從WCF服務設置WPF控制

Question

我正在嘗試使用WCF服務設置名稱（文本框）值。 我在WPF應用程序中托管服務。 我最初使用MVVM模型從MainWindow.cs設置文本框值，並且它起作用了。 但是后來我將一些屬性設為靜態，以便通過服務合同訪問它們。 似乎仍然可以設置Model屬性的屬性，但不能在文本框中更改值。 誰能指導我？

Model.cs

 public class Model : INotifyPropertyChanged
{

    public event PropertyChangedEventHandler PropertyChanged;
    protected virtual void OnPropertyChanged(string propertyName)
    {
        PropertyChangedEventHandler handler = PropertyChanged;
        if (handler != null) handler(this, new PropertyChangedEventArgs(propertyName));
    }
    protected bool SetField<T>(ref T field, T value, string propertyName)
    {
        if (EqualityComparer<T>.Default.Equals(field, value)) return false;
        field = value;
        OnPropertyChanged(propertyName);
        MessageBox.Show(field.ToString());

        return true;
    }

    // props
    private static string testname;
    public  static string TestName
    {
        get { return testname; }
        set {
            Model m = new Model();
            m.SetField(ref testname, value, "TestName");
        }
    }    


}

WCF InameService.cs

 public class nameService : InameService
{
    public void setMyName(string name)
    {
        Model.TestName = name;

    }


}

MainWindow.xaml

<Grid Name="GridName">

    <TextBox Name="TextName" HorizontalAlignment="Left" Height="23" Margin="193,140,0,0" TextWrapping="Wrap" Text="{Binding TestName, Mode=TwoWay, UpdateSourceTrigger=PropertyChanged}" VerticalAlignment="Top" Width="120" />

</Grid>

MainWindow.xaml.cs

public partial class MainWindow : Window
{
    public MainWindow()
    {
        ServiceHost host = new ServiceHost(typeof(nameService));
        InitializeComponent();
        host.Open();

        Model s = new Model();
        //this.DataContext = s.NameValue.TestName;
        Model.TestName = "Alicia";
        this.TextName.DataContext = s;

    }
}

Answer 1

感謝內森（Nathan）的幫助。 以下是答案：

我將ViewModel更改為Singleton Class，並在創建實例時實例化了復合Model對象。

`class ViewModel {私有靜態volatile ViewModel實例； 私有靜態對象_mutex = new object（）;

    private ViewModel() { }


    private  Model model;        

    public  Model NameValue
    {
        get { return model; }
        set { model = value; }
    }        


    public static ViewModel Instance
    {
        get
        {
            if (instance == null)
            {
                lock (_mutex)
                {
                    if (instance == null)
                    {
                        instance = new ViewModel();
                        instance.model = new Model();
                    }
                }
            }

            return instance;
        }
    }
}`

然后更改MainWindow.xaml.cs

try
        {
            ViewModel s = ViewModel.Instance;

            s.NameValue.TestName = "Alicia";
            this.DataContext = s;
            this.TextName.DataContext = s;
        }
        catch (Exception e)
        {
            MessageBox.Show("Error" + e.Message);
        }

在服務合同類中進行了類似的更改。 我希望這會幫助一些試圖獲得價值的人

Answer 2

不要使用靜態屬性，因為它們無法綁定到它們。 可以使用靜態對象代替，也可以將Model對象傳遞給服務，例如在構造函數中，然后將該實例用於更新。

public class nameService : InameService
{

    private Model model; 

    public nameService(Model m) 
    {
       model = m;
    }

    public void setMyName(string name)
    {
        model.TestName = name;
    }
}

public class Model : INotifyPropertyChanged
{

    public event PropertyChangedEventHandler PropertyChanged;
    protected virtual void OnPropertyChanged(string propertyName)
    {
        PropertyChangedEventHandler handler = PropertyChanged;
        if (handler != null) handler(this, new PropertyChangedEventArgs(propertyName));
    }
    protected bool SetField<T>(ref T field, T value, string propertyName)
    {
        if (EqualityComparer<T>.Default.Equals(field, value)) return false;
        field = value;
        OnPropertyChanged(propertyName);
        MessageBox.Show(field.ToString());

        return true;
    }

    // props
    private string testname;
    public  string TestName
    {
        get { return testname; }
        set {
            Model m = new Model();
            m.SetField(ref testname, value, "TestName");
        }
    }    
}