使用Mutex运行应用程序的单个实例

为了仅允许正在运行的应用程序的单个实例使用互斥锁。 代码如下。 这是做到这一点的正确方法吗？ 代码中是否有缺陷？

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");

        }            
    }

---

我这样做了一次，我希望这会有所帮助：

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;
}

---

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());
  }
}

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

---

我使用这个：

    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();
        }
    }    

链接地址: http://www.djcxy.com/p/51157.html

上一篇: Run single instance of an application using Mutex

下一篇: How do I prevent my app from launching over and over again