python运行一个程序,点击屏幕其他地方程序停止运行 python点击屏幕某个图标

这是我弹出与托盘图标关联的上下文菜单的方式

class MyClass(QObject):
def __init__():
self._testSignal.connect(self._test_show)
self.myTrayIcon.activated.connect(lambda reason: self._update_menu_and_show(reason))
def show():
self._testSignal.emit()
@pyqtSlot()
def _test_show():
self._trayIcon.activated.emit(QSystemTrayIcon.Context)
@QtCore.pyqtSlot()
def _update_menu_and_show(reason):
if reason in (QSystemTrayIcon.Trigger, QSystemTrayIcon.Context):
mySystemTrayIcon._update_menu()
# Trigger means user initiated, Context used for simulated
# if simulated seems like we have to tell the window to explicitly show
if reason == QSystemTrayIcon.Context:
mySystemTrayIcon.contextMenu().setWindowFlags(QtCore.Qt.WindowStaysOnTopHint|QtCore.Qt.FramelessWindowHint)
pos = mySystemTrayIcon.geometry().bottomLeft()
mySystemTrayIcon.contextMenu().move(pos)
mySystemTrayIcon.contextMenu().show()
...
class MySystemTrayIcon(QSystemTrayIcon):
def _update_menu(self):
# logic to populate menu
self.setContextMenu(menu)
...
MyClass().show()

似乎您必须在上下文菜单上设置WindowStaysOnTopHint才能显示它.

此解决方案特定于mac,因为它假定任务栏位于顶部.

副作用是,即使用户单击其他位置,上下文菜单也始终位于顶部.我在上下文菜单上放置了一个事件过滤器,它注册的唯一有用的事件是QEvent.Leave.