Datagrid Wpf C#

Question

Ich möchte Daten in Datagrid in wpf.What zeigt die Art, Daten dynamisch anzuzeigen.

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(); 

} 

Answer 1

Sie können jede Art eingestellt, die IEnumerable als ItemsSource für DataGrid implementiert.

Zum Beispiel ObservableCollection<T>:

XAML:

<DataGrid Name="dataGrid"></DataGrid> 

Code-Behind:

public MainWindow() 
{ 
    FillDataGrid(); 

} 

private void FillDataGrid() 
{ 
    ObservableCollection<Person> coll = new ObservableCollection<Person>(); 
    for (int start = 0; start < 10; start++) 
    { 
     coll.Add(new Person() { IdPerson = start, Name = "Bill" + start.ToString(), SurName = "ONeill" }); 
    } 
    dataGrid.ItemsSource = coll; 
} 

Modellklasse:

public class Person 
{ 
    public int IdPerson { get; set; } 
    public string Name { get; set; } 
    public string SurName { get; set; } 
} 

OR mit D ataTable:

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; 
} 

Answer 2

Random rm = new Random(); 
    Random rm2 = new Random(); 
    string code = "034" + rm2.Next(0, 7); 
    List<string> numList = new List<string>(); 
    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) ; 

     numList.Add(num); 
    } 
    dataGrid1.ItemsSource = numList; 

Aktualisieren Sie Ihren Code mit diesem. Es sollte funktionieren.

Hoffe das hilft !!

Answer 3

Zuerst können Sie alle num Werte einer Liste oder einer Observable Collection zuweisen. Dann können Sie diese Liste oder Sammlung als Artikelquelle für Ihr Datagrid festlegen.

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;