Subject: [xsl] changing select in apply-templates From: "Prasad Akella" <avlnprasad@xxxxxx> Date: Sun, 20 Feb 2005 16:33:03 +0100 |
hi, i had problems with xsl:for each where i would like to change the select in run time. i was not able to solve with it and thus changed my problem. its like this i have a stylesheet A.xsl in which i am doing some xforms transformation using A.xml as the data source. A.xsl - - - - - (generates form using xforms controls) <xsl:stylesheet> <xsl:template match = /> uses A.xml - - - - some xforms statements <xsl:apply-templates select="Exam/Question" mode="content"/> </xsl:template> <xsl:template match="Question" mode="content"> ------some statements to perform some actions to display data </xsl:template> </xsl:stylesheet> This is my second xsl which would take the above xsl as input and transform all the xforms controls into equivalent xhtml controls B.xsl - - - - - - - (generates equivalent xhtml controls from the above A.xsl) templates matching all controls of xforms to write equivalent xhtml controls my problem is here that i am not able to make this xsl take A.xml as input i am using something like <xsl:variable name="tempxml" select="document('A.xml')"/> i would like to pass this as $tempxml//Question for my above apply-templates written in A.xsl i am not able to do that and thus am not able to render data. i am not able to control my <xsl:template mode="content"> in the above xsl as its also being transformed. kindly sugggest as to how i can over come this problem i would like my <xsl:apply-templates select="$tempxml//Question" match="content/"> to be used instead of the above xsl:apply-templates this in such a way that in my B.xsl, the data source is taken and the data is rendered accordingly. with regards, prasad Akella __________________________________________________________ Mit WEB.DE FreePhone mit hoechster Qualitaet ab 0 Ct./Min. weltweit telefonieren! http://freephone.web.de/?mc=021201
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