Subject: Re: [xsl] getting the node position in source xml in a variable From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Wed, 27 Feb 2002 15:11:58 +0000 |
Hi Gurvinder, > I have one more query... you said that i am not passing the result > tree fragment what does it mean... i thought when i select nodes and > pass it as parameter is is passing result tree fragment No - when you select nodes and pass them using the select attribute of xsl:with-param, then you pass those nodes as a *node set*. It's only if you use the *content* of the xsl:with-param that you create a result tree fragment. For example, if you did: <xsl:call-template name="lookup"> <xsl:with-param name="name">test</xsl:with-param> <xsl:with-param name="nodes"> <xsl:copy-of select="/test/Level2" /> </xsl:with-param> </xsl:call-template> With this call, both $name and $nodes are result tree fragments. It doesn't matter too much with $name, because it's a result tree fragment whose root node only has one child - a text node with the value 'test' - although a result tree fragment is a little more unweildy for the processor than a plain string would be. On the other hand, the $nodes parameter is set to a result tree fragment whose root node has a number of Level2 element children. Because that result tree fragment has structure, you usually have to convert it back into a node set (e.g. using msxsl:node-set()) to do anything useful with it. This means that in general you should use the select attribute to set the values of variable-binding elements, such as xsl:variable, xsl:param and xsl:with-param. Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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