Subject: RE: Copying unparsed entity using XSL From: "Richard Lander" <rlander@xxxxxxxxxxxxxxxxxxx> Date: Thu, 11 Nov 1999 12:28:59 -0500 |
Hello, I'm not sure of the second step, but the first one is likely: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE Job SYSTEM "doc.dtd" [ <!ENTITY pcl0001 SYSTEM "pcl0001.pcl" NDATA BINARY> <!ATTLIST Print file ENTITY #REQUIRED> ]> <Job> <Print file="pcl0001"/> </Job> Richard. -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxx]On Behalf Of Bovone Stefano Sent: Thursday, November 11, 1999 10:09 AM To: 'XSL-List@xxxxxxxxxxxxxxxx' Subject: Copying unparsed entity using XSL I have a xml document like this: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE Job SYSTEM "doc.dtd" [ <!ENTITY pcl0001 SYSTEM "pcl0001.pcl" NDATA BINARY> ]> <Job> <Print/> </Job> I would like to obtain the identity trasformation using something like: <xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template> but I want that the Entity: <!ENTITY pcl0001 SYSTEM "pcl0001.pcl" NDATA BINARY> is in the output file. Is this possible using XSLT (and XT or LotusXSL or ..) ? Thanks in advance. Bye. 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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