Subject: AW: [xsl] OR expr with node sets From: "Lensch, Thomas" <Lensch@xxxxxxxx> Date: Tue, 18 Oct 2005 12:39:42 +0200 |
Hi, now i've found the problem causing the error. The error's cause isn't the xsl:if (in fact both "foo or bar" and "foo | bar" works!), but one of the following statements where i tried to union "dok/zonen/textsuche/jpk/jpk-titel/div/normfassungen | dok/notindexed/jpk/jpk-titel" in a template-parameter. Clearly, this doesn't work in the called template. But i must say XALAN's error messages are sometimes very awful and don't lead to the bug's cause. Thanks for your suggestions. They lead me to find the bug. Regards, Thomas. -----Urspr|ngliche Nachricht----- Von: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx] Gesendet: Dienstag, 18. Oktober 2005 11:50 An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Betreff: RE: [xsl] OR expr with node sets 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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