Subject: Re: Re: Re: How to distinguish b/n a scalar and a node-set having From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Fri, 3 Nov 2000 14:43:01 -0400 (EST) |
David, This is a good idea and it actually works in MSXML 3: count(node-set(nodeSet) | node-set(nodeSet)) returns 1. count(node-set(Scalar) | node-set(Scalar)) returns 2. Unfortunately, Saxon always returns 1. Also, I've heard that the standard node-set() function in XSLT 1.1 will throw error when passed a scalar argument. Thanks once again, Dimitre Novatchev. -------------------- Original message -------------------- Date: Wed, 1 Nov 2000 09:59:15 -0400 (EST) From: David_Marston@xxxxxxxxx Subject: Re: Re: How to distinguish b/n a scalar and a node-set having a single text node Dimitre Novatchev wrote: >> I am trying to determine whether the value of a parameter is a >> scalar [boolean/number/string] >> or a node-set, and to do so without using extension functions (I >> consider node-set() to be "standard", as it will be in XSLT 1.1). >... >>The only unsolved case remains when $pValue contains exactly one >>text node. Depending on what node-set(string) does, you may be able to build something based on unions. If $pValue is a node-set, then ($pValue | $pValue) is a node-set with the same count, because it's a union of two references to the same node-set. Unfortunately, putting a "scalar" string in either a union or count() is an error. If node-set($pValue) for a scalar $pValue produces a different node-set on each invocation, then the union of two of them will yield a node-set with a count() of 2. If node-set($pValue) for $pValue being a node-set already simply passes through the same node-set, then the union of two of them will yield a node-set with a count() of 1. .................David Marston __________________________________________________ Do You Yahoo!? >From homework help to love advice, Yahoo! Experts has your answer. http://experts.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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