PyQt: Show menu in a system tray application

Question

First of all, I\'m an experienced C programmer but new to python. I want to create a simple application in python using pyqt. Let\'s imagine this application it is as simple

Answer 1

I think I would prefer the following as it doesn't seem to depend upon QT's internal garbage collection decisions.

import sys
from PyQt4 import QtGui

class SystemTrayIcon(QtGui.QSystemTrayIcon):
    def __init__(self, icon, parent=None):
        QtGui.QSystemTrayIcon.__init__(self, icon, parent)
        self.menu = QtGui.QMenu(parent)
        exitAction = self.menu.addAction("Exit")
        self.setContextMenu(self.menu)

def main():
    app = QtGui.QApplication(sys.argv)
    style = app.style()
    icon = QtGui.QIcon(style.standardPixmap(QtGui.QStyle.SP_FileIcon))
    trayIcon = SystemTrayIcon(icon)

    trayIcon.show()
    sys.exit(app.exec_())

if __name__ == '__main__':
    main()

Answer 2

With a pyqt5 connected event:

class SystemTrayIcon(QtWidgets.QSystemTrayIcon):

    def __init__(self, icon, parent=None):
        QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)
        menu = QtWidgets.QMenu(parent)
        exitAction = menu.addAction("Exit")
        self.setContextMenu(menu)    
        menu.triggered.connect(self.exit)

    def exit(self):
        QtCore.QCoreApplication.exit()

Answer 3

I couldn't get any of the above answers to work in PyQt5 (the exit in the system tray menu, wouldn't actually exit), but i managed to combine them for a solution that does work. I'm still trying to determine if exitAction should be used further somehow.

import sys
from PyQt5 import QtWidgets, QtCore, QtGui

class SystemTrayIcon(QtWidgets.QSystemTrayIcon):

    def __init__(self, icon, parent=None):
        QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)
        menu = QtWidgets.QMenu(parent)
        exitAction = menu.addAction("Exit")
        self.setContextMenu(menu)
        menu.triggered.connect(self.exit)

    def exit(self):
        QtCore.QCoreApplication.exit()

def main(image):
    app = QtWidgets.QApplication(sys.argv)
    w = QtWidgets.QWidget()
    trayIcon = SystemTrayIcon(QtGui.QIcon(image), w)
    trayIcon.show()
    sys.exit(app.exec_())


if __name__ == '__main__':
    on='icon.ico'
    main(on)

Answer 4

Here is the PyQt5 version (was able to implement the Exit action of demosthenes's answer). Source for porting from PyQt4 to PyQt5

import sys
from PyQt5 import QtCore, QtGui, QtWidgets
# code source: https://stackoverflow.com/questions/893984/pyqt-show-menu-in-a-system-tray-application  - add answer PyQt5
#PyQt4 to PyQt5 version: https://stackoverflow.com/questions/20749819/pyqt5-failing-import-of-qtgui
class SystemTrayIcon(QtWidgets.QSystemTrayIcon):

    def __init__(self, icon, parent=None):
        QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)
        menu = QtWidgets.QMenu(parent)
        exitAction = menu.addAction("Exit")
        self.setContextMenu(menu)

def main(image):
    app = QtWidgets.QApplication(sys.argv)

    w = QtWidgets.QWidget()
    trayIcon = SystemTrayIcon(QtGui.QIcon(image), w)

    trayIcon.show()
    sys.exit(app.exec_())

if __name__ == '__main__':
    on=r''# ADD PATH OF YOUR ICON HERE .png works
    main(on)

Answer 5

Well, after some debugging I found the problem. The QMenu object it is destroyed after finish __init__ function because it doesn't have a parent. While the parent of a QSystemTrayIcon can be an object for the QMenu it has to be a Qwidget. This code works (see how QMenu gets the same parent as the QSystemTrayIcon which is an QWidget):

import sys
from PyQt4 import QtGui

class SystemTrayIcon(QtGui.QSystemTrayIcon):

    def __init__(self, icon, parent=None):
        QtGui.QSystemTrayIcon.__init__(self, icon, parent)
        menu = QtGui.QMenu(parent)
        exitAction = menu.addAction("Exit")
        self.setContextMenu(menu)

def main():
    app = QtGui.QApplication(sys.argv)

    w = QtGui.QWidget()
    trayIcon = SystemTrayIcon(QtGui.QIcon("Bomb.xpm"), w)

    trayIcon.show()
    sys.exit(app.exec_())

if __name__ == '__main__':
    main()

Answer 6

Here is the code with Exit action implemented

import sys
from PyQt4 import QtGui, QtCore

class SystemTrayIcon(QtGui.QSystemTrayIcon):
    def __init__(self, icon, parent=None):
       QtGui.QSystemTrayIcon.__init__(self, icon, parent)
       menu = QtGui.QMenu(parent)
       exitAction = menu.addAction("Exit")
       self.setContextMenu(menu)
       QtCore.QObject.connect(exitAction,QtCore.SIGNAL('triggered()'), self.exit)

    def exit(self):
      QtCore.QCoreApplication.exit()

def main():
   app = QtGui.QApplication(sys.argv)

   w = QtGui.QWidget()
   trayIcon = SystemTrayIcon(QtGui.QIcon("qtLogo.png"), w)

   trayIcon.show()
   sys.exit(app.exec_())

if __name__ == '__main__':
    main()