日期:2014-05-18  浏览次数:21231 次

怎么只能运行一个程序
如何做到只能运行一个程序?并且在第二次打开程序的时候将第一次打开的程序窗口激活

我做的是个TXT编辑器

我想如果我直接打开一个文本文件(之前已经打开一个程序了)但是这个文本文件是在第一次打开的程序窗口中显示的

怎么弄啊?

------解决方案--------------------
同时只有一个窗体实例,再次打开激活已打开窗体的范例
------解决方案--------------------
C# code

        [DllImport("User32.dll")]
        private static extern bool ShowWindowAsync(IntPtr hWnd, int cmdShow);
        [DllImport("User32.dll")]
        private static extern bool SetForegroundWindow(IntPtr hWnd);
        [DllImport("User32.dll", EntryPoint = "FindWindow")]
        private extern static IntPtr FindWindow(string lpClassName, string lpWindowName);
        //添加api
        //获取当前运行的实例
        public static Process GetRunningInstance()
        {
            Process current = Process.GetCurrentProcess();
            Process[] processes = Process.GetProcessesByName(current.ProcessName);

            foreach (Process process in processes)
            {
                if (process.Id != current.Id)
                    if (Assembly.GetExecutingAssembly().Location.Replace("/", "\\") == current.MainModule.FileName)
                        return process;
            }
            return null;
        }
        public static void HandleRunningInstance(Process instance)
        {
            if (instance != null && instance.MainWindowHandle.ToInt32() != 0)
            {
                string windowTile = instance.MainWindowTitle.Trim();
                if (windowTile == "系统标题")
                    ShowWindowAsync(instance.MainWindowHandle, 3);
                else
                    ShowWindowAsync(instance.MainWindowHandle, 1);
                SetForegroundWindow(instance.MainWindowHandle);
            }
            else
            {
                IntPtr hwnd = FindWindow(null, "系统标题");
                if (hwnd.ToInt32() == 0)
                {
                    hwnd = FindWindow(null, "系统标题");
                    ShowWindowAsync(hwnd, 1);
                    SetForegroundWindow(hwnd);
                }
                else
                {
                    ShowWindowAsync(hwnd, 3);
                    SetForegroundWindow(hwnd);
                }
            }
        }

------解决方案--------------------
http://topic.csdn.net/u/20091118/18/9f3f7c0a-20fa-4006-9ccd-823734b10353.html