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Sivasankari SubramanyanSubject: xquery variable substitution
Author: Sivasankari Subramanyan
Date: 02 Sep 2008 06:14 AM
This is my code

for $i in doc("episode.xml")//episode
return <episode>{ $i/@name }
{
let $xmfi := concat( $i/parent::node()/@show, ".xml" )
for $y in doc($xmfi)//photos[@title=$i/@name]
return $y
}
</episode>

the episode.xml contains an attribute called show and the value of that attribute is the name of my photo galleries, episode.xml

When I use $xmfi the content coming in $y is coming as a string and I am not able to access the content using xpath.

Any help will be appreciated


Documentgossip-girl.xml
Gossip girl xml

Unknownepisode.xml
Episode xml file
Posttop
Minollo I.Subject: xquery variable substitution
Author: Minollo I.
Date: 02 Sep 2008 07:57 AM
You seem to be already doing the right thing:
...
for $y in doc($xmfi)//photos[@title=$i/@name]
return $y

Assuming $xmfi contains a valid URI, that will return all <photos> elements satisfying the @title=$i/@name condition in the document referenced by $xmfi. Be careful about relative paths.
   
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