Run only single instance of flutter desktop application - dart

Question

I am working on flutter desktop app. I want to execute only single instance of app. But currently it allows me to run more than one instance. How can I allow only one .exe file of this application to run?

Answer 1

This is the customization in default flutter windows application properties, so we have to code in C++ for that purpose. A single window application instance can be achieved using a Mutex:
HANDLE hMutexHandle=CreateMutex(NULL, TRUE, L"my.mutex.name");
HWND handle=FindWindowA(NULL, "Test Application");
if (GetLastError() == ERROR_ALREADY_EXISTS)
{
WINDOWPLACEMENT place = { sizeof(WINDOWPLACEMENT) };
GetWindowPlacement(handle, &place);
switch(place.showCmd)
{
case SW_SHOWMAXIMIZED:
ShowWindow(handle, SW_SHOWMAXIMIZED);
break;
case SW_SHOWMINIMIZED:
ShowWindow(handle, SW_RESTORE);
break;
default:
ShowWindow(handle, SW_NORMAL);
break;
}
SetWindowPos(0, HWND_TOP, 0, 0, 0, 0, SWP_SHOWWINDOW | SWP_NOSIZE | SWP_NOMOVE);
SetForegroundWindow(handle);
return 0;
}
Opening the win32_window.cpp file and adding this code snippet at the start in CreateAndShow() method will restrict the application to a single instance.

Answer 2

I am getting a compilation error.
The solution is this line at the end of the if block.
ReleaseMutex(hMutexHandle);

Answer 3

open windows/runnner/win32_cpp:
// add this function above CreateAndShow
bool CheckOneInstance()
{
HANDLE m_hStartEvent = CreateEventW( NULL, FALSE, FALSE, L"Global\\yourpackage" );
if(m_hStartEvent == NULL)
{
CloseHandle( m_hStartEvent );
return false;
}
if (GetLastError() == ERROR_ALREADY_EXISTS) {
CloseHandle( m_hStartEvent );
m_hStartEvent = NULL;
// already exist
// send message from here to existing copy of the application
return false;
}
// the only instance, start in a usual way
return true;
}
bool Win32Window::CreateAndShow(const std::wstring& title,
const Point& origin,
const Size& size) {
//Add the check
if( !CheckOneInstance()){
return false;
}
Destroy();
....
}