Subject: RE: [xsl] XSLT does not handles charcters less than x0020 hex From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Mon, 16 Apr 2001 12:26:42 +0100 |
> Michael Kay wrote: > > > I suggest outputting a processing instruction such as > <?char x1f?> and > > writing your own ContentHandler to do the final serialization. > > I am very new in XML. Could you explain more on how to do that? > > If possible, I like to have it done only with XSL to keep it > portable You can't solve this problem solely with XSLT, because XSLT output will never contain characters that are invalid in XML. I'm sorry, I don't have time to explain how to write a SAX2 ContentHandler, and if you need to ask the question, then it's probably not the right approach for you. An alternative way of post-processing the output is using a text editor or Perl. The ContentHandler solution will be portable between most Java XSLT processors. Mike Kay XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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