WPF 如何从 DataGrid 转换为 DataTable？

Question

您好，我关注这个guide 是为了了解如何使用 DataGrid。

我遇到的问题是如何将 DataGrid 中的数据转换为 DataTable？

我正在尝试使用的代码是：

DataTable dt = ((DataView)dg.ItemsSource).ToTable();

但它给了我一个错误说明：

无法转换类型的对象 'System.Collections.Generic.List`1[WPFProject.Person]' 输入 'System.Data.DataView'。

我的代码与示例非常相似，除了我使用 Person 类而不是 user 并且我创建了 Person 类型的列表以便将数据插入到数据网格中。

public class Person
    {
        public bool CheckBox { get; set; }
        public int ID { get; set; }
        public string Name { get; set; }
    }

谢谢。

Answer 1

你不能那样转换它。只有当您的DataGrid 的ItemSource 不是DataView 时，您的代码才有效。您需要明确地编写代码（您可以轻松地将任何 collection<T> 谷歌搜索到 DataTable）将您的 collection 转换为 DataTable。看看下面的例子：

XAML

  <StackPanel>
    <DataGrid ItemsSource="{Binding list}" x:Name="myGrid"/>
    <Button Content="convert back" Click="Button_Click_1" />
</StackPanel>

逻辑：

private void Button_Click_1(object sender, RoutedEventArgs e)
    {
        var list = new List<MyClass>(myGrid.ItemsSource as IEnumerable<MyClass>);
        var dataTable = ToDataTable(list);
        if (dataTable != null)
        {

        }
    }

    public static DataTable ToDataTable<T>(List<T> items)
    {
        DataTable dataTable = new DataTable(typeof(T).Name);

        //Get all the properties
        PropertyInfo[] Props = typeof(T).GetProperties(BindingFlags.Public | BindingFlags.Instance);
        foreach (PropertyInfo prop in Props)
        {
            //Setting column names as Property names
            dataTable.Columns.Add(prop.Name);
        }
        foreach (T item in items)
        {
            var values = new object[Props.Length];
            for (int i = 0; i < Props.Length; i++)
            {
                //inserting property values to datatable rows
                values[i] = Props[i].GetValue(item, null);
            }
            dataTable.Rows.Add(values);
        }
        //put a breakpoint here and check datatable
        return dataTable;
    }

输出

Answer 2

将 DataGrid 转换为 DataTable

1.DataGrid Itemsource与CandidateClass列表绑定。

2.从DataGrid的Itemsource中获取Item，然后序列化为Json，反序列化为DataTable。

  var Item = ListDataGrid.ItemsSource as IList<CandidateClass>;
  var json = JsonConvert.SerializeObject(Item);
  DataTable dt = (DataTable)JsonConvert.DeserializeObject(json, (typeof(DataTable)));

注意：如果DataGrid是基于类对象的列表形成的。这个方法很有帮助。

Answer 3

复制过去的方法1

public static DataTable DataGridtoDataTable(DataGrid dg)
        {
dg.SelectAllCells();
            dg.ClipboardCopyMode = DataGridClipboardCopyMode.IncludeHeader;
            ApplicationCommands.Copy.Execute(null, dg);
            dg.UnselectAllCells();
            String result = (string)Clipboard.GetData(DataFormats.CommaSeparatedValue);
            string[] Lines = result.Split(new string[] { "\r\n", "\n" }, 
            StringSplitOptions.None);
            string[] Fields;
            Fields = Lines[0].Split(new char[] { ',' });
            int Cols = Fields.GetLength(0);
            DataTable dt = new DataTable();
            //1st row must be column names; force lower case to ensure matching later on.
            for (int i = 0; i < Cols; i++)
                dt.Columns.Add(Fields[i].ToUpper(), typeof(string));
            DataRow Row;
            for (int i = 1; i < Lines.GetLength(0)-1; i++)
            {
                Fields = Lines[i].Split(new char[] { ',' });
                Row = dt.NewRow();
                for (int f = 0; f < Cols; f++)
                {
                    Row[f] = Fields[f];
                }
                dt.Rows.Add(Row);
            }
            return dt;
            
