[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: most efficient way to get XML source's parent dir
On Feb 11, 2009, at 10:24 AM, Michael Kay wrote:
Well, I tried to simplify the example. In reality, I am using the unparsed-text function which has no base-uri argument () and the resolver is provided with a full java.net.URI so I needed a way to create the correct path in the XPath. I should have been more clear. Am I still missing something? best, -Rob
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