c# - 使用 Mutex 运行应用程序的单个实例

Question

为了只允许运行应用程序的单个实例，我正在使用互斥锁。代码如下。这是正确的方法吗？代码中是否有任何缺陷？

当用户第二次尝试打开应用程序时如何显示已经运行的应用程序。目前（在下面的代码中），我只是显示另一个实例已经在运行的消息。

    static void Main(string[] args)
    {
        Mutex _mut = null;

        try
        {
            _mut = Mutex.OpenExisting(AppDomain.CurrentDomain.FriendlyName);
        }
        catch
        {
             //handler to be written
        }

        if (_mut == null)
        {
            _mut = new Mutex(false, AppDomain.CurrentDomain.FriendlyName);
        }
        else
        {
            _mut.Close();
            MessageBox.Show("Instance already running");

        }            
    }

score 19 · Accepted Answer

我这样做过一次，希望对您有所帮助：

bool createdNew;

Mutex m = new Mutex(true, "myApp", out createdNew);

if (!createdNew)
{
    // myApp is already running...
    MessageBox.Show("myApp is already running!", "Multiple Instances");
    return;
}

score 9 · Accepted Answer

static void Main() 
{
  using(Mutex mutex = new Mutex(false, @"Global\" + appGuid))
  {
    if(!mutex.WaitOne(0, false))
    {
       MessageBox.Show("Instance already running");
       return;
    }

    GC.Collect();                
    Application.Run(new Form1());
  }
}

来源： http: //odetocode.com/Blogs/scott/archive/2004/08/20/401.aspx

score 3 · Accepted Answer

我用这个：

    private static Mutex _mutex;

    private static bool IsSingleInstance()
    {
        _mutex = new Mutex(false, _mutexName);

        // keep the mutex reference alive until the normal 
        //termination of the program
        GC.KeepAlive(_mutex);

        try
        {
            return _mutex.WaitOne(0, false);
        }
        catch (AbandonedMutexException)
        {
            // if one thread acquires a Mutex object 
            //that another thread has abandoned 
            //by exiting without releasing it

            _mutex.ReleaseMutex();
            return _mutex.WaitOne(0, false);
        }
    }


    public Form1()
    {
        if (!isSingleInstance())
        {
            MessageBox.Show("Instance already running");
            this.Close();
            return;
        }

        //program body here
    }

    private void Form1_FormClosing(object sender, FormClosingEventArgs e)
    {
        if (_mutex != null)
        {
            _mutex.ReleaseMutex();
        }
    }

score 1 · Accepted Answer

看看这个问题

这篇文章有一个链接：被误解的互斥锁，其中解释了互斥锁的用法。

score 1 · Accepted Answer

查看此页面上显示的代码示例

简而言之，您使用重载 Mutex ctor(bool, string, out bool)，它通过 out 参数告诉您，您是否拥有 Named Mutex。如果您是第一个实例，则在调用 ctor 后，此输出参数将包含 true - 在这种情况下，您将照常进行。如果此参数为 false，则表示另一个实例已经获得所有权/正在运行，在这种情况下，您会显示错误消息“另一个实例已在运行”。然后优雅地退出。

score 1 · Accepted Answer

使用具有超时和安全设置的应用程序。我使用了我的自定义类：

private class SingleAppMutexControl : IDisposable
    {
        private readonly Mutex _mutex;
        private readonly bool _hasHandle;

        public SingleAppMutexControl(string appGuid, int waitmillisecondsTimeout = 5000)
        {
            bool createdNew;
            var allowEveryoneRule = new MutexAccessRule(new SecurityIdentifier(WellKnownSidType.WorldSid, null),
                MutexRights.FullControl, AccessControlType.Allow);
            var securitySettings = new MutexSecurity();
            securitySettings.AddAccessRule(allowEveryoneRule);
            _mutex = new Mutex(false, "Global\\" + appGuid, out createdNew, securitySettings);
            _hasHandle = false;
            try
            {
                _hasHandle = _mutex.WaitOne(waitmillisecondsTimeout, false);
                if (_hasHandle == false)
                    throw new System.TimeoutException();
            }
            catch (AbandonedMutexException)
            {
                _hasHandle = true;
            }
        }

        public void Dispose()
        {
            if (_mutex != null)
            {
                if (_hasHandle)
                    _mutex.ReleaseMutex();
                _mutex.Dispose();
            }
        }
    }

并使用它：

    private static void Main(string[] args)
    {
        try
        {
            const string appguid = "{xxxxxxxx-xxxxxxxx}";
            using (new SingleAppMutexControl(appguid))
            {

                Console.ReadLine();
            }
        }
        catch (System.TimeoutException)
        {
            Log.Warn("Application already runned");
        }
        catch (Exception ex)
        {
            Log.Fatal(ex, "Fatal Error on running");
        }
    }

score 0 · Accepted Answer

0

还有这个。是否使用互斥锁来防止同一程序的多个实例安全运行？

于 2009-05-04T12:04:31.313 回答

c# - 使用 Mutex 运行应用程序的单个实例

7 回答 7

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