我有一个简单的控件,其中包含一个DataSrid,ItemsSource绑定到DataTable。当我填充DataTable时,我可以看到在DataGrid中添加了行,但是没有显示数据。我没有为此DataGrid使用任何特殊样式(采用默认样式),唯一的设置是将AutoGenerateColumn设置为True。
在XAML中
<DataGrid AutoGenerateColumns="True" ItemsSource="{Binding TableResult}"/>
在视图模型中
private DataTable tableResult = new DataTable();
public DataTable TableResult
{
get { return tableResult; }
set { tableResult = value; OnPropertyChanged("TableResult"); }
}
private void FillTable()
{
DataColumn c = new DataColumn();
c.ColumnName = "Col1";
this.TableResult.Columns.Add(c);
c = new DataColumn();
c.ColumnName = "Col2";
this.TableResult.Columns.Add(c);
DataRow row1 = this.TableResult.NewRow();
row1["Col1"] = "Blue";
row1["Col2"] = "12";;
this.TableResult.Rows.Add(row1);
DataRow row2 = this.TableResult.NewRow();
row2["Col1"] = "Red";
row2["Col2"] = "18";
this.TableResult.Rows.Add(row2);
DataRow row3 = this.TableResult.NewRow();
row3["Col1"] = "Yellow";
row3["Col2"] = "27";
this.TableResult.Rows.Add(row3);
}
答案 0 :(得分:2)
填充DataGrid数据的工作示例:
public MyViewModel()//constructor of MyViewModel
{
FillMyDataGrid();
}
private DataTable employeeDataTable;
public DataTable EmployeeDataTable
{
get { return employeeDataTable; }
set
{
employeeDataTable = value;
OnPropertyChanged("EmployeeDataTable");
}
}
public FillMyDataGrid()
{
var _ds = new DataSet("Test");
employeeDataTable = new DataTable();
employeeDataTable = _ds.Tables.Add("DT");
for (int i = 0; i < 50; i++)
{
//employeeDataTable.Columns.Add(i.ToString() + ".");
employeeDataTable.Columns.Add(i.ToString());
}
for (int i = 0; i < 2; i++)
{
var theRow = employeeDataTable.NewRow();
for (int j = 0; j < 50; j++)
{
theRow[j] = "a";
}
employeeDataTable.Rows.Add(theRow);
}
}
XAML代码:
<DataGrid ItemsSource="{Binding EmployeeDataTable}"/>
<强>更新强>
我已经测试了您的示例并且它可以正常工作,但您应该调用您的方法
填充FillTable()
的构造函数中的DataTable
。例如:
public class YourViewModel
{
public YourViewModel()
{
FillTable();
}
}
答案 1 :(得分:0)
您还必须定义列:例如:
<DataGrid AutoGenerateColumns="False" ItemsSource="{Binding Source=TableResult}" >
<DataGrid.Columns>
<DataGridTextColumn Header="Name" Binding="{Binding col1}"/>
<DataGridTextColumn Header="Surname" Binding="{Binding col2}"/>
<DataGridTextColumn Header="Phone" Binding="{Binding col3}" />
</DataGrid.Columns>
</DataGrid>
确保您已在视图模型中实施INotifyPropertyChanged
答案 2 :(得分:0)
你需要调用DefaultView,这里是FillTable()方法的改变,
private void FillTable()
{
DataColumn c = new DataColumn();
c.ColumnName = "Col1";
this.TableResult.Columns.Add(c);
c = new DataColumn();
c.ColumnName = "Col2";
this.TableResult.Columns.Add(c);
DataRow row1 = this.TableResult.NewRow();
row1["Col1"] = "Blue";
row1["Col2"] = "12"; ;
this.TableResult.Rows.Add(row1);
DataRow row2 = this.TableResult.NewRow();
row2["Col1"] = "Red";
row2["Col2"] = "18";
this.TableResult.Rows.Add(row2);
DataRow row3 = this.TableResult.NewRow();
row3["Col1"] = "Yellow";
row3["Col2"] = "27";
this.TableResult.Rows.Add(row3);
dataGrid1.ItemsSource = this.tableResult.DefaultView;
}
答案 3 :(得分:0)
您的ItemsSource需要是对象列表,而不是DataRow / Column。在任何情况下,在ViewModel上引用这些类型的东西都会破坏MVVM!
所以:
public List<Result> TableResult {get;set;} // raise property change on set
和
public class Result
{
public string Color {get;set;}
public int Number {get;set;}
}
编辑:由于您事先对表本身一无所知,我建议您使用一个简单的字符串列表来生成行和列。
这样的事情:
public List<List<string>> TableResult { get; set; } // raise property changed
和
List<string> columns = new List<string>() { "Col1", "Col2", "Col3", "Col4" };
List<string> row1 = new List<string>() { "a", "b", "c", "d" };
List<string> row2 = new List<string>() { "w", "x", "y", "z" };
List<List<string>> table = new List<List<string>>(){columns,row1,row2};
TableResult = table;
然后,您必须创建转换器以将行转换为DataTable:
public class ListToTables : IValueConverter
{
public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
{
var list = value as List<List<string>>;
var headers = list.First();
DataTable dt = new DataTable();
foreach (string header in headers)
{
dt.Columns.Add(header);
}
foreach (List<string> row in list.Skip(1))
{
int index = 0;
DataRow r = dt.NewRow();
foreach (string col in row)
{
r[index++] = col;
}
dt.Rows.Add(r);
}
return dt;
}
public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
{
throw new NotSupportedException();
}
}
在您的XAML上,将其添加到您的资源(例如UserControl.Resources):
<local:ListToTables x:Key="listToTables"></local:ListToTables>
其中&#34;本地&#34;是转换器的添加命名空间,并且:
<DataGrid ItemsSource="{Binding TableResult,Converter={StaticResource listToTables}}" >
用于您的DataGrid。
编辑决赛:
如果您认为MVVM允许,您可以保留DataTable,只需用以下代码替换代码的后半部分:
DataTable tempDataTable = new DataTable();
DataColumn c = new DataColumn();
c.ColumnName = "Col1";
tempDataTable.Columns.Add(c);
c = new DataColumn();
c.ColumnName = "Col2";
tempDataTable.Columns.Add(c);
DataRow row1 = tempDataTable.NewRow();
row1["Col1"] = "Blue";
row1["Col2"] = "12";;
tempDataTable.Rows.Add(row1);
DataRow row2 = tempDataTable.NewRow();
row2["Col1"] = "Red";
row2["Col2"] = "18";
tempDataTable.Rows.Add(row2);
DataRow row3 = tempDataTable.NewRow();
row3["Col1"] = "Yellow";
row3["Col2"] = "27";
tempDataTable.Rows.Add(row3);
this.DataTable = tempDataTable;
您必须将DataTable设置为引发属性更改。添加它不会做任何事情;)
答案 4 :(得分:0)
如果我添加以下内容,您的代码将适用于我
<a>
答案 5 :(得分:0)
我刚刚对此进行了角力,发现了另一个非常可靠的答案,该答案包含了此处的部分但并非全部其他内容。 Xaml(假设datacontext是在页面级别设置的):
<DataGrid ItemsSource="{Binding Findings}" />
ViewModel中的代码(假设使用数据创建了DataTable)
Findings = null;
Findings = dataTable.DefaultView
关键是首先将绑定属性显式设置为null。我发现在WinForms中也是如此。