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.
-----Ursprüngliche Nachricht-----
Von: Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com
[mailto:Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com]
Gesendet: Dienstag, 18. Oktober 2005 11:50
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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
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