Sorry for beeing not precise enough:
"
" 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.
I guess i'm still mixed up with the xslt syntax. So any help is appreciated.
Georg.
2006/7/7, Georg Hohmann <georg(_dot_)hohmann(_at_)gmail(_dot_)com>:
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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