[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: New Line using XSLT
Using separator=" & # x a ; " (without the spaces) should work in theory, but it's a bit vulnerable to corruption if there's a complex pipeline processing the XSLT code: the entity reference can get turned into a real newline, which then gets turned into a single space by the process of attribute value normalization next time the XML is parsed. So codepoints-to-string(10) is probably more robust. Or if you prefer, use a variable <xsl:variable name="NL" select="codepoints-to-string(10)"/> <xsl:value-of select="..." separator="{$NL}"/> Michael Kay Saxonica > On 2 Mar 2016, at 10:27, Heiko Niemann kontakt@xxxxxxxxxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > Hi, > > when outputting text with new line(s) I favor using a value-of element > plus seperator with any number of nested sequence elements. For 'new line' > as seperator I prefer using codepoints-to-string(10), which worked for me > as the most reliable way. I bet there are pros and cons, but at the moment > this works best for me and seems to be the most flexible. > > Sample: > > <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > version="2.0"> > <xsl:output method="text"/> > <xsl:template match="/"> > <xsl:value-of separator="{codepoints-to-string(10)}"> > <xsl:for-each select="1 to 10"> > <xsl:sequence select="."/> > </xsl:for-each> > </xsl:value-of> > </xsl:template> > </xsl:stylesheet> > > One con would be that this reqires XSLT 2.0/XPath 2.0. > > > Heiko > > > > >> Hi, >> >> I want to generate a "text" file in which every row should be in separate >> line. >> I have use "
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" and <xsl:text> >> </xsl:text> >> >> But all the above seems not to work. >> >> Please suggest. >> >> -- >> Vishnu Singh | http://marklogicgd.blogspot.in/
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