如何在xQuery中获得没有子节点的节点? [英] How to get node without children in xQuery?

问题描述

因此，我实际上有两个要加入的元素节点.我希望顶层节点保持不变，但子节点将被那些交叉引用替换.

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.

给出:

<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>

我想遍历列表"并交叉引用内容"中的元素以得到此结果:

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>

我想这样做，而不必知道列表"上可能有哪些属性.换句话说，我不想像

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 }

推荐答案

使用:

$list1/item[@foo = $list2/item/@foo]

$list1/item[@foo = $list2/item/@foo]

这将选择$list1中所有<item>元素，这些元素的foo属性值等于$ list2中<item>元素之一的foo属性.

This selects all <item> elements in $list1 the value of whose foo attribute is equal to the foo attribute of one of the <item> elements in $list2.

要复制<list>元素的所有属性，请执行以下操作:

In order to copy all attributes of the <list> element, do something like this:

  for $attr in /whateverIsthePathLeadingToList/list/@*
    return 
      attibute {name($attr)} {$attr}

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