The xsl:copy-of instruction is only useful where you want to do an exact
copy. The standard way to do a transformation where the output is nearly the
same as the input is to start with the identity template rule:
<xsl:template match="*">
<xsl:copy>
<xsl:copy-of select="*"/>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
and then override it for elements that need to be modified:
<xsl:template match="MenuData">
<xsl:copy>
<xsl:value-of select="normalize-space(.)"/>
</xsl:copy>
</xsl:template>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: António Mota [mailto:amsmota(_at_)gmail(_dot_)com]
Sent: 25 January 2005 19:12
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Removing line-feeds on a imported tree via document
Here's another interesting one. In my XSLT i have at some point
<xsl:copy-of select="$xmldoc/Menus/Menu" />
working ok, EXCEPT sometimes i get things like
<Menu>
<Menu_K>X00066</Menu_K>
<Titulo>(Cada Utente)</Titulo>
<MenuData>SELECT Utente_K, NomeLogin
FROM Utente
WHERE UtenteTipo='U'</MenuData>
<MenuTipo>TXT</MenuTipo>
</Menu>
Notice the line breaks on MenuData, that is causing me problems later
on. So i want to get rid of then. I've tryed using
<xsl:copy-of select="normalize-space($xmldoc/Menus/Menu)" />
because i don't understand nothing of this, but soon realise the
error, so i went back and define a template
<xsl:template match="MenuData">
<xsl:copy><xsl:value-of
select="normalize-space(.)"/></xsl:copy>
</xsl:template>
but did no good, cause it seems that only match the original MenuDatas
and not the imported ones.
Any sugestions? Thanks.
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