PyQt: Show menu in a system tray application

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Question :

PyQt: Show menu in a system tray application

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 as when it is run it has to put an icon in the system tray and it has offer an option in its menu to exit the application.

This code works, it shows the menu (I don’t connect the exit action and so on to keep it simple)

import sys
from PyQt4 import QtGui

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

    trayIcon = QtGui.QSystemTrayIcon(QtGui.QIcon("Bomb.xpm"), app)
    menu = QtGui.QMenu()
    exitAction = menu.addAction("Exit")
    trayIcon.setContextMenu(menu)

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

if __name__ == '__main__':
    main()

But this doesn’t:

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()
        exitAction = menu.addAction("Exit")
        self.setContextMenu(menu)

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

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

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

if __name__ == '__main__':
    main()

I probably miss something. There are no errors but in the second case when I click with the right button it doesn’t show the menu.

Asked By: Nextorlg

||

Answer #1:

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()
Answered By: Nextorlg

Answer #2:

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()
Answered By: Gerard

Answer #3:

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()
Answered By: demosthenes

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)
Answered By: MagTun

Answer #5:

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()
Answered By: Jonathan

Answer #6:

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)
Answered By: Roochiedoor

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