Oxygen XML Forum

Posted: **Sun Nov 08, 2009 2:34 pm**

I'm a newbie to XQuery and XPath. I'm trying to compare two XML structures. The problem I now have: What exactly makes an XPath (string?)?

I have the following code:

Code: Select all

declare variable $x1 := <top>

                           <rec>

                             <d1>25</d1>

                             <d2>10</d2>

                           </rec>

                        </top>;

declare variable $x2 := <top>

                           <rec>

                             <d1>30</d1>

                             <d2>10</d2>

                           </rec>

                        </top>;

for $node in $x1/rec

return for $elem in $node/*

       return if ($elem = $x2/rec/*[local-name()=local-name($elem)])

              then (<match>{$elem}</match>)

              else (<diff>{$elem}</diff>)

which produces

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<?xml version="1.0" encoding="UTF-8"?>

<diff>

   <d1>25</d1>

</diff>

<match>

   <d2>10</d2>

</match>

So far so good, but it took me a while to get there. Specifically, in my original trials, I had something like

Code: Select all

       return if ($elem = $x2/rec/local-name($elem))

              then (<match>{$elem}</match>)

              else (<diff>{$elem}</diff>)

What confuses me:

Checking with the debugger, I find local-name(elem) = 'd1'
but also $x2/rec/local-name($elem) evaluates to 'd1' (rather than what I expected: 30

Hence, what I picture for myself as an XPath string seems to be the wrong concept. But what is the right one?

Thanks for your help! Stefan

Posted: **Mon Nov 09, 2009 6:43 pm**

Hello,

Stefan_E wrote:I'm a newbie to XQuery and XPath. I'm trying to compare two XML structures. The problem I now have: What exactly makes an XPath (string?)?

Maybe you should go through an XPath tutorial?

Stefan_E wrote:What confuses me:
Checking with the debugger, I find local-name(elem) = 'd1'

but also $x2/rec/local-name($elem) evaluates to 'd1' (rather than what I expected: 30
Hence, what I picture for myself as an XPath string seems to be the wrong concept. But what is the right one?

The result of an expression like $x2/rec/local-name($elem) is the result of the function local-name() which is the element name: d1 or d2 or rec or top etc.

Regards,
Sorin

Posted: **Mon Nov 09, 2009 10:25 pm**

Hi Sorin,

thanks for the link to the tutorial; I'm also using the O'Reilly book on XQuery, but I lack the explanation of the concept:

my background is perl, so I could expect local-name() to evaluate to d1 and then the whole path string to read <evaluation of $x2> . '/rec/' . <evaluation of local-name>, hence /top/rec/d1, which would then again be taken as an XPath string, leading to the element <d1>30</d1>
or, as I observe, I only get <evaluation of local-name>; but then I'd expect that somebody throws an error (or at least a warning) because of the unused part $x2/rec/

... so, I'm still a bit searching the concept behind the behavior - in order to make my next XQuery writing considerably faster

Regards, Stefan

Oxygen XML Forum

What makes an XPath String?

What makes an XPath String?

Re: What makes an XPath String?

Re: What makes an XPath String?