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Re: [xsl] How to get and display part of XML document in IE5/6?
Subject: Re: [xsl] How to get and display part of XML document in IE5/6? From: Joerg.Heinicke@xxxxxx Date: Sat, 15 Jun 2002 06:19:52 +0200 (MEST) |
Hello Quinghui, in what framework do you want to use XInclude and XPointer? I think it's not possible to do it client-side. At least IE has no XInclude-transformer. And I don't know any browser which has one, but maybe Mozilla has still more secrets. On the other hand a two-step transformation is also not possible I think. But where is the problem to do it in one step/transformation process? <xsl:template match="/"> <xsl:apply-templates select="id('ebnf')"/> </xsl:template> <xsl:template match="the-element-which-has-ebnf-as-id"> <!-- transform it to what you want --> <html> <head><title>test</title></head> <body> This is the element which has 'ebnf' as ID. </body> </html> </xsl:template> Regards, Joerg > I want to get and display part of a linked XML document, I can make > use of XLink and XPointer to refer to the target, e.g. > http://www.w3.org/TR/1998/REC-xml-19980210.xml#xpointer(id("ebnf")) > but how can I display the content in normal explorer(e.g. IE5/IE6)? > I have not found the solutions or samples to solve this problem. > > In another way, I try to write stylesheet to extract the target to be a > separate document. then I can display it in explorer with another > stylesheet. but XSL does not provide such function to extract the > part of XML document, without stripping the tag. > > Can you give me some suggestion? > > Thanks > > Qinghui -- GMX - Die Kommunikationsplattform im Internet. http://www.gmx.net XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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