CODEWITHSUNDEEP
c#windows-phone-7mvvm
Vijay
Vijay

Reputation: 3502

I can not navigate to the next page in windows phone 7

I am trying to navigate form one page to another page depends upon the login.

If already user logged in, Welcome page should open. Else Log in Page should open.

I am trying like this. The Splash Page is the start up page.

This is a Splash Screen Xaml.cs:

namespace NewExample.Views
{
    public partial class SplashPage : PhoneApplicationPage
    {
        public SplashPage()
        {
            InitializeComponent();
            this.DataContext = new SplashPageViewModel();
        }
    }
}

This is Splash Screen View Model: Here I am check the user already logged in or not.

namespace NewExample.ViewModel
{
    public class SplashPageViewModel
    {
        public static bool isLogin = false;

        public SplashPageViewModel()
        {
            var rootFrame = (App.Current as App).RootFrame;
            if (isLogin)
                rootFrame.Navigate(new Uri("/Views/WelcomePage.xaml", UriKind.Relative));
            else
                rootFrame.Navigate(new Uri("/Views/LoginPage.xaml", UriKind.Relative));
        }
    }
}

But it is not working. The Splash Page only showing. This is not navigating to another page. Please help me to resolve this problem.

If I write the codes in Splash Page Xaml.cs, it's working.

 protected override void OnNavigatedTo(System.Windows.Navigation.NavigationEventArgs e)
        {
            var rootFrame = (App.Current as App).RootFrame;
            if (isLogin)
                rootFrame.Navigate(new Uri("/Views/WelcomePage.xaml", UriKind.Relative));
            else
                rootFrame.Navigate(new Uri("/Views/LoginPage.xaml", UriKind.Relative));
        }

But I want to use this in ViewModel.

Upvotes: 0

Views: 68

Answers (1)

Ivan Zub
Ivan Zub

Reputation: 805

If I understand right what you're trying to accomplish - here is a complete guide at nokia.developer portal.

And if you want to perform UI actions from a ViewModel wrap them with the call to Dispatcher.BeginInvoke .

Upvotes: 1

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