Subject: Re: [xsl] Error when params are in Xpath expression From: Dmitri Snytkine <d.snytkine@xxxxxxxxx> Date: Tue, 12 May 2009 08:25:46 -0400 |
OK, so how do I find an element named tr_string that has attribute @str with a value $str I have a param declared like this <xsl:param name="str" select="'recent_albums'"/> Then I need to use $str in the Xpath This does not work: <xsl:value-of select="document('')/*/trans:tr/tr_string[@str=$str]"/> On Tue, May 12, 2009 at 7:13 AM, Michael Kay <mike@xxxxxxxxxxxx> wrote: > > Variable references such as $toLang can be used in place of literal values. > They can't be used in place of arbitrary fragments of XPath expression text > (they aren't macros). If $toLang is an element name, then you can find > elements of that name using *[name()=$toLang]. > > Regards, > > Michael Kay > http://www.saxonica.com/ > http://twitter.com/michaelhkay > > > >> -----Original Message----- >> From: Dmitri Snytkine [mailto:d.snytkine@xxxxxxxxx] >> Sent: 12 May 2009 12:06 >> To: xsl-list >> Subject: [xsl] Error when params are in Xpath expression >> >> <!-- get translation of a string --> >> <xsl:template name="translateString"> >> <xsl:param name="str" select="'recent_albums'"/> >> <xsl:param name="toLang"/> >> <xsl:param name="fallback" select="'en'"/> >> >> <xsl:choose> >> <xsl:when test="true()"> >> >> <xsl:value-of >> select="document('')/*/trans:tr/tr_string[@str=$str]/$toLang"/> >> </xsl:when> >> <xsl:otherwise> >> <xsl:value-of select="'not found'"/> >> </xsl:otherwise> >> </xsl:choose> >> </xsl:template> >> >> >> I get errors when running this template. >> If I manually replace the $str and $toLang, then I am getting >> the correct result. >> >> <xsl:value-of >> select="document('')/*/trans:tr/tr_string[@str='recent_images']/ru"/> >> >> What is wrong with my syntax that uses params? Are params >> allowed in Xpath expression?
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