Question

我知道这是一个非常简单和基本的问题，但我想知道如何解决这个问题。我在从datagrdiview序列化对象时遇到了问题。

此行在

中对我不起作用

if (dataGridView1 != null) 
    _example.Details = dataGridView1.DataBindings;   // Not Working.

请告诉我我在这里做的错误是什么......！谢谢。

private void button3_Click(object sender, EventArgs e)  //Read
{
            XmlSerializer _serializer = new XmlSerializer(typeof(XMLExample));
            XMLExample _example = new XMLExample();

            // Read file.
            using (TextReader textReader = new StreamReader(@"C:\test\testserialization.xml"))
            {
                _example = (XMLExample)_serializer.Deserialize(textReader);
                textReader.Close();
            }

            textBox1.Text = _example.ID;
            textBox2.Text = _example.Initial;

            dataGridView1.DataSource = _example.Details;
        }

        private void button2_Click(object sender, EventArgs e) //Write
        {
            XmlSerializer _serializer = new XmlSerializer(typeof(XMLExample));
            XMLExample _example = new XMLExample();

            _example.ID = textBox1.Text;
            _example.Initial = textBox2.Text;

            List<Detail> _source = new List<Detail>();

            for (int i = 0; i < 10; i++)
            {
                _source.Add(new Detail
                {
                    FirstName = "Name_" + i,
                    LastName = "Surname_" + i,
                    Section = "section_"+i,
                });
            }

            dataGridView1.DataSource = _source;

            if (dataGridView1 != null && dataGridView1.DataSource != null) 
               _example.Details = (List<Detail>)dataGridView1.DataSource;

            using (TextWriter textWriter = new StreamWriter(@"C:\test\testserialization.xml"))
            {
                _serializer.Serialize(textWriter, _example);
                textWriter.Close();
            }
        }

班级档案：

[Serializable]
public class StudentInfo : INotifyPropertyChanged
{
    public event PropertyChangedEventHandler PropertyChanged;

    public string UserID { get; set; } // Textbox1
    public string Department { get; set; } // Textbox2
    public List<Detail> Details { get; set; }
}

[Serializable]
public class Detail // Datagridview1
{
    [XmlElement("FirstName")] //Datagridview
    public string FirstName { get; set; }
    [XmlElement("LastName")] //Datagridview
    public string LastName { get; set; }
    [XmlElement("Section")] //Datagridview
    public string Section { get; set; }
}

Answer 1

试试这个：

if (dataGridView1 != null && dataGridView1.DataSource != null) test.Details = (List<Detail>)dataGridView1.DataSource;

将代码List<Detail>的示例代码添加到DataGridView对象的属性DataSource（在构造函数中添加此代码）：

 List<Detail> _source = new List<Detail>();
            for (int i = 0; i < 10; i++)
            {
                _source.Add(new Detail
                {
                    FirstName = "Name_" + i,
                    LastName = "Surname_" + i,
                    Section = "Section_" + i
                });
            }

            dataGridView1.DataSource = _source;

Answer 2

您必须序列化不是dataGridView和DataTable或DataSource

示例：

XmlSerializer ser = new XmlSerializer(typeof(DataTable));
DataTable dt = new DataTable("data");
TextWriter writer = new StreamWriter(Application.StartupPath+"\\"+fname+".xml");
ser.Serialize(writer, dt);
writer.Close();

将datagridview序列化为xml文件无法正常工作

2 个答案: