[bruno le hyaric <bruno.lehyaric@gm...>]

Once again, thanks a lot Michael!

Sorry, I was a bit tired when I wrote my first message and I forgot to
add the error message and stack trace which said that I can't use
document() function outside of a XSLT context...

And effectively, my XQuery works fine with the doc() function! Thanks!

Regards,

Bruno.


Message: 2
Date: Wed, 28 Oct 2009 23:02:06 -0000
From: "Michael Kay" <mike@...>
Subject: Re: [saxon] How to use XQuery from s9api
To: "'Mailing list for the SAXON XSLT and XQuery processor'"
       <saxon-help@...>
Message-ID: <1DAA5366E75A456B8BD24538A0F2AF1A@...>
Content-Type: text/plain;       charset="us-ascii"


First, some general advice about problem solving.

(a) generally, if an error occurs, there will be an error message. The
message is designed to help you understand the cause of the error. If it
doesn't help you, then it will probably help someone else who knows the
software better than you do. If you want help understanding an error, then
showing them the error message is always a useful first step.

(b) it's also useful to know when the error occurs. XQuery distinguishes
between static errors and dynamic errors. The stack trace you are outputting
will distinguish the two cases. Again, if you want help, providing this kind
of information would be useful.

(c) a more specific point: unless you take steps to do otherwise (which you
haven't done in this program) Saxon will output error information to the
System.err output destination. In a console application, this will appear on
the console. In a GUI application, unless you take steps to display it in
some window, it won't appear at all. The first thing to do before trying to
solve an exception like this is to make sure that either (a) the System.err
output is visible, or (b) you direct the error information somewhere else.

I'm afraid that from the information given, I have absolutely no idea why
your query is failing. The most obvious reason would be that it is trying to
read a source file that does not exist, or exists at a different location
from the one specified.

I've just noticed that you call the document() function rather than doc().
The document() function is defined in XSLT but not in XQuery. That's another
possible cause.

A third possible cause: your call to document() uses a relative URI. This is
interpreted relative to the base URI of the query. But you haven't specified
a base URI for the query. To do this, call setBaseURI() on the
XQueryCompiler.

Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay

>
> I'm still working on my search engine for XML documents...
> When I make a request my search engine find relevant elements
> from an index data structure and return a list of XML element URIs:
>
> ex:
>
> document("./path/to/document.xml")//path/to/element
>
> So I think to use XQueries to fetch all relevant elements in
> one result XML tree.
> I try this Java code but an exception rise and I don't know
> how to handle it...
>
>
>               String xQuery = "for $n in
> document('./collection/d001.xml')//p\n" +
>                                               "return $n";
>
>               Processor processor = new Processor(false);
>               XQueryCompiler compiler = processor.newXQueryCompiler();
>               XQueryExecutable xqueryExecutable = null;
>               try {
>                       xqueryExecutable = compiler.compile(xQuery);
>               } catch (SaxonApiException e) {
>                       // TODO Auto-generated catch block
>                       e.printStackTrace();
>               }
>               XQueryEvaluator xqueryEvaluator =
> xqueryExecutable.load();
>               XdmValue xdmValue = null;
>               try {
>                       xdmValue = xqueryEvaluator.evaluate();
>               } catch (SaxonApiException e) {
>                       // TODO Auto-generated catch block
>                       e.printStackTrace();
>               }
>               XdmSequenceIterator iterator = xdmValue.iterator();
>               while(iterator.hasNext()) {
>                       System.out.println(iterator.next().toString());
>               }
>               System.out.println("End of execution...");
>
>
> Best regards,
>
> Bruno.
>
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