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Changing DataGrid cell value programmatically in WPF

I have a view DataGridthat populates it with the contents of a list at runtime. I want to change (for example) the value of the cell in row: 2 and column: 3.

How can i do this?

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2 answers

It is not that difficult.

You need helper functions;

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 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;
}

public static DataGridCell GetCell(this DataGrid grid, int row, int column)
{
    DataGridRow rowContainer = grid.GetRow(row);
    return grid.GetCell(rowContainer, column);
}

Use yourDataGrid.GetCell (5, 3) .Content = "all you need";

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For example, if you have a TextBox on your grid

<TextBox  Grid.Row="2" Grid.Column="3" Text="{Binding ...}"/>

Full sample

<Grid x:Name="RootElement">
      <Grid.ColumnDefinitions>
        <ColumnDefinition Width="*"/>
        <ColumnDefinition Width="*"/>
        <ColumnDefinition Width="*"/>
      </Grid.ColumnDefinitions>

      <Grid.RowDefinitions>
        <RowDefinition/>
        <RowDefinition/>
        <RowDefinition/>
      </Grid.RowDefinitions>

      ...
      <TextBox  Grid.Row="2" Grid.Column="3" Text="{Binding ...}"/>
      ...
    </Grid>

You can use these helpers; you have this link http://techiethings.blogspot.fr/2010/05/get-wpf-datagrid-row-and-cell.html .

, stackoverflow, ,

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;
}

public static DataGridCell GetCell(this DataGrid grid, int row, int column)
{
    DataGridRow rowContainer = grid.GetRow(row);
    return grid.GetCell(rowContainer, column);
}
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