx >Subject: Re: How to declare a param which name is in a XML file? > > >Don Bruey wrote: >> >> See the doc for <xsl:param> http://www.w3.org. Briefly, ><xsl:param> is not >> valid where you have it. It can only be a top-level element >or a child of >> <xsl:template>. >> > >He was talking about a <param> tag in his XML document, not an ><xsl:param> tag. > >Gary > > > 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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