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

这是我的两分钱

 static class Program
    {
        [STAThread]
        static void Main()
        {
            bool createdNew;
            using (new Mutex(true, "MyApp", out createdNew))
            {
                if (createdNew) {
                    Application.EnableVisualStyles();
                    Application.SetCompatibleTextRenderingDefault(false);
                    var mainClass = new SynGesturesLogic();
                    Application.ApplicationExit += mainClass.tray_exit;
                    Application.Run();
                }
                else
                {
                    var current = Process.GetCurrentProcess();
                    foreach (var process in Process.GetProcessesByName(current.ProcessName).Where(process => process.Id != current.Id))
                    {
                        NativeMethods.SetForegroundWindow(process.MainWindowHandle);
                        break;
                    }
                }
            }
        }
    }

其他回答

这就是我最终处理这个问题的方式。注意,调试代码仍然在那里进行测试。这段代码在App.xaml.cs文件的OnStartup中。(WPF)

        // Process already running ? 
        if (Process.GetProcessesByName(Process.GetCurrentProcess().ProcessName).Length > 1)
        {

            // Show your error message
            MessageBox.Show("xxx is already running.  \r\n\r\nIf the original process is hung up you may need to restart your computer, or kill the current xxx process using the task manager.", "xxx is already running!", MessageBoxButton.OK, MessageBoxImage.Exclamation);

            // This process 
            Process currentProcess = Process.GetCurrentProcess();

            // Get all processes running on the local computer.
            Process[] localAll = Process.GetProcessesByName(Process.GetCurrentProcess().ProcessName);

            // ID of this process... 
            int temp = currentProcess.Id;
            MessageBox.Show("This Process ID:  " + temp.ToString());

            for (int i = 0; i < localAll.Length; i++)
            {
                // Find the other process 
                if (localAll[i].Id != currentProcess.Id)
                {
                    MessageBox.Show("Original Process ID (Switching to):  " + localAll[i].Id.ToString());

                    // Switch to it... 
                    SetForegroundWindow(localAll[i].MainWindowHandle);

                }
            }

            Application.Current.Shutdown();

        }

这可能有我还没有发现的问题。如果我遇到了,我会更新我的答案。

基于命名互斥的方法不是跨平台的,因为命名互斥在Mono中不是全局的。基于进程枚举的方法没有任何同步,可能会导致不正确的行为(例如,同时启动的多个进程可能都根据时间自行终止)。在控制台应用程序中不需要基于windows系统的方法。这个解决方案建立在Divin的答案之上,解决了所有这些问题:

using System;
using System.IO;

namespace TestCs
{
    public class Program
    {
        // The app id must be unique. Generate a new guid for your application. 
        public static string AppId = "01234567-89ab-cdef-0123-456789abcdef";

        // The stream is stored globally to ensure that it won't be disposed before the application terminates.
        public static FileStream UniqueInstanceStream;

        public static int Main(string[] args)
        {
            EnsureUniqueInstance();

            // Your code here.

            return 0;
        }

        private static void EnsureUniqueInstance()
        {
            // Note: If you want the check to be per-user, use Environment.SpecialFolder.ApplicationData instead.
            string lockDir = Path.Combine(
                Environment.GetFolderPath(Environment.SpecialFolder.CommonApplicationData),
                "UniqueInstanceApps");
            string lockPath = Path.Combine(lockDir, $"{AppId}.unique");

            Directory.CreateDirectory(lockDir);

            try
            {
                // Create the file with exclusive write access. If this fails, then another process is executing.
                UniqueInstanceStream = File.Open(lockPath, FileMode.Create, FileAccess.Write, FileShare.None);

                // Although only the line above should be sufficient, when debugging with a vshost on Visual Studio
                // (that acts as a proxy), the IO exception isn't passed to the application before a Write is executed.
                UniqueInstanceStream.Write(new byte[] { 0 }, 0, 1);
                UniqueInstanceStream.Flush();
            }
            catch
            {
                throw new Exception("Another instance of the application is already running.");
            }
        }
    }
}

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…

我喜欢一个解决方案,以允许多个实例,如果exe是从其他路径调用。我修改了CharithJ溶液方法一:

   static class Program {
    [DllImport("user32.dll")]
    private static extern bool ShowWindow(IntPtr hWnd, Int32 nCmdShow);
    [DllImport("User32.dll")]
    public static extern Int32 SetForegroundWindow(IntPtr hWnd);
    [STAThread]
    static void Main() {
        Process currentProcess = Process.GetCurrentProcess();
        foreach (var process in Process.GetProcesses()) {
            try {
                if ((process.Id != currentProcess.Id) && 
                    (process.ProcessName == currentProcess.ProcessName) &&
                    (process.MainModule.FileName == currentProcess.MainModule.FileName)) {
                    ShowWindow(process.MainWindowHandle, 5); // const int SW_SHOW = 5; //Activates the window and displays it in its current size and position. 
                    SetForegroundWindow(process.MainWindowHandle);
                    return;
                }
            } catch (Exception ex) {
                //ignore Exception "Access denied "
            }
        }

        Application.EnableVisualStyles();
        Application.SetCompatibleTextRenderingDefault(false);
        Application.Run(new Form1());
    }
}

我在这里找不到一个简单的解决方案,所以我希望有人会喜欢这个:

更新2018-09-20

把这段代码放在Program.cs中:

using System.Diagnostics;

static void Main()
{
    Process thisProcess = Process.GetCurrentProcess();
    Process[] allProcesses = Process.GetProcessesByName(thisProcess.ProcessName);
    if (allProcesses.Length > 1)
    {
        // Don't put a MessageBox in here because the user could spam this MessageBox.
        return;
    }

    // Optional code. If you don't want that someone runs your ".exe" with a different name:

    string exeName = AppDomain.CurrentDomain.FriendlyName;
    // in debug mode, don't forget that you don't use your normal .exe name.
    // Debug uses the .vshost.exe.
    if (exeName != "the name of your executable.exe") 
    {
        // You can add a MessageBox here if you want.
        // To point out to users that the name got changed and maybe what the name should be or something like that^^ 
        MessageBox.Show("The executable name should be \"the name of your executable.exe\"", 
            "Wrong executable name", MessageBoxButtons.OK, MessageBoxIcon.Error);
        return;
    }

    // Following code is default code:
    Application.EnableVisualStyles();
    Application.SetCompatibleTextRenderingDefault(false);
    Application.Run(new MainForm());
}