Re: [xsl] Create XML from XPath expressions

Subject: Re: [xsl] Create XML from XPath expressions
From: "Heiko Niemann" <kontakt@xxxxxxxxxxxxxxxx>
Date: Tue, 18 Aug 2009 15:12:51 +0200
> Hello,
> There are many open questions.
> E.g. the desired <item> element: Is it supposed to be created from the
> <Path> information?


> Or: How can I know that <Path>/item/street/@type</Path> is an
> attribute of the immediately preceding <Path>/item/street</Path> and
> not of some earlier <item>?

There will be indices if there are several siblings with the same name. In
this case there was just one item element.

> You could first process each <Difference> element by parsing the
> <Path> to create elements and attributes, like
> <item><street>2020 Washington Ave.</street></item>
> <item><street type="business"/></item>
> <item><zip>90210</zip></item>
> <item><city>Los Angeles</city></item>
> As far as I can see, this is the only data that is specified by the
> source document. Combining this into a single <item> is an assumption
> and could be handled according to the assumed logic in a second step.

What would the second step look like? I guess parsing will cause some pain
when there are indices. :( Anyhow, I was looking for a way (an easy way)
to transform the Path data into a document structure, since I need the
whole branch and not just a single element, which I would get if I applied
the (evaluated) XPath on the original document (e.g. copy-of). At least I
have not found a way yet to say: find this node/nodeset using this xpath
expression and then return the whole branch which leads to that

> - Michael M|ller-Hillebrand
> PS: Which software creates such a diff report?

It's a component of a BPM software.


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