Subject: Re: [xsl] difficulty using xsl:analyze-string From: "Mukul Gandhi gandhi.mukul@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Wed, 18 Jul 2018 10:06:05 -0000 |
Thanks, David for the clarifications. My modified XSLT transform is following, <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:output method="xhtml" doctype-system=" http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd" doctype-public="-//W3C//DTD XHTML 1.0 Transitional//EN"/> <xsl:template match="/"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>test</title> </head> <body> <xsl:analyze-string select="unparsed-text('test.txt')" regex="\r\n"> <xsl:matching-substring> <br/> </xsl:matching-substring> <xsl:non-matching-substring> <xsl:value-of select="."/> </xsl:non-matching-substring> </xsl:analyze-string> </body> </html> </xsl:template> </xsl:stylesheet> Now I'm getting the result from transform, that I desired. Its worth mentioning about the regex in the xsl:analyze-string above. I'm now reading the input for xsl:analyze-string from a text file (the line delimiter is \r\n on windows), and don't have to write \\ in the regex. And that makes me understand your point, "that example is matching a newline but you wanted to match the two characters \n". On Wed, Jul 18, 2018 at 12:24 PM, David Carlisle d.p.carlisle@xxxxxxxxx < xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > I've some other questions as below, related to xsl:analyze-string, > > > > 1) The 1st example here, https://www.w3.org/TR/xslt20/#regex-examples > > > > specifies regex as \n. That probably got me wrong. > > that example is matching a newline but you wanted to match the two > characters \n > > > > > 2) The output I receive, has <br> instead of <br/>. I specified <br/> in > my > > stylesheet. The output I'm getting might be probably because, I have > > <xsl:output method="html"/> in the stylesheet. > > yes /> is a syntax error in html4 and specified as invalid but ignored in > html5 > > > When I change output spec to following, <xsl:output method="xhtml"/> I > get > > <br></br> in the output. > > > you would get <br/> for an empty br in the xhtml namespace in xhtml output > but you output br in no-namespace. > > > > > When I specify, xsl:output method="xhtml", the following is the complete > > output I get, > > > > <?xml version="1.0" encoding="UTF-8"?><html> > > <head> > > <title>test</title> > > </head> > > <body>hello world > > <br></br>experimenting with XSLT > > <br></br>how are you > > </body> > > </html> > > > > Shouldn't an XHTML document begin with something like, > > > > <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" > > "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> > > > > <html xmlns="http://www.w3.org/1999/xhtml"> > > to get that (which is a very last-century kind of document markup:-) > you need to output elements in the http://www.w3.org/1999/xhtml > namespace and specify that dtd in the attributes of xsl:output. -- Regards, Mukul Gandhi
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