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我正在用c#读取一个XML文件的标签，我已经读取到文件末尾的代码，但我想逐个标签地读取它。事实上，我想要一个保存每个标签的列表。这是我的xml：

<XML>
<Groups>
 <Group Name="Group1" ID="D7EBC5D6-0E6D-499E-B528-34BE14382755">
    <Item ID="9A4FA56F-EAA0-49AF-B7F0-8CA09EA39167"/>
    <Item ID="351FEF76-B826-426F-88C4-DBAAA60F886B"/>
    <Item ID="96A4CBFC-04CD-4D27-ADE6-585C05E4DBC9"/>
    <Item ID="D8876943-5861-4D62-9249-C5FEF88219FA"/>
</Group>
<Group Name="Group2" ID="CC012258-14AC-44E9-BA0F-78AE7C569FCB">
    <Item ID="9A4FA56F-EAA0-49AF-B7F0-8CA09EA39167"/>
    <Item ID="351FEF76-B826-426F-88C4-DBAAA60F886B"/>
</Group>
</Groups>
 <Items>
<Item>
    <GUID>9A4FA56F-EAA0-49AF-B7F0-8CA09EA39167</GUID>
    <Type>button</Type>
    <Title>Save</Title>
    <Value>submit</Value>
    <Name>btnsave</Name>
    <MaxLen>5</MaxLen>
</Item>    
<Item>
    <GUID>351FEF76-B826-426F-88C4-DBAAA60F886B</GUID>
    <Type>text</Type>
    <Title>Name:</Title>
    <Name>txtname</Name>
    <Value>Name</Value>
    <MaxLen>2</MaxLen>
</Item>
</Items>
</XML>

还有我的 C# 代码：

    public Guid GroupsGuid;
    public Guid ItemsGuid;
    List<Guid> ItemsIndex = new List<Guid>();
    public string Datastring;        

    public override void LoadFromXML(string XMLFileAddress)
    {
        XmlTextReader reader = new XmlTextReader(XMLFileAddress);                      
        while (reader.Read())
        {
            switch (reader.Name)
            {
                case "Groups":
                    while (reader.Read())
                    {
                        if (reader.NodeType == XmlNodeType.EndElement)
                            break;
                        switch (reader.Name)
                        {
                            case "Group":
                                GroupsGuid=Guid.Parse(reader.GetAttribute("ID"));
                                Datastring += "<" + reader.Name + " name = " + reader.GetAttribute("Name") + " ID = " + GroupsGuid + "/>";
                                while (reader.Read())
                                {
                                    if (reader.NodeType == XmlNodeType.EndElement)
                                        break;
                                    switch (reader.Name)
                                    {
                                        case "Item":
                                            ItemsGuid = Guid.Parse(reader.GetAttribute("ID"));
                                            Datastring += "<" + reader.Name + " ID = " + reader.GetAttribute("ID") + "/>";                                                
                                            ItemsIndex.Add(GroupsGuid);
                                            break;
                                    }
                                }
                                break;
                        }
                    }
                    reader.Close();
                    break;
            }
        }
    }

我想在我的程序中将标签视为对象。先感谢您。

4

2 回答 2

7

这可能会有所帮助：

XDocument doc = XDocument.Load(your file);
var result = doc.Descendants().ToList();

编辑：要获得结果（根据您的评论），请尝试以下操作：

XDocument doc = XDocument.Load(your file);

var groupItems = doc.Descendants("Group")
                    .SelectMany(i => i.Elements("Item"))
                    .Attributes("ID")
                    .Select(j => (string)j)
                    .ToList();

var nodes =       doc.Descendants("Items")
                     .Elements("Item")
                     .Where(i => groupItems.Contains((string)i.Element("GUID")))
                     .ToList();
于 2013-04-29T11:48:47.563 回答
2 
         XmlDocument doc = new XmlDocument();

        doc.Load(path);// write xml path

        XmlNodeList elemList = doc.GetElementsByTagName("ID");


        for (int i = 0; i < elemList.Count; i++)
        {
            string ID = elemList[i].Attributes["ID"].Value;

           listBox1.Items.Add(ID);


        }
于 2014-08-12T19:42:13.027 回答