CODEWITHSUNDEEP
silverlightsilverlight-4.0
happysmile
happysmile

Reputation: 7777

cannot create an instance in xaml in silverlight 4.0

In the xaml code i get an error telling cannot create an instance of this AllEmployeeViewModel class file, actually this class file exists in the solution folder when i type scr: the intellsene shows me the class file

<UserControl.Resources>
    <scr:AllEmployeeViewModel  x:Key="empName"></scr:AllEmployeeViewModel>
</UserControl.Resources>
<Grid x:Name="MainGrid" Background="White" Width="400"
    Height="407" DataContext="{Binding Source={StaticResource empName}}" >
<Grid  x:Name="grdAllEmp" DataContext="{Binding Path=EmployeeClass}">
    <sdk:DataGrid AutoGenerateColumns="True" Height="274" 
                    HorizontalAlignment="Left" Margin="8,8,0,0" 
                    Name="dgEmployee" VerticalAlignment="Top" Width="385"
                    ItemsSource="{Binding}"/>
    <Button Content="Get All Employees" Height="23" 
            HorizontalAlignment="Left" Margin="12,288,0,0" 
            Name="btnAllEmplloyees" VerticalAlignment="Top" Width="381" 
            Command="{Binding  Path=DataContext.GetEmployees,ElementName=MainGrid}"/>
</Grid>

i am trying to bind the data to grid, if i ignore the compile time error and run its gives an error key not found.

please let me know the solutionif you know,working on this issue from past 2days

any help would be great thanks.

Upvotes: 1

Views: 4827

Answers (5)

Zeeshan Ajmal
Zeeshan Ajmal

Reputation: 880

I was getting the same error, i will explain it to you hopefully it will help you. In my ViewModel's constructor i was executing some code, all code was withing below if condition except,

       If(!IsInDesignMode)
       {
             //  my code
       }

       //   Problamatic method execution point

Except a method that i wanted to execute every time, but it turns out that you can only execute code if above condition is satisfied else your view model instance will not be created.

So to avoid this you have to do like that:

       If(!IsInDesignMode)
       {
             //  my code
            //   Problamatic method execution point
       }

Put all your code inside that condition and everything will be fine.
Note: I was using MVVMLight library along with Model-View-ViweModel pattern.

Upvotes: 0

Binu Augustine
Binu Augustine

Reputation: 41

I too had the same issue. cannot create instance of viewmodel

Just copy this code and place it in ViewModel 

public bool IsDesignTime
{
    get
    {
      return (Application.Current == null) || 
             (Application.Current.GetType() == typeof(Application));
    }
}




//Constructor    
public ViewModelClass()
{
    if(IsDesignTime == false)
    {
       //Your Code 
    }
}

Upvotes: 4

happysmile
happysmile

Reputation: 7777

Just add this line in MainPage.xaml.cs page

InitializeComponent();
        if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode(this))
        {
            //Code that throws the exception
        }

it works fine.

Upvotes: 2

iCollect.it Ltd
iCollect.it Ltd

Reputation: 93621

You have a binding to EmployeeClass. That has to be a collection of some type for this to work, but the name EmployeeClass sounds like a single object and not a collection.

You really needed to post your View Model code as we had to guess this.

I put together a quick example and if the ViewModel contains:

public ObservableCollection<EmployeeClass> Employees { get; set; }

and I populate them with a few sample EmployeeClass objects, 

public AllEmployeeViewModel()
{
    this.Employees  = new ObservableCollection<EmployeeClass>();
    this.Employees.Add(new EmployeeClass() { Name = "One" });
    this.Employees.Add(new EmployeeClass() { Name = "Two" });

and I change the binding to:

<Grid  x:Name="grdAllEmp" DataContext="{Binding Path=Employees}">

It looks like this (no other changes): enter image description here

Upvotes: 0

Luke Woodward
Luke Woodward

Reputation: 65054

Does your class AllEmployeeViewModel have a zero-argument constructor? WIthout that, Silverlight cannot create an instance of your class.

Upvotes: 0

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