获取 DataGrid WPF 中选定的行项

If you're using the MVVM pattern you can bind a SelectedRecord property of your VM with SelectedItem of the DataGrid, this way you always have the SelectedValue in you VM. Otherwise you should use the SelectedIndex property of the DataGrid.

You can use the SelectedItem property to get the currently selected object, which you can then cast into the correct type. For instance, if your DataGrid is bound to a collection of Customer objects you could do this:

Customer customer = (Customer)myDataGrid.SelectedItem;

Alternatively you can bind SelectedItem to your source class or ViewModel.

<Grid DataContext="MyViewModel">
<DataGrid ItemsSource="{Binding Path=Customers}"
SelectedItem="{Binding Path=SelectedCustomer, Mode=TwoWay}"/>
</Grid>

public IEnumerable<DataGridRow> GetDataGridRows(DataGrid grid)
{
var itemsSource = grid.ItemsSource as IEnumerable;
if (null == itemsSource) yield return null;
foreach (var item in itemsSource)
{
var row = grid.ItemContainerGenerator.ContainerFromItem(item) as DataGridRow;
if (null != row) yield return row;
}
}


private void DataGrid_Details_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
try
{
var row_list = GetDataGridRows(DataGrid_Details);
foreach (DataGridRow single_row in row_lis)
{
if (single_row.IsSelected == true)
{
MessageBox.Show("the row no."+single_row .GetIndex ().ToString ()+ " is selected!");
}
}


}
catch { }
}

Well I will put similar solution that is working fine for me.

 private void DataGrid1_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
try
{
if (DataGrid1.SelectedItem != null)
{
if (DataGrid1.SelectedItem is YouCustomClass)
{
var row = (YouCustomClass)DataGrid1.SelectedItem;


if (row != null)
{
// Do something...


//  ButtonSaveData.IsEnabled = true;


//  LabelName.Content = row.Name;


}
}
}
}
catch (Exception)
{
}
}

if I select the second row -

 Dim jason As DataRowView




jason = dg1.SelectedItem


noteText.Text = jason.Item(0).ToString()

noteText will be 646. This is VB, but you get it.

private void Fetching_Record_Grid_MouseDoubleClick_1(object sender, MouseButtonEventArgs e)
{
IInputElement element = e.MouseDevice.DirectlyOver;
if (element != null && element is FrameworkElement)
{
if (((FrameworkElement)element).Parent is DataGridCell)
{
var grid = sender as DataGrid;
if (grid != null && grid.SelectedItems != null && grid.SelectedItems.Count == 1)
{
//var rowView = grid.SelectedItem as DataRowView;
try
{
Station station = (Station)grid.SelectedItem;
id_txt.Text =  station.StationID.Trim() ;
description_txt.Text =  station.Description.Trim();
}
catch
{


}
}
}
}
}

This is pretty simple in this DataGrid dg and item class is populated in datagrid and listblock1 is a basic frame.

private void DataGrid_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
try
{
var row_list = (Item)dg.SelectedItem;
listblock1.Content = "You Selected: " + row_list.FirstName + " " + row_list.LastName;
}
catch { }


}
public class Item
{
public string FirstName { get; set; }
public string LastName { get; set; }
}

Just discovered this one after i tried Fara's answer but it didn't work on my project. Just drag the column from the Data Sources window, and drop to the Label or TextBox.

use your Model class to get row values selected from datagrid like,

        XDocument xmlDoc = XDocument.Load(filepath);


if (tablet_DG.SelectedValue == null)
{
MessageBox.Show("select any record from list..!", "select atleast one record", MessageBoxButton.OKCancel, MessageBoxImage.Warning);
}
else
{
try
{
string tabletID = "";


