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C# WPF,如何创建单个实例,并在c中启动另一个实例时显示主窗口#_C#_Wpf_Mutex_Single Instance - Fatal编程技术网

C# WPF,如何创建单个实例,并在c中启动另一个实例时显示主窗口#

C# WPF,如何创建单个实例,并在c中启动另一个实例时显示主窗口#,c#,wpf,mutex,single-instance,C#,Wpf,Mutex,Single Instance,作为标题状态,我想制作一个单实例程序,并在启动另一个实例时显示MainWindow。 我已经显示了一条消息,只允许一个实例 public class MyApplication { static Mutex mutex = new Mutex(true, "FirstInstance"); [STAThread] public static void Main(string[] args) { if (mutex.WaitOne(TimeSpan.Zero

作为标题状态,我想制作一个单实例程序,并在启动另一个实例时显示MainWindow。 我已经显示了一条消息,只允许一个实例

public class MyApplication
{
static Mutex mutex = new Mutex(true, "FirstInstance");
    [STAThread]
    public static void Main(string[] args)
    {
        if (mutex.WaitOne(TimeSpan.Zero, true))
        {
            App app = new App();
            app.InitializeComponent();
            app.Run();
        }
        else
        {
            MessageBox.Show("only one instance at a time");
        }
    }
}
这很好,但我想显示主窗口而不是消息,所以我尝试了

static Application app;//change app to static
if (mutex.WaitOne(TimeSpan.Zero, true))
        {
            app = new App();
            app.InitializeComponent();
            app.Run();
        }  
else
{
    app.MainWindow.WindowState = WindowState.Normal;
}
我得到“System.NullReferenceException:对象引用未设置为对象的实例”。似乎应用程序中的主窗口(是静态的)是空的,我不明白为什么

所以我尝试了这篇文章 但WndProc方法在WPF中不存在

如果你能帮助我,我将不胜感激。
谢谢

如果我没有误解你的问题,试着这样做

[STAThread]
    public static void Main(string[] args)
    {
        Task task = new Task(() => { Thread.Sleep(200); MessageBox.Show("what a marvelous engineering"); });
        task.Start();
        //If you want application not to run untill task is complete then just use wait
        task.Wait();

            App app = new App();
            app.InitializeComponent();
            app.Run();
    }

但我想知道你是否希望在主窗口实例化之前完成C#工作,为什么不在App App=new App()之前完成其他工作,让代码按顺序运行。

我创建了一个示例WPF应用程序,只更改了App.xaml.cs文件。如果“单实例”窗口已打开,此代码将查找与当前进程名称匹配的任何进程,如果该进程有一个窗口,则显示该窗口:

public partial class App : Application
{
    // signals to restore the window to its normal state
    private const int SW_SHOWNORMAL = 1;

    // create the mutex
    private const string MUTEXNAME = "FirstInstance";
    private readonly Mutex _mutex = new Mutex(true, MUTEXNAME);

    public App()
    {
        if (!_mutex.WaitOne(TimeSpan.Zero))
        {
            ShowExistingWindow();
            Shutdown();
        }
    }

    [DllImport("User32.dll")]
    private static extern bool SetForegroundWindow(IntPtr hWnd);

    [DllImport("user32.dll")]
    private static extern bool ShowWindow(IntPtr hWnd, int nCmdShow);

    // shows the window of the single-instance that is already open
    private void ShowExistingWindow()
    {
        var currentProcess = Process.GetCurrentProcess();
        var processes = Process.GetProcessesByName(currentProcess.ProcessName);
        foreach (var process in processes)
        {
            // the single-instance already open should have a MainWindowHandle
            if (process.MainWindowHandle != IntPtr.Zero)
            {
                // restores the window in case it was minimized
                ShowWindow(process.MainWindowHandle, SW_SHOWNORMAL);

                // brings the window to the foreground
                SetForegroundWindow(process.MainWindowHandle);

                return;
            }
        }
    }
}

仅供参考,这在调试模式下不起作用,因为.vshost成为进程名称的一部分。如果需要它在调试模式下工作,则需要迭代所有进程,而不是调用Process.getProcessByName。

因为首先需要创建应用程序实例和Mainwindow实例。