Subject: AW: [xsl] Omitting default namespace in the output - XSLT 2.0 From: "Huditsch Roman" <Roman.Huditsch@xxxxxxxxxxxxx> Date: Mon, 20 Dec 2004 14:53:42 +0100 |
I forgot to mention that the default namespace in my source document is prefixed with "ln". So this explaines my match pattern "ln:table". wbr, Roman _______________________________________ Roman Huditsch IT and Electronic Publishing LexisNexis ARD Orac Marxergasse 25 1030 Vienna Austria ph: +43-1-534 52-1514 f: +43-1-534 52-140 e-mail roman.huditsch@xxxxxxxxxxxxx www.lexisnexis.at > -----Urspr|ngliche Nachricht----- > Von: Huditsch Roman [mailto:Roman.Huditsch@xxxxxxxxxxxxx] > Gesendet: Montag, 20. Dezember 2004 14:52 > An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Betreff: [xsl] Omitting default namespace in the output - XSLT 2.0 > > Hi, > > Given an example input like: > > <?xml version="1.0" encoding="UTF-8"?> > <norm xmlns="myDefaultNamespace"> > <table> > <row> > <cell>My Table</cell> > </row> > </table> > </norm> > > I searched for an easy way to get output data, which is not > associated to my default namespace any more, with the help of > <xsl:copy-of> in XSLT 2.0 I hoped that the attribute > "copy-namespaces" set to "no" would help me here, but > unfortunately I had no luck with Saxon 8.1.1 > > My output still looks like > > <table> > <row xmlns="myDefaultNamespace"> > <cell>My Table</cell> > </row> > </table> > > > XSLT: > ===== > > <xsl:template match="ln:table"> > <table> > <xsl:copy-of select="node() | @*" > copy-namespaces="no"/> > </table> > </xsl:template> > > > Thanks in advance for your input! > > wbr, > Roman > _______________________________________ > > Roman Huditsch > IT and Electronic Publishing > LexisNexis ARD Orac > Marxergasse 25 > 1030 Vienna > Austria > ph: +43-1-534 52-1514 > f: +43-1-534 52-140 > e-mail roman.huditsch@xxxxxxxxxxxxx > www.lexisnexis.at
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