Simplifing OR condition

knifi · May 25th, 2009, 04:11 PM

I have check whether h1 or h2 or span comes as ancestor ..is this considered efficient.

test="ancestor::*[local-name() = 'h1' or local-name() = 'h2' or local-name() = 'span']"

samjudson · May 26th, 2009, 04:02 AM

It is quite hard to answer questions about 'efficiency' without a) indepth knowledge about the XSLT processor in question, b) knowledge of the XML document being processed and c) knowledge of the complete XSLT document.

It is quite possible that this is the most efficient way to do what you want, but I couldn't say that from what you have shown us so far.

For example if the XML document has a huge tree, and the size of the ancestor axis is large then this might be a very inefficient way of doing the check. It may be better to find all h1, h2 and span element and then process their children.

mhkay · May 26th, 2009, 04:44 AM

I agree with Sam. One point, though, is that ancestor::h1 is likely to be a faster test than ancestor::*[local-name()='h1']. It's also cleaner: the whole point about namespaces is that the name "h1" only means anything in the context of a particular namespace.

If your stylesheet is unacceptably slow, it's unlikely that this expression is the problem. (And if it's not, why are you worried?)