除去没有孩子的节点的父节点

问题描述

我必须与从XML文件中删除特定节点的问题。

I have a question related to removing specific nodes from xml file.

下面是我的XML的示例：

Here is my sample of XML:

<?xml version="1.0" encoding="UTF-8"?>
<root>
  <nodeA attribute="1">
    <nodeB attribute="table">
      <nodeC attribute="500"></nodeC>
      <nodeC attribute="5"></nodeC>
    </nodeB>
    <nodeB attribute="3">
      <nodeC attribute="4"></nodeC>
      <nodeC attribute="5"></nodeC>
      <nodeC attribute="5"></nodeC>
    </nodeB>
    <nodeB attribute="placeHolder">
    <nodeB attribute="toRemove">
      <nodeB attribute="glass"></nodeB>
        <nodeE attribute="7"></nodeE>
      <nodeB attribute="glass"></nodeB>
      <nodeB attribute="glass"></nodeB>
    </nodeB>
    </nodeB>
    <nodeB attribute="3">
      <nodeC attribute="4"></nodeC>
      <nodeC attribute="5"></nodeC>
      <nodeC attribtue="5"></nodeC>
     </nodeB>
    <nodeB attribute="placeHolder">
    <nodeB attribute="toRemove">
      <nodeB attribute="glass"></nodeB>
        <nodeE attribute="7"></nodeE>
      <nodeB attribute="glass"></nodeB>
      <nodeB attribute="glass"></nodeB>
    </nodeB>
    </nodeB>
  </nodeA>
</root>

我想删除节点节点B =的文档，删除不删除该节点的儿童。从那以后，我需要做同样的事情节点B的属性=占位符。结果的一部分将看起来像：

I would like to remove node nodeB="toRemove" without removing childrens of this node. After that I need to do same thing with nodeB attribute="placeHolder". Part of result would look like that:

     <nodeB attribute="3">
      <nodeC attribute="4"></nodeC>
      <nodeC attribute="5"></nodeC>
      <nodeC attribtue="5"></nodeC>
     </nodeB>
     <nodeB attribute="glass"></nodeB>
        <nodeE attribute="7"></nodeE>
     <nodeB attribute="glass"></nodeB>
     <nodeB attribute="glass"></nodeB>

我一直想这样的代码才达到的：

I have been trying code like this to achive that:

        XmlNodeList nodeList = doc.SelectNodes("//nodeB[@attribute=\"toRemove\"]");

        foreach (XmlNode node in nodeList)
        {
            foreach (XmlNode child in node.ChildNodes)
            {
                node.ParentNode.AppendChild(child);
            }
            node.ParentNode.RemoveChild(node);
        }
        doc.Save(XmlFilePathSource);

我能够找到节点与期望的属性的文档，删除或占位符，但是我不能动的孩子这种由一个水平节点上。你能帮助我在这种情况下？它可以使用LINQ，的XDocument，XmlReader中解决方案，但我更喜欢使用XmlDocument的工作。
感谢您的帮助，您可以提供我提前

I am able to locate node with desired attribute toRemove or placeHolder, however I am not able to move children of this nodes up by one level. Could you help me in this case? It can be solution with Linq, XDocument, XmlReader but I prefer working with XmlDocument.Thank you for any help you could provide me in advance.

编辑：

在本情况下，我用稍微修改代码（保持顺序）查野人写道波纹管。一旦删除

In this case I have used slightly modified code(to preserve order) that Chuck Savage wrote bellow. Once to remove

  <nodeB attribute="toRemove"> </nodeB>

，然后做同样的

and then do the same with

  <nodeB attribute="placeHolder"></nodeB>

下面是稍微修改代码

  XElement root = XElement.Load(XmlFilePathSource);
  var removes = root.XPathSelectElements("//nodeB[@attribute=\"toRemove\"]");
  foreach (XElement node in removes.ToArray())
  {
    node.Parent.AddAfterSelf(node.Elements());
    node.Remove();
  }
  root.Save(XmlFilePathSource);

XSLT通过@MiMo提供的方法是非常有用的，以及在这种情况下。

xslt approach provided by @MiMo is very useful as well in this case.

推荐答案

使用LINQ到XML和你的XPath，

Using Linq-to-XML and your XPath,

XElement root = XElement.Load(XmlFilePathSource); // or .Parse(string)
var removes = root.XPathSelectElements("//nodeB[@attribute=\"toRemove\"]");
foreach (XElement node in removes.ToArray())
{
    node.AddBeforeSelf(node.Elements());
    node.Remove();
}
root.Save(XmlFilePathSource);

请注意：XPath是提供 System.Xml.XPath

Note: XPath is available in System.Xml.XPath

注2：您可以向/从XmlDocument的使用的的，因为你喜欢的XmlDocument。

Note2: You can convert to/from XmlDocument using these extensions since you prefer XmlDocument.

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