从 XML 文件填充文本框

Posted 2023-05-07

【中文标题】从 XML 文件填充文本框

【英文标题】：Populate Text boxes from XML File

【发布时间】：2016-03-04 12:12:01

【问题描述】：

我将两个文本框保存为 .XML 文件。现在我想将它们加载到我的文本框中以便能够编辑它们。我也继承了 Person 类。有什么帮助吗？谢谢

 public partial class TeacherForm : Form

    public TeacherForm()
    
        InitializeComponent();
    

    private void tSavebtn_Click(object sender, EventArgs e)

    
        try
        
            Teacher teacher = new Teacher();
            teacher.Name1 = tNametxtBox.Text;
            teacher.ID1 = tIDtxtBox.Text;
            TeacherXML.Save(teacher);
        
        catch (Exception ex)
        
            MessageBox.Show(ex.Message);
        

---

 public class TeacherXML

    private const string DataRepositoryFolder = "data\\Teachers\\";
    private const string FileExtension = ".xml";

    public static bool Save(Teacher teacher)
    
        XmlSerializer xs = new XmlSerializer(typeof(Teacher));

        string targetPath = GetRepositoryFilePath(teacher.ID1);

        using (StreamWriter sw = new StreamWriter(targetPath, false, Encoding.UTF8))
        
            xs.Serialize(sw, teacher);
        
        return true;
    

编辑：

---

  public static Teacher Load(string ID)
    
        string targetPath = GetRepositoryFilePath(ID);

        if (File.Exists(targetPath))
        
            XmlSerializer xs = new XmlSerializer(typeof(Teacher));
            using (StreamReader sr = new StreamReader(targetPath))
            
                return xs.Deserialize(sr) as Teacher;
            
        

        return null;
    

我也创建了那个位.. 我现在需要的是从示例 OpenFileDialog 中打开一个文件，使用 BtnClick 操作选择文件并加载文本框。谢谢

【参考方案1】：

在 Button_Click 上加载：

将事件处理程序添加到您的按钮（这里我称之为button1）：

button1.Click += button1_Click;

在你的button1_Click 你可以这样做：

private void button1_Click_1(object sender, EventArgs e)

    OpenFileDialog ofd = new OpenFileDialog();
    ofd.Filter = "XML Files|*.xml";
    if(ofd.ShowDialog()== System.Windows.Forms.DialogResult.OK)
    
        try
        
            XmlSerializer xs = new XmlSerializer(typeof(Teacher));
            using (FileStream sr = new FileStream(ofd.FileName, FileMode.Open))
            
                var teacher = xs.Deserialize(sr) as Teacher;

                tNametxtBox.Text = teacher.Name1;
                tIDtxtBox.Text = teacher.ID1;
            
        
        catch(Exception ex)
        
            MessageBox.Show(ex.Message);
        
    

【讨论】：

我需要 btnClick 操作上的 OpenFileDialog...这不在 btnClick 操作中...我想 XD 谢谢