How can i change viewModel dynamically if i have one viewModel for each user control
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In main window this is my data context
DataContext="Binding Source=StaticResource VMLocator, Path=MainVM"
and this is my grid binding
<Grid>
<ContentControl Content="Binding" Style="StaticResource ChatContentStyle"/>
</Grid>
and that's how i switch them
<DataTemplate x:Key="LoginTemplate">
<views:LoginView/>
</DataTemplate>
<DataTemplate x:Key="ChatTemplate">
<views:ChatView/>
</DataTemplate>
<Style x:Key="ChatContentStyle" TargetType="ContentControl">
<Setter Property="ContentTemplate" Value="StaticResource LoginTemplate"/>
<Style.Triggers>
<DataTrigger Binding="Binding UserMode" Value="x:Static enums:UserModes.Chat">
<Setter Property="ContentTemplate" Value="StaticResource ChatTemplate"/>
</DataTrigger>
</Style.Triggers>
</Style>
I want to change the viewModel exact like I'm changing the view
wpf mvvm
asked Nov 11 at 5:54
ישראל קוסייב
83
add a comment |
up vote
0
down vote
favorite
In main window this is my data context
DataContext="Binding Source=StaticResource VMLocator, Path=MainVM"
and this is my grid binding
<Grid>
<ContentControl Content="Binding" Style="StaticResource ChatContentStyle"/>
</Grid>
and that's how i switch them
<DataTemplate x:Key="LoginTemplate">
<views:LoginView/>
</DataTemplate>
<DataTemplate x:Key="ChatTemplate">
<views:ChatView/>
</DataTemplate>
<Style x:Key="ChatContentStyle" TargetType="ContentControl">
<Setter Property="ContentTemplate" Value="StaticResource LoginTemplate"/>
<Style.Triggers>
<DataTrigger Binding="Binding UserMode" Value="x:Static enums:UserModes.Chat">
<Setter Property="ContentTemplate" Value="StaticResource ChatTemplate"/>
</DataTrigger>
</Style.Triggers>
</Style>
I want to change the viewModel exact like I'm changing the view
wpf mvvm
asked Nov 11 at 5:54
ישראל קוסייב
83
Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In main window this is my data context
DataContext="Binding Source=StaticResource VMLocator, Path=MainVM"
and this is my grid binding
<Grid>
<ContentControl Content="Binding" Style="StaticResource ChatContentStyle"/>
</Grid>
and that's how i switch them
<DataTemplate x:Key="LoginTemplate">
<views:LoginView/>
</DataTemplate>
<DataTemplate x:Key="ChatTemplate">
<views:ChatView/>
</DataTemplate>
<Style x:Key="ChatContentStyle" TargetType="ContentControl">
<Setter Property="ContentTemplate" Value="StaticResource LoginTemplate"/>
<Style.Triggers>
<DataTrigger Binding="Binding UserMode" Value="x:Static enums:UserModes.Chat">
<Setter Property="ContentTemplate" Value="StaticResource ChatTemplate"/>
</DataTrigger>
</Style.Triggers>
</Style>
I want to change the viewModel exact like I'm changing the view
wpf mvvm
asked Nov 11 at 5:54
ישראל קוסייב
83
In main window this is my data context
DataContext="Binding Source=StaticResource VMLocator, Path=MainVM"
and this is my grid binding
<Grid>
<ContentControl Content="Binding" Style="StaticResource ChatContentStyle"/>
</Grid>
and that's how i switch them
<DataTemplate x:Key="LoginTemplate">
<views:LoginView/>
</DataTemplate>
<DataTemplate x:Key="ChatTemplate">
<views:ChatView/>
</DataTemplate>
<Style x:Key="ChatContentStyle" TargetType="ContentControl">
<Setter Property="ContentTemplate" Value="StaticResource LoginTemplate"/>
<Style.Triggers>
<DataTrigger Binding="Binding UserMode" Value="x:Static enums:UserModes.Chat">
<Setter Property="ContentTemplate" Value="StaticResource ChatTemplate"/>
</DataTrigger>
</Style.Triggers>
</Style>
I want to change the viewModel exact like I'm changing the view
wpf mvvm
wpf mvvm
asked Nov 11 at 5:54
ישראל קוסייב
83
asked Nov 11 at 5:54
ישראל קוסייב
83
edited Nov 11 at 8:13
asked Nov 11 at 5:54
ישראל קוסייב
83
asked Nov 11 at 5:54
ישראל קוסייב
83
asked Nov 11 at 5:54
ישראל קוסייב
83
83
Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33
add a comment |
Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33
Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33
add a comment |
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Make your ContentControl bind against a property that provides either the login viewmodel or the chat viewmodel. Set the DataType for each DataTemplate to the respective viewmodel type. (The style trigger then would have no use anymore and can be deleted.) Depending on user action or program logic, swap the login/chat viewmodel in that property...
– elgonzo
Nov 11 at 6:15
MVVM rule: Each View has exactly one ViewModel which represents an abstraction of that View.
– Sir Rufo
Nov 11 at 7:52
@SirRufo I would say, that View is a representation of ViewModel
– Rekshino
Nov 12 at 8:15
@Rekshino Yes of course, the View is the concrete representation of the ViewModel, which is an abstraction of the View
– Sir Rufo
Nov 12 at 8:30
@SirRufo ..representation of VieModel or it's part, I would say. :)
– Rekshino
Nov 12 at 8:33