将DataTable转换为XML文件,反之亦然

Question

我在读取XML文件时遇到问题DataTable.最初,我正在写一个DatatableXML文件并保存它.现在,当我想将XML文件读回时DataTable,它没有发生.

以下代码用于编写文件:

private void saveAsToolStripMenuItem_Click(object sender, EventArgs e) 
{
    if (myDT_For_DGV.Rows.Count != 0)
    {
        saveFileDialog1.ShowDialog();
        saveFileDialog1.FileName = "checkOutFile.xml";
        myDT_For_DGV.TableName = "CheckOutsAndIns";
        myDT_For_DGV.WriteXml(saveFileDialog1.FileName, true);
    }
    else
    {
        MessageBox.Show("Please add licences to DataGridView, you havent added any licences as of now", "Alert");
    }
}

并且为了阅读文件:

private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
    //write code to open file
    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        //myFile = openFileDialog1.FileName;
        System.IO.MemoryStream xmlStream = new System.IO.MemoryStream();

        xmlStream.Position = 0;

        myDT_For_DGV.ReadXml(openFileDialog1.FileName);
        //MessageBox.Show(openFileDialog1.FileName);
    }
}

Answer 1

使用更高级别的DataSet可能更容易,如下所示:

DataSet dataSet = new DataSet();
DataTable dataTable = new DataTable();

dataSet.Tables.Add(dataTable);
// Save to disk
dataSet.WriteXml(@"C:\MyDataset.xml");

// Read from disk
dataSet.ReadXml(@"C:\MyDataset.xml");

Answer 2

我修复了它，问题是，表名是在保存时分配的，但不是在读取时分配的。所以全局分配表名，这样读写就没有任何问题了。

所以代码将是

myDT_For_DGV.TableName = "CheckOutsAndIns";

if (openFileDialog1.ShowDialog() == DialogResult.OK) 
       {              
              myDT_For_DGV.ReadXml(@openFileDialog1.FileName);
            //MessageBox.Show(openFileDialog1.FileName);

        }

//TO WRITE TO XML
if (myDT_For_DGV.Rows.Count != 0)
        {
            saveFileDialog1.ShowDialog();
            saveFileDialog1.FileName = "checkOutFile.xml";
            myDT_For_DGV.WriteXml(saveFileDialog1.FileName, true);
        }