Subject: RE: [xsl] OR expr with node sets From: <Jarno.Elovirta@xxxxxxxxx> Date: Tue, 18 Oct 2005 12:49:45 +0300 |
Hi, > > The above uses an union expression, not an or expression. > There's a difference. > > > > <xsl:if test="foo | bar"> > > > > collects two node-sets, creates an union and then casts the > combined node-set to a boolean. > > > > <xsl:if test="foo or bar"> > > > > collects two node-sets, casts both of them into booleans > and then makes an OR comparison between the resulting booleans. > > > > That's true according to the way things are specified, > although the end > result is always the same so an actual implementation may well do the > same thing in both those cases (and in both cases not > generate the whole > set, but stop looking as soon as it finds any node, as it > knows that it > is in a boolean context). Naturally, could have specified that "a naive implementation following the spec... ", but I wanted to highlight that "|" is not an OR operator. Xalan throwing an exception in the case the original poster described is clearly wrong, can't say why it fails, though. Cheers, Jarno -- Kevin Energy: DJ Kevin Energy + MC Sharkey @ Enchanted Australia
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Re: [xsl] OR expr with node sets, David Carlisle | Thread | [xsl] No Context Item Available Err, UlyLee |
Re: [xsl] OR expr with node sets, David Carlisle | Date | [xsl] Re: xsl-list Digest 18 Oct 20, Bijo Alex Thomas |
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