如何将属性添加到节点的所有特定子级 [英] How to add attribute to all particular children of a node

问题描述

我有一个下面的节点，我想在其中向所有add节点添加属性.

I have the following node in which I want to add attribute to all add nodes.

<test>
  <add>x1</add>
  <c><add>x2</add></c>
  <b att1="x">x</b>
</test>

我尝试过

functx:add-attributes($test, xs:QName('att1'), 1)

它可以将属性添加到test节点.但是

It can add the attribute to the test node. But

当我尝试

functx:add-attributes($test/add, xs:QName('att1'), 1)

它将属性添加到第一个添加节点，但仅返回具有添加属性的添加节点.然后，当我尝试使用$test//add时，它将引发错误.

It added the attribute to the first add node but returns only add node with added attribute. Then when I tried with $test//add it throws error.

当我尝试

for $add in $test//add 
   return functx:add-attributes($add, xs:QName('att1'), 1)

它分别返回两个添加节点.现在，如何重构原始节点以将属性仅添加到指定的节点.

It returns two add nodes individually. Now, how to restructure the original node to add the attributes to only the specified nodes.

推荐答案

首先，让我指出，仅用于内存使用和更新数据库内容的方式有所不同.对于后者，您可以执行以下操作:

First, let me point out that there is a difference in how this is done for just in-memory use versus updating the content of the database. For the latter, you could do:

for $add in $test//add
return
  xdmp:node-insert-child(
    $add, 
    attribute atta1 { 1 }
  )

要在内存中进行更改，这就是functx所做的，您将制作原始文件的副本，并在构建副本时对其进行更改.这称为递归下降，是一种很常见的模式.不久前，我写了一篇博客文章，内容显示了如何实现递归下降，但是本质上，您将执行一个typeswitch，当它遇到"add"元素时，将创建新属性.您可以使用functx函数.遵循以下原则(未经测试):

To change it in memory, which is what functx does, you'll be making a copy of the original, making changes in the copy as you build it. This is called recursive descent and is a pretty common pattern. I wrote a blog post a while ago that shows how to implement recursive descent, but essentially you'll do a typeswitch that, when it encounters an "add" element, creates the new attribute. You can use the functx function for that. Something along these lines (untested):

declare function local:change($node) 
{ 
  typeswitch($node) 
    case element(add) return 
      functx:add-attributes($node, xs:QName('att1'), 1)
    case element() return 
      element { fn:node-name($node) } { 
        $node/@*, 
        $node/node() ! local:change(.)
      } 
    default return $node 
};

此代码假定add元素内部没有add元素；如果愿意的话，那么您想为第一种情况做类似第二种情况的操作.

This code assumes that an add element won't have add elements inside of it; if you will, then you'd want to do something like the second case for the first.

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