Windows中的活动窗口和QWidget :: activateWindow()

Question

状态的Qt文档QWidget::activateWindow():

在Windows上,如果您在应用程序当前不是活动应用程序时调用它,则它不会使其成为活动窗口.它将更改任务栏条目的颜色,以指示窗口已以某种方式更改.这是因为Microsoft不允许应用程序中断用户当前在另一个应用程序中执行的操作.

但是,Skype似乎无视这一规则.如果Skype正在运行但不是活动应用程序,我可以从开始菜单启动它,它将现有实例带到前台,激活它并获取输入焦点.

我怎么能这样做？

Answer 1

(NOTE: This is specific to how QtSingleApplication works)

The solution is stupidly simple for my issue. Simply call AllowSetForegroundWindow(ASF_ANY); at the beginning of the application, and the original process will thus be allowed to bring itself to the foreground by use of SetForegroundWindow(). No strange hacks, just one line of code to add and no need to modify QtSingleApplication either.