Not sure exactly from this question what is variable and what is fixed.
The simple answer is to get all attributes use @*
Taking a *guess* at how you want to link things
for $list in //list
return
<list>{ $list/@* ,
for $item in $list/*
return
//stuff/it...@foo eq $item/foo]
}
</list>
return
From: [email protected]
[mailto:[email protected]] On Behalf Of Mike
Brevoort
Sent: Friday, June 11, 2010 4:36 PM
To: General Mark Logic Developer Discussion
Subject: [MarkLogic Dev General] How to get a node without children
I posted this question up at StackOverflow as well:
http://stackoverflow.com/questions/3026038/how-to-get-node-without-child
ren-in-xquery
Hopefully this has a simple answer, "xQuery/xPath 201" perhaps. So I
have two nodes of elements that I'm essentially trying to join. I want
the top level node to stay the same but the child nodes to be replaced
by those cross referenced.
Given:
<stuff>
<item foo="foo" boo="1"/>
<item foo="bar" boo="2" />
<item foo="baz" boo="3"/>
<item foo="blah boo="4""/>
</stuff>
<list a="1" b="2">
<foo>bar</foo>
<foo>baz</foo>
</list>
I want to loop through "list" and cross reference elements in "stuff"
for this result:
<list a="1" b="2">
<item foo="bar" boo="2" />
<item foo="baz" boo="3"/>
</list>
I want to do this without having to know about what attributes might be
on "list". In other words I don't want to have to explicitly call them
out like
attribute a { $list/@a }, attribute b { $list/@b }
Thanks!
Mike
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