Re: [xsl] difficulty using xsl:analyze-string

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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