如何使用C#从XML文件中删除节点

Question

可能重复:
如何从XmlNodeList中删除XmlNode

嗨,我如何从XML文件中删除一组节点.这是一段代码片段.

string path = @"C:\Documents and Settings\e454935\Desktop\NUnitSettings.xml";
FileStream fs = new FileStream(path, FileMode.Open, FileAccess.Read, FileShare.ReadWrite);
System.Xml.XmlDocument xmldoc = new System.Xml.XmlDocument();
xmldoc.Load(fs);
fs.Close();
xmldoc.DocumentElement.RemoveChild(xmldoc.DocumentElement.ChildNodes[1]);
FileStream WRITER = new FileStream(path, FileMode.Truncate, FileAccess.Write, FileShare.ReadWrite);
xmldoc.Save(WRITER);
WRITER.Close(); 

我尝试使用以下代码来删除一个节点并获得"对象引用未设置为对象的实例".在

xmldoc.DocumentElement.RemoveChild(xmldoc.DocumentElement.ChildNodes[1]);

这是一个示例XML文件,

<?xml version="1.0"?>
<Xml1>
  <Settings>
    <Setting name="DisplayFormat" value="Full" />
    <Setting name="File1" value="a" />
    <Setting name="File1" value="b" />
    <Setting name="File1" value="c" />
    <Setting name="File1" value="d" />
  </Settings>
</Xml1>

实际上从这个文件我想删除四个File1节点,其值为"a,b,c,d",然后我想添加一个节点,

<Setting name="File1" value="e" />

我怎样才能做到这一点.？

Answer 1

您可以使用Linq to XML来执行此操作:

XDocument doc = XDocument.Load("input.xml");
var q = from node in doc.Descendants("Setting")
        let attr = node.Attribute("name")
        where attr != null && attr.Value == "File1"
        select node;
q.ToList().ForEach(x => x.Remove());
doc.Save("output.xml");

Answer 2

从XML删除节点

            XmlDocument doc = new XmlDocument();
            doc.Load(path);
            XmlNodeList nodes = doc.SelectNodes("//Setting[@name='File1']");
            for (int i = nodes.Count - 1; i >= 0; i--)
            {
                nodes[i].ParentNode.RemoveChild(nodes[i]);
            }
            doc.Save(path);

在XML中为节点添加属性

    XmlDocument originalXml = new XmlDocument();
    originalXml.Load(path);
    XmlNode menu = originalXml.SelectSingleNode("//Settings");
    XmlNode newSub = originalXml.CreateNode(XmlNodeType.Element, "Setting", null);
    XmlAttribute xa = originalXml.CreateAttribute("name");
    xa.Value = "qwerty";
    XmlAttribute xb = originalXml.CreateAttribute("value");
    xb.Value = "555";
    newSub.Attributes.Append(xa);
    newSub.Attributes.Append(xb);
    menu.AppendChild(newSub);
    originalXml.Save(path);

Answer 3

使用XPath定位要删除的节点可能更容易.这个stackoverflow线程可能会给你一些想法.

在您的情况下,您将使用此表达式找到所需的四个节点:

XmlDocument doc = new XmlDocument();
doc.Load(fileName);
XmlNodeList nodes = doc.SelectNodes("//Setting[@name='File1']");