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。有帮助吗?感谢
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;
}
修改
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..现在我需要的是从OpenBileDialog中打开一个文件,用BtnClick操作选择文件并加载文本框。感谢
答案 0 :(得分:3)
加载Button_Click:
为您的按钮添加事件处理程序(此处我称之为button1
):
button1.Click += button1_Click;
在button1_Click
你可以做到:
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);
}
}
}