On Wed, Jul 18, 2018 at 11:35 AM, Liam R. E. Quin liam(_at_)w3(_dot_)org <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Remember that \ isn't special in XML or XPath strings, so your string
contains literal backslashes followed by lower case n. In a regular
expression \ is special however, so to match a literal backslash-n you
want \\n insteaf of \n.
Thanks, Liam. Your suggestion solves my problem.
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.
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.
When I change output spec to following, <xsl:output method="xhtml"/> I
get <br></br> in the output.
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";>
--
Regards,
Mukul Gandhi
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