c# - 将 XDocument 解析为字典

Question

关于如何将整个 XML 文档解析为字典，我需要朝着正确的方向努力。我的计划是让键成为路径，每个嵌套类型由“->”分隔。例如：

<Foo>
    <Bar>3</Bar>
    <Foo>
        <Bar>10</Bar>
    </Foo>
</Foo>

如果我想获得一个值，我只需使用以下方法从字典中获取它：

string value = Elements["Foo->Bar"];

我不确定如何递归地遍历每个元素。任何帮助表示赞赏。

Answer 1

直截了当的解决方案：

    private static string GetElementPath(XElement element)
    {
        var parent = element.Parent;
        if(parent == null)
        {
            return element.Name.LocalName;
        }
        else
        {
            return GetElementPath(parent) + "->" + element.Name.LocalName;
        }
    }

    static void Main(string[] args)
    {
        var xml = @"
            <Foo>
                <Bar>3</Bar>
                <Foo>
                    <Bar>10</Bar>
                </Foo>
            </Foo>";
        var xdoc = XDocument.Parse(xml);
        var dictionary = new Dictionary<string, string>();
        foreach(var element in xdoc.Descendants())
        {
            if(!element.HasElements)
            {
                var key = GetElementPath(element);
                dictionary[key] = (string)element;
            }
        }
        Console.WriteLine(dictionary["Foo->Bar"]);
    }

Answer 2

public static void parse()
    {
        Stack<String> stck = new Stack<string>();
        List<String> nodes = new List<string>();
        Dictionary<String, String> dictionary = new Dictionary<string, string>();

        using (XmlReader reader = XmlReader.Create("path:\\xml.xml"))
        {

            while (reader.Read())
            {
                if (reader.NodeType == XmlNodeType.Element)
                {

                    stck.Push(reader.Name);

                }
                if (reader.NodeType == XmlNodeType.Text)
                {
                    StringBuilder str = new StringBuilder();
                    if (stck.Count > 0)
                        nodes = stck.ToList<String>();
                    //List<String> _nodes ;
                    nodes.Reverse(0,nodes.Count);
                    foreach (String node in nodes)
                        str.Append(node + " --> ");

                    dictionary.Add(str.ToString(), reader.Value);
                    str.Clear();
                }
                if (reader.NodeType == XmlNodeType.EndElement)
                {

                    stck.Pop();

                }
            }
        }

        foreach (KeyValuePair<String, String> KVPair in dictionary)
            Console.WriteLine(KVPair.Key + " : " + KVPair.Value);
        Console.Read();
    }

Answer 3

您需要查看 xpath，使用的路径格式与您想要的有点不同，但是如果需要，您可以创建一个 wrap 函数

string GetByPath(string mypath)
{
  //process path, and reformat it to xpath
  //then get by xpath
}

http://support.microsoft.com/kb/308333

http://www.codeproject.com/Articles/9494/Manipulate-XML-data-with-XPath-and-XmlDocument-C

Answer 4

另一种选择，但不确定 XML 文档的真实结构

    string data = "<Foo><Bar>3</Bar><Foo><bar>123</bar></Foo></Foo>";

    XDocument xdoc = XDocument.Parse(data);
    Dictionary<string, string> dataDict = new Dictionary<string, string>();

    foreach (XElement xelement in doc.Descendants().Where(x => x.HasElements == false))
    {
        int keyInt = 0;
        string keyName = xelement.Name.LocalName;

        while (dataDict.ContainsKey(keyName))
            keyName = xelement.Name.LocalName + "->" + keyInt++;

        dataDict.Add(keyName, xelement.Value);
    }