数据网格Wpf C#
我想在wpf中显示datagrid中的数据。动态显示数据的方法是什么数据网格Wpf C#,c#,wpf,datagrid,C#,Wpf,Datagrid,我想在wpf中显示datagrid中的数据。动态显示数据的方法是什么 Random rm = new Random(); Random rm2 = new Random(); string code = "034" + rm2.Next(0, 7); string num = code + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(
Random rm = new Random();
Random rm2 = new Random();
string code = "034" + rm2.Next(0, 7);
string num = code + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9);
for (int i = 1; i <= 10000; i++)
{
code = "034" + rm2.Next(0, 7);
num = code + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) ;
dataGrid1.ItemsSource = num.ToString();
}
Random rm=new Random();
随机rm2=新随机();
字符串代码=“034”+rm2.Next(0,7);
string num=code+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9);
对于(int i=1;i您可以将实现IEnumerable
的任何类型设置为ItemsSource
对于DataGrid
例如,observetecollection
:
XAML:
<DataGrid Name="dataGrid"></DataGrid>
或使用数据表:
public MainWindow()
{
FillDataGrid();
}
private DataTable employeeDataTable;
private void FillDataGrid()
{
employeeDataTable = new DataTable();
employeeDataTable = _ds.Tables.Add("DT");
for (int i = 0; i < 80; i++)
{
employeeDataTable.Columns.Add(i.ToString());
}
for (int i = 0; i < 100; i++)
{
var theRow = employeeDataTable.NewRow();
for (int j = 0; j < 80; j++)
{
theRow[j] = "a";
}
employeeDataTable.Rows.Add(theRow);
}
gridEmployees.ItemsSource = employeeDataTable.DefaultView;
}
public主窗口()
{
FillDataGrid();
}
私有数据表employeeDataTable;
私有void FillDataGrid()
{
employeeDataTable=新数据表();
employeeDataTable=_ds.Tables.Add(“DT”);
对于(int i=0;i<80;i++)
{
employeeDataTable.Columns.Add(i.ToString());
}
对于(int i=0;i<100;i++)
{
var theRow=employeeDataTable.NewRow();
对于(int j=0;j<80;j++)
{
theRow[j]=“a”;
}
employeeDataTable.Rows.Add(theRow);
}
gridEmployees.ItemsSource=employeeDataTable.DefaultView;
}
您可以将实现IEnumerable
的任何类型设置为ItemsSource
forDataGrid
例如,observetecollection
:
XAML:
<DataGrid Name="dataGrid"></DataGrid>
或使用数据表:
public MainWindow()
{
FillDataGrid();
}
private DataTable employeeDataTable;
private void FillDataGrid()
{
employeeDataTable = new DataTable();
employeeDataTable = _ds.Tables.Add("DT");
for (int i = 0; i < 80; i++)
{
employeeDataTable.Columns.Add(i.ToString());
}
for (int i = 0; i < 100; i++)
{
var theRow = employeeDataTable.NewRow();
for (int j = 0; j < 80; j++)
{
theRow[j] = "a";
}
employeeDataTable.Rows.Add(theRow);
}
gridEmployees.ItemsSource = employeeDataTable.DefaultView;
}
public主窗口()
{
FillDataGrid();
}
私有数据表employeeDataTable;
私有void FillDataGrid()
{
employeeDataTable=新数据表();
employeeDataTable=_ds.Tables.Add(“DT”);
对于(int i=0;i<80;i++)
{
employeeDataTable.Columns.Add(i.ToString());
}
对于(int i=0;i<100;i++)
{
var theRow=employeeDataTable.NewRow();
对于(int j=0;j<80;j++)
{
theRow[j]=“a”;
}
employeeDataTable.Rows.Add(theRow);
}
gridEmployees.ItemsSource=employeeDataTable.DefaultView;
}
Random rm=new Random();
随机rm2=新随机();
字符串代码=“034”+rm2.Next(0,7);
List numList=新列表();
string num=code+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9);
对于(int i=1;iRandom rm=new Random();
随机rm2=新随机();
字符串代码=“034”+rm2.Next(0,7);
List numList=新列表();
string num=code+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9)+rm.Next(0,9);
对于(int i=1;i首先,您可以将所有num
值分配给列表或可观察的集合。然后您可以将此列表或集合设置为datagrid的itemsource
List<string> numLst = new List<string>();
for (int i = 1; i <= 10000; i++)
{
code = "034" + rm2.Next(0, 7);
num = code + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) ;
numLst.Add(num);
}
dataGrid1.ItemsSource = numLst;
List numLst=new List();
对于(int i=1;i首先,您可以将所有num
值分配给列表或可观察的集合。然后您可以将此列表或集合设置为datagrid的itemsource
List<string> numLst = new List<string>();
for (int i = 1; i <= 10000; i++)
{
code = "034" + rm2.Next(0, 7);
num = code + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) + rm.Next(0, 9) ;
numLst.Add(num);
}
dataGrid1.ItemsSource = numLst;
List numLst=new List();
对于(int i=1;我@ArsalanAhmed可以自由提问。如果你觉得我的回答对你有帮助,那么你可以将我的回答标记为一个答案,以简化将来对其他人的搜索。请阅读此@ArsalanAhmed,可以自由提问。如果你觉得我的回答对你有帮助,那么你可以将我的回答标记为一个答案,以简化将来的搜索其他人的h。请看这个