在。net下使用c#和WPF(而不是Windows窗体或控制台),创建一个只能作为单个实例运行的应用程序的正确方法是什么?
我知道它与某种叫做互斥的神秘事物有关,我很少能找到有人费心停下来解释其中一个是什么。
代码还需要通知已经运行的实例,用户试图启动第二个实例,如果存在命令行参数,还可能传递任何命令行参数。
在。net下使用c#和WPF(而不是Windows窗体或控制台),创建一个只能作为单个实例运行的应用程序的正确方法是什么?
我知道它与某种叫做互斥的神秘事物有关,我很少能找到有人费心停下来解释其中一个是什么。
代码还需要通知已经运行的实例,用户试图启动第二个实例,如果存在命令行参数,还可能传递任何命令行参数。
当前回答
这是我用的。它结合了进程枚举来执行切换和互斥,以防止“活动点击器”:
public partial class App
{
[DllImport("user32")]
private static extern int OpenIcon(IntPtr hWnd);
[DllImport("user32.dll")]
private static extern bool SetForegroundWindow(IntPtr hWnd);
protected override void OnStartup(StartupEventArgs e)
{
base.OnStartup(e);
var p = Process
.GetProcessesByName(Process.GetCurrentProcess().ProcessName);
foreach (var t in p.Where(t => t.MainWindowHandle != IntPtr.Zero))
{
OpenIcon(t.MainWindowHandle);
SetForegroundWindow(t.MainWindowHandle);
Current.Shutdown();
return;
}
// there is a chance the user tries to click on the icon repeatedly
// and the process cannot be discovered yet
bool createdNew;
var mutex = new Mutex(true, "MyAwesomeApp",
out createdNew); // must be a variable, though it is unused -
// we just need a bit of time until the process shows up
if (!createdNew)
{
Current.Shutdown();
return;
}
new Bootstrapper().Run();
}
}
其他回答
这里有一个解决方案:
Protected Overrides Sub OnStartup(e As StartupEventArgs)
Const appName As String = "TestApp"
Dim createdNew As Boolean
_mutex = New Mutex(True, appName, createdNew)
If Not createdNew Then
'app is already running! Exiting the application
MessageBox.Show("Application is already running.")
Application.Current.Shutdown()
End If
MyBase.OnStartup(e)
End Sub
你可以使用Mutex类,但是你很快就会发现你需要自己实现传递参数的代码。当我读Chris Sell的书时,我学到了一个用WinForms编程的技巧。这个技巧使用了框架中已经可用的逻辑。我不知道你怎么想,但当我了解到可以在框架中重用的东西时,这通常是我采取的路线,而不是重新发明轮子。当然,除非它不能做到我想要的一切。
当我进入WPF时,我想到了一种使用相同代码的方法,但在WPF应用程序中。基于您的问题,这个解决方案应该能够满足您的需求。
首先,我们需要创建应用程序类。在这个类中,我们将重写OnStartup事件并创建一个名为Activate的方法,该方法将在稍后使用。
public class SingleInstanceApplication : System.Windows.Application
{
protected override void OnStartup(System.Windows.StartupEventArgs e)
{
// Call the OnStartup event on our base class
base.OnStartup(e);
// Create our MainWindow and show it
MainWindow window = new MainWindow();
window.Show();
}
public void Activate()
{
// Reactivate the main window
MainWindow.Activate();
}
}
Second, we will need to create a class that can manage our instances. Before we go through that, we are actually going to reuse some code that is in the Microsoft.VisualBasic assembly. Since, I am using C# in this example, I had to make a reference to the assembly. If you are using VB.NET, you don't have to do anything. The class we are going to use is WindowsFormsApplicationBase and inherit our instance manager off of it and then leverage properties and events to handle the single instancing.
public class SingleInstanceManager : Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase
{
private SingleInstanceApplication _application;
private System.Collections.ObjectModel.ReadOnlyCollection<string> _commandLine;
public SingleInstanceManager()
{
IsSingleInstance = true;
}
protected override bool OnStartup(Microsoft.VisualBasic.ApplicationServices.StartupEventArgs eventArgs)
{
// First time _application is launched
_commandLine = eventArgs.CommandLine;
_application = new SingleInstanceApplication();
_application.Run();
return false;
}
protected override void OnStartupNextInstance(StartupNextInstanceEventArgs eventArgs)
{
// Subsequent launches
base.OnStartupNextInstance(eventArgs);
_commandLine = eventArgs.CommandLine;
_application.Activate();
}
}
基本上,我们使用VB位来检测单个实例并进行相应的处理。OnStartup将在第一个实例加载时被触发。当应用程序再次运行时,OnStartupNextInstance被触发。如您所见,我可以通过事件参数获得在命令行上传递的内容。我将值设置为一个实例字段。您可以在这里解析命令行,也可以通过构造函数和对Activate方法的调用将它传递给应用程序。
第三,是时候创建我们的入口点了。我们将利用SingleInstanceManager,而不是像通常那样更新应用程序。
public class EntryPoint
{
[STAThread]
public static void Main(string[] args)
{
SingleInstanceManager manager = new SingleInstanceManager();
manager.Run(args);
}
}
好吧,我希望您能够理解所有内容,能够使用这个实现并使其成为您自己的实现。
永远不要使用命名互斥来实现单实例应用程序(至少在生产代码中不要这样做)。恶意代码可以很容易地DoS(拒绝服务)你的屁股…
我添加了一个sendMessage方法到NativeMethods类。
显然,如果应用程序没有显示在任务栏中,postmessage方法不会工作,但是使用sendmessage方法解决了这个问题。
class NativeMethods
{
public const int HWND_BROADCAST = 0xffff;
public static readonly int WM_SHOWME = RegisterWindowMessage("WM_SHOWME");
[DllImport("user32")]
public static extern bool PostMessage(IntPtr hwnd, int msg, IntPtr wparam, IntPtr lparam);
[DllImport("user32.dll", CharSet = CharSet.Auto)]
public static extern IntPtr SendMessage(IntPtr hWnd, int Msg, IntPtr wParam, IntPtr lParam);
[DllImport("user32")]
public static extern int RegisterWindowMessage(string message);
}
这么简单的问题有这么多答案。稍微改变一下这里是我对这个问题的解决方案。
Creating a Mutex can be troublesome because the JIT-er only sees you using it for a small portion of your code and wants to mark it as ready for garbage collection. It pretty much wants to out-smart you thinking you are not going to be using that Mutex for that long. In reality you want to hang onto this Mutex for as long as your application is running. The best way to tell the garbage collector to leave you Mutex alone is to tell it to keep it alive though out the different generations of garage collection. Example:
var m = new Mutex(...);
...
GC.KeepAlive(m);
我从这个网页上获得了灵感:http://www.ai.uga.edu/~mc/SingleInstance.html