Run single instance of an application using Mutex
In order to allow only a single instance of an application running I'm using mutex. The code is given below. Is this the right way to do it? Are there any flaws in the code?
How to show the already running application when user tries to open the application the second time. At present (in the code below), I'm just displaying a message that another instance is already running.
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();
}
}
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