XSLTransformati on problem

[avnrao]

Hi,

sorry if this is cross post.. I have a simple XSLTransformati on which I am
trying to load it in XSLTransform class as indicated below

loading xsl :

XslTransform tran = new XslTransform();
tran.Load(@"C:\ AV\DotNet Projects\Replac eString.xsl");

but the problem is its always giving the following errors when I load..
1. '' is an invalid QName - I have kind of resolved this by using { } for
accessing params.
2. boolean(contain s({$str},{$from }) is an invalid XPath expression. - i am
not sure why this is an invalid xpath.

please do help.

ReplaceString.x sl :
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:styleshe et version="1.0"
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:call-template name="replace-string">
<xsl:with-param name="str" select="Str"></xsl:with-param>
<xsl:with-param name="from" select="From"></xsl:with-param>
<xsl:with-param name="to" select="To"></xsl:with-param>
</xsl:call-template>
</xsl:template>
<xsl:template name="replace-string">
<xsl:param name="str"/>
<xsl:param name="from"/>
<xsl:param name="to"/>
<xsl:if test="contains( {$str},{$from}) ">
<xsl:call-template name="">
<xsl:with-param name="str" select="substri ng-before({$str},{ $from})"/>
<xsl:with-param name="from" select="substri ng-after({$str},{$ from})"/>
<xsl:with-param name="to" select="$to"/>
</xsl:call-template>
</xsl:if>
<xsl:value-of select="concat( {$str},{$to})"/>
</xsl:template>
</xsl:stylesheet>

thank you,
Av.

[Das Duck]

Well this is an XSL question rather than a C#-question.

But to my knowledge the solution is very simple:
contains, substring-before, substring-after and concat are does not handle
params inclosed in {}. instead you would simply do this: concat($str, $to).

You would only use { and } in situations like this:
<xsl:element name="{concat($ str, $to}">stuff in element</xsl:element>

What { and } does is to tell the XSL parser that everything in between is
XSL-commands of a kind. Why is that needed? Well in the displayed example
the parser would expect a valid QName and "concat($st rm $to)" isn't - so by
using {} we tell the parser to use the result of the commands in the
brackets instead.

/rap

"avnrao" <av*@newsgroups .com> wrote in message
news:Ot******** ******@TK2MSFTN GP09.phx.gbl...

Hi,

sorry if this is cross post.. I have a simple XSLTransformati on which I am
trying to load it in XSLTransform class as indicated below

loading xsl :

XslTransform tran = new XslTransform();
tran.Load(@"C:\ AV\DotNet Projects\Replac eString.xsl");

but the problem is its always giving the following errors when I load..
1. '' is an invalid QName - I have kind of resolved this by using { } for
accessing params.
2. boolean(contain s({$str},{$from }) is an invalid XPath expression. - i am
not sure why this is an invalid xpath.

please do help.

ReplaceString.x sl :
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:styleshe et version="1.0"
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:call-template name="replace-string">
<xsl:with-param name="str" select="Str"></xsl:with-param>
<xsl:with-param name="from" select="From"></xsl:with-param>
<xsl:with-param name="to" select="To"></xsl:with-param>
</xsl:call-template>
</xsl:template>
<xsl:template name="replace-string">
<xsl:param name="str"/>
<xsl:param name="from"/>
<xsl:param name="to"/>
<xsl:if test="contains( {$str},{$from}) ">
<xsl:call-template name="">
<xsl:with-param name="str" select="substri ng-before({$str},{ $from})"/>
<xsl:with-param name="from" select="substri ng-after({$str},{$ from})"/>
<xsl:with-param name="to" select="$to"/>
</xsl:call-template>
</xsl:if>
<xsl:value-of select="concat( {$str},{$to})"/>
</xsl:template>
</xsl:stylesheet>

thank you,
Av.