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问题描述
是否有一种简单的方法可以打印所有Messenger.Default.Send()进行调试?不想覆盖它们.
Is there a simple way to print all Messenger.Default.Send() to debug? Don't want to override them.
推荐答案
是的,所以我最后用Messenger自己的包装器结束了.没有明智的解决方案,只有简单的包装即可.万一其他人会需要它:
Yep, so i ended up with own wrapper around Messenger. No smart solutions, just straightforward wrapping. In case, anybody else would need it:
public static class MvvmLightMessenger
{
public static void Register<TMessage>(object recipient, Action<TMessage> action)
{
Messenger.Default.Register(recipient,action);
}
public static void Register<TMessage>(object recipient, bool receiveDerivedMessagesToo, Action<TMessage> action)
{
Messenger.Default.Register(recipient, receiveDerivedMessagesToo, action);
}
public static void Register<TMessage>(object recipient, object token, Action<TMessage> action)
{
Messenger.Default.Register(recipient, token, action);
}
public static void Register<TMessage>(object recipient, object token, bool receiveDerivedMessagesToo, Action<TMessage> action)
{
Messenger.Default.Register(recipient, token, receiveDerivedMessagesToo, action);
}
public static void Send<TMessage>(TMessage message)
{
Debug.WriteLine("{!} Message: " + message);
Messenger.Default.Send<TMessage>(message);
}
public static void Send<TMessage, TTarget>(TMessage message)
{
Debug.WriteLine("{!} Message: " + message + " to target: " + typeof(TTarget));
Messenger.Default.Send<TMessage, TTarget>(message);
}
public static void Send<TMessage>(TMessage message, object token)
{
Debug.WriteLine("{!} Message: " + message + " token: " + token);
Messenger.Default.Send<TMessage>(message, token);
}
public static void Unregister<TMessage>(object recipient)
{
Messenger.Default.Unregister<TMessage>(recipient);
}
public static void Unregister<TMessage>(object recipient, Action<TMessage> action)
{
Messenger.Default.Unregister<TMessage>(recipient, action);
}
public static void Unregister<TMessage>(object recipient, object token)
{
Messenger.Default.Unregister<TMessage>(recipient, token);
}
public static void Unregister<TMessage>(object recipient, object token, Action<TMessage> action)
{
Messenger.Default.Unregister<TMessage>(recipient, token, action);
}
}
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