/*here i have used my model class named as TabletMode*/


var row_list = (TabletModel)tablet_DG.SelectedItem;
tabletID= row_list.TabletID;


var items = from item in xmlDoc.Descendants("Tablet")
where item.Element("TabletID").Value == tabletID
select item;


foreach (var item in items)
{
item.SetElementValue("Instance",row_list.Instance);
item.SetElementValue("Database",row_list.Database);
}


xmlDoc.Save(filepath);
MessageBox.Show("Details Updated..!"
+ Environment.NewLine + "TabletId: " +row_list.TabletID + Environment.NewLine
+ "Instance:" + row_list.Instance + Environment.NewLine + "Database:" + row_list.Database, "", MessageBoxButton.YesNoCancel, MessageBoxImage.Information);
}


catch (Exception ex)
{
MessageBox.Show(ex.StackTrace);
}
}

You can also:

DataRowView row = dataGrid.SelectedItem as DataRowView;
MessageBox.Show(row.Row.ItemArray[1].ToString());

@Krytox answer with MVVM

    <DataGrid
Grid.Column="1"
Grid.Row="1"
Margin="10" Grid.RowSpan="2"
ItemsSource="{Binding Data_Table}"
SelectedItem="{Binding Select_Request, Mode=TwoWay}" SelectionChanged="DataGrid_SelectionChanged"/>//The binding






#region View Model
private DataRowView select_request;
public DataRowView Select_Request
{
get { return select_request; }
set
{
select_request = value;
OnPropertyChanged("Select_Request"); //INotifyPropertyChange
OnSelect_RequestChange();//do stuff
}
}

There are a lot of answers here that probably work in a specific context, but I was simply trying to get the text value of the first cell in a selected row. While the accepted answer here was the closest for me, it still required creating a type and casting the row into that type. I was looking for a simpler solution, and this is what I came up with:

MessageBox.Show(((DataRowView)DataGrid.SelectedItem).Row[0].ToString());

This gives me the first column in the selected row. Hopefully this helps someone else.

I got something like this:

private void DataGrid_SelectionChanged(object sender, SelectionChangedEventArgs e) {


DataGrid dg = (DataGrid)sender;


DataRowView selectedRow = dg.SelectedItem as DataRowView;


if(selectedRow != null) {
txtXXX.Text = selectedRow["xxx"].ToString();
txtYYY.Text = selectedRow["yyy"].ToString();
txtZZZ.Text = selectedRow["zzz"].ToString();
}


}

Your front code:

• set your model namespace xmlns:viewModel="clr-namespace:CPL3_workstation.MVVM.ModelViews.Tables"
• set xmlns:d="http://schemas.microsoft.com/expression/blend/2008", because d:DataContext="{d:DesignInstance...}

<DataGrid d:DataContext = "{d:DesignInstance viewModel:TestViewModel, IsDesignTimeCreatable = True}"
ItemsSource = "{Binding Rows}"
SelectedItem = "{Binding Selector}">
<DataGrid.Columns>
<DataGridTextColumn
Binding = "{Binding Content1}"
Header = "Column 1" />
<DataGridTextColumn
Binding = "{Binding Content2}"
Header = "Column 2" />
<DataGridTextColumn
Binding = "{Binding Content3}"
Header = "Column 3" />
</DataGrid.Columns>
</DataGrid>

Your ViewModel and the Model below:

namespace CPL3_workstation.MVVM.ModelViews.Tables
{
public class TestViewModel
{
public IEnumerable<TestModel> Rows { get; set; }


public TestModel Selector
{
get => selector;
set => selector = value;
}


private TestModel selector;


public TestViewModel()
{
Rows = new List<TestModel>
{
new TestModel{ Content1 = "one", Content2 = "two", Content3 = "three" },
new TestModel{ Content1 = "one", Content2 = "two", Content3 = "three" },
new TestModel{ Content1 = "one", Content2 = "two", Content3 = "three" },
new TestModel{ Content1 = "one", Content2 = "two", Content3 = "three" }
};
}
}


public class TestModel
{
public string Content1 { get; set; }
public string Content2 { get; set; }
public string Content3 { get; set; }
}
}

As you can see, the SelectedItem has a binding to the Selector property of the type of your model.