Avoid creating empty row in datagrid

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I'm having two classes Cricket, Football and List of Observable collections of type object. Based on certain condition I want to add the object of type Cricket/Football to Observable Collection. I'm not assigning any data i.e just creating and instance of class Cricket/Football and adding that instance to Observable Collections and binding to UI. My expectation is, as I'm not assigning any data to the instance of Cricket/Football, only header has to create in the datagrid. But what I found was a row with the default value of the variables defined under the respective class along with the row header as I'm creating the instance of that class. How shall I avoid creating void row where my datagrid header is unaffected.

HTML

<DataGrid SelectionMode="Single" VerticalAlignment="Stretch" ItemsSource="{Binding itemSource, UpdateSourceTrigger=PropertyChanged}" 
CanUserReorderColumns="False" CanUserAddRows="False" IsReadOnly="True" CanUserDeleteRows="False" CanUserResizeColumns="False" 
HorizontalGridLinesBrush="Black" VerticalGridLinesBrush="Black" HorizontalScrollBarVisibility="Disabled" VerticalScrollBarVisibility="Disabled" />

What I have tried:

I googled for the above mentioned problem and all I was found is to set CanUserAddRows property to false

Posted 25-Jul-16 4:11am

partha143

Updated 25-Jul-16 4:14am

v2

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partha143 25-Jul-16 10:12am

My Model :
<pre lang="c#">
public class Cricket : INotifyPropertyChanged
{
private string batsmen;
private int scorecore;

public string Batsmen
{
get { return batsmen; }
set
{
batsmen = value;
OnPropertyChanged("Batsmen");
}
}

public int Scorecore
{
get { return scorecore; }
set
{
scorecore = value;
OnPropertyChanged("Scorecore");
}
}
}
</pre>

my View Model :
<pre lang="c#">
Cricket cricket = new Cricket();
MyList.Add(cricket);

ObservableCollection<object> myList;
public ObservableCollection<object> MyList
{
get { return myList; }
set
{
myList = value;
OnPropertyChanged("MyList");
}
}
</pre>

MayurDighe 26-Jul-16 14:04pm

Dear parth143,

Please use "Improve question" widget to provide additional information in question itself.

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