MVVM Light:RelayCommand:定义它是懒惰还是构造函数？

Question

有几个关于如何在ViewModel中定义RelayCommand的示例:

选项1(懒惰):

/// <summary>
/// Gets the LogOnCommand.
/// </summary>
/// <value>The LogOnCommand.</value>
public RelayCommand<LogOnUser> LogOnCommand
{
    get
    {
        if (this.logOnCommand == null)
        {
            this.logOnCommand = new RelayCommand<LogOnUser>(
                action =>
                {
                    // Action code...
                },
                g => g != null);
        }

        return this.logOnCommand;
    }
}

选项2(在构造函数中)

/// <summary>
/// Initializes a new instance of the <see cref="LogOnFormViewModel"/> class.
/// </summary>
public LogOnFormViewModel()
{
    this.logOnCommand = new RelayCommand<LogOnUser>(
                action =>
                {
                    // Action code...
                },
                g => g != null);
}

/// <summary>
/// Gets the LogOnCommand.
/// </summary>
/// <value>The LogOnCommand.</value>
public RelayCommand<LogOnUser> LogOnCommand {get; private set;}

什么是最好/最清晰的设计？

Answer 1

这取决于你喜欢什么样的风格.大多数人不喜欢在财产获取者中拥有一堆逻辑,如果他们可以避免它.

就个人而言,我更喜欢使用真正的方法来调用而不是匿名方法.我的ViewModel看起来像这样.

public class MyViewModel : ViewModelBase
{
    public RelayCommand<CommandParam> MyCommand { get; private get; }

    public MyViewModel()
    {
        CreateCommands();
    }

    private void CreateCommands()
    {
        MyCommand = new RelayCommand<CommandParam>(MyCommandExecute);
    }

    private void MyCommandExecute(CommandParam parm)
    {
        // Action code...
    }
}

请注意,如果您未使用enable命令,则无需调用设置该命令的ctor重载.