无法在 C# 中读取 XML 节点答案

【问题标题】：Unable to read XML Nodes in C#无法在 C# 中读取 XML 节点
【发布时间】：2020-01-13 21:50:40
【问题描述】：

我正在为我的工作场所创建一个知识库，当我单击 .NET 表单列表框中的一个名称时，它会依次填充多个文本框。

如何在单击更改时强制执行，但它不会在 Windows MessageBox 中显示任何数据。我已经测试了 Windows MessageBox 以确保正在启动更改

消息框确保更改已启动
点击更改时自动填充文本

String client_file_location = @"REDACTED UNC PATH"; // Clients

            XmlDocument config = new XmlDocument();

            FileInfo config_file = new FileInfo(client_file_location);
            config.Load(client_file_location);

            string selected_item = client_list_box.Text;


            XDocument xml = XDocument.Load(client_file_location);

            var nodes = (from n in xml.Descendants("clients")
                         where n.Element("client").Attribute("client_name").Value == selected_item
                         select new
                         {
                             Company = (string)n.Element("Company").Value,
                             knowledge = (string)n.Element("Knowledge").Value
                         }).ToList();

            foreach(var n in nodes) 
            {
                System.Windows.Forms.MessageBox.Show(n.Company);
                new_client_name.Text = n.Company;
                knowledge_base_location.Text = n.knowledge;
            }

<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<!--Configuration File for Empired Program: The Curator-->
<clients>
  <client client_name="Test #1">
    <Company>Test #1</Company>
    <Knowledge>http://test.location/</Knowledge>
    <ClientFile>Test #1.xml</ClientFile>
  </client>
</clients>

应该填写“new_client_name”和“knowledge_base_location”框，但没有输入任何内容

【问题讨论】：

标签： c# xml winforms

【解决方案1】：

尝试以下：

           var nodes = (from n in xml.Descendants("client")
                         where (string)n.Attribute("client_name") == selected_item
                         select new
                         {
                             Company = (string)n.Element("Company"),
                             knowledge = (string)n.Element("Knowledge")
                         }).ToList();

【讨论】：

【解决方案2】：

只需使用列表尝试序列化概念。我的示例代码：

        //Write XML
            List<UO> lstUO = new List<UO>();
            using (StreamWriter writer = new StreamWriter(FilePath,false))
            {
                XmlSerializer serializer = new XmlSerializer(typeof(List<UO>));
                serializer.Serialize(writer, lstUO);
                writer.Close();
            }

            //Read XML

            using (FileStream stream = File.OpenRead(FilePath))
            {
                XmlSerializer serializer = new XmlSerializer(typeof(List<UO>));
                List<UO> dezerializedList = (List<UO>)serializer.Deserialize(stream);
                stream.Close();
            }

            public class UO
            {
                public string input { get; set; }
                public string input1 { get; set; }
            }

【讨论】：