Subject: [xsl] Re: XSLT to remove characters and whitespaces From: "Georg Hohmann" <georg.hohmann@xxxxxxxxx> Date: Fri, 7 Jul 2006 14:22:07 +0200 |
"
" replaces linefeed+carriage-returns "	" replaces tabulators "normalize-space" removes all unnecessary whitespaces
This replacements do their job in another stylesheet. So it is more a problem of copying elements the right way than replacing chars. I know the copy-elements-stylesheet from the the xslt-cookbook but i don't know how to customize it for my needs (and i don't have the need to copy any attributs).
I though the provided stylesheet would do the following: Select each node (*), create element with same name (name={name()}) transform the content, select next node ... But the output looks like <tag1>content1</tag1>content1 ...
So maybe the usage of * is wrong. So i tried node() instead but than i get an error for the usage of name() for the element name that says i'm generating in invalid qname.
Hello,
i have a xml file with some content in it which contains some unwanted carriage returns and whitespaces. Now I'm trying to write a stylesheet which makes an exact copy of the source file but without the returns and whitespaces. I thought this should work:
<?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output name="stripped" method="xml" version="1.0" encoding="UTF-8" indent="yes"/> <xsl:strip-space elements="*"/> <xsl:template match="/"> <xsl:result-document format="stripped" href="result.xml"> <xsl:apply-templates/> </xsl:result-document> </xsl:template> <xsl:template match="*"> <xsl:element name="{name()}"> <xsl:value-of select="normalize-space(translate(translate(., '
', ' '), '	', ' '))"/> <xsl:apply-templates/> </xsl:element> </xsl:template> </xsl:stylesheet>
But the output is a mess in parts. What am I doing wrong?
Regards, Georg
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