        }

================================================ ==========================================

方法二

public DataTable DataGridtoDataTable(DataGrid dg)
        {

            DataTable dt = new DataTable();

            for (int i = 0; i <= dg.Columns.Count - 1; i++)
            {
                dt.Columns.Add(dg.Columns[i].Header.ToString(), typeof(string));
            }
            DataRow Row;

            for (int i = 0; i <= dg.Items.Count - 1; i++)
            {
                Row = dt.NewRow();

                for (int k = 0; k <= dg.Columns.Count - 1; k++)
                {
                    Row[dg.Columns[k].Header.ToString()] = gettabelcell(i, k,dg);

                }
                dt.Rows.Add(Row);
            }

            return dt;

        }

public String gettabelcell(int x, int y,DataGrid dg)
        {
            try
            {
                String myString = dg.GetCell(x, y).ToString();

                if (myString.Contains(":"))
                {
                    String[] s = myString.Split(':');
                    String item = s[1];
                    return item.Trim();
                }
                else
                {
                    return "";
                }
            }
            catch
            {
                return "";
            }
        }

这是课

static class ExtensionHelpers
   {

       public static T GetVisualChild<T>(Visual parent) where T : Visual
       {
           T child = default(T);
           int numVisuals = VisualTreeHelper.GetChildrenCount(parent);
           for (int i = 0; i < numVisuals; i++)
           {
               Visual v = (Visual)VisualTreeHelper.GetChild(parent, i);
               child = v as T;
               if (child == null)
               {
                   child = GetVisualChild<T>(v);
               }
               if (child != null)
               {
                   break;
               }
           }
           return child;
       }

       public static DataGridCell GetCell(this DataGrid grid, int row, int column)
       {
           DataGridRow rowContainer = grid.GetRow(row);
           return grid.GetCell(rowContainer, column);
       }
       public static DataGridRow GetSelectedRow(this DataGrid grid)
       {
           return (DataGridRow)grid.ItemContainerGenerator.ContainerFromItem(grid.SelectedItem);
       }
       public static DataGridRow GetRow(this DataGrid grid, int index)
       {
           DataGridRow row = (DataGridRow)grid.ItemContainerGenerator.ContainerFromIndex(index);
           if (row == null)
           {
               // May be virtualized, bring into view and try again.
               grid.UpdateLayout();
               grid.ScrollIntoView(grid.Items[index]);
               row = (DataGridRow)grid.ItemContainerGenerator.ContainerFromIndex(index);
           }
           return row;
       }

       public static DataGridCell GetCell(this DataGrid grid, DataGridRow row, int column)
       {
           if (row != null)
           {
               DataGridCellsPresenter presenter = GetVisualChild<DataGridCellsPresenter>(row);

               if (presenter == null)
               {
                   grid.ScrollIntoView(row, grid.Columns[column]);
                   presenter = GetVisualChild<DataGridCellsPresenter>(row);
               }

               DataGridCell cell = (DataGridCell)presenter.ItemContainerGenerator.ContainerFromIndex(column);
               return cell;
           }
           return null;
       }
   }

Answer 4

你不能像在 winforms 中那样在 WPF 中做到这一点。在 Wpf 中，您可以像这样从 dataTable 设置 datagrid ItemsSource。

dg.ItemsSource = dataTable.AsDataView();

在您的情况下，您想从 datagrid 获取 dataTable。 所以你可以试试下面的代码。

DataView view = (DataView) dg.ItemsSource;
DataTable dataTable = view.Table.Clone();
foreach (var dataRowView in view)
{
    dataTable.ImportRow(dataRowView.Row);
}
var dataTableFromDataGrid = dataTable;