How to get node without children in xQuery?
- by mbrevoort
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 }
