PyQt的：在系统托盘中的应用显示菜单

Question

首先，我是一个经验丰富的C程序员，但新的Python。我想创建一个使用PyQt的Python中的简单应用。让我们想象一下这个应用程序是当它运行它把一个图标在系统托盘中那样简单，它有提供在其菜单中的选项退出应用程序。

此代码的工作，它显示菜单（我不连接exit操作等，以保持它的简单）

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()

但是，这并不：

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()

我可能会错过一些东西。有没有错误，但在第二种情况下，当我点击与正确的按钮，它不显示菜单。

Answer 1

好，一些调试后我发现这个问题。该QMenu对象是完成__init__功能后销毁，因为它没有父。虽然QSystemTrayIcon的父可以是用于QMenu对象它必须是一个QWidget。此代码的工作（见QMenu如何得到同一父这是一个QWidget的所述QSystemTrayIcon）：

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 2

我想我宁愿以下，因为它似乎并不依赖于Qt的内部垃圾收集的决定。

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 3

下面是与退出动作的代码中实现

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()

Answer 4

下面是PyQt5版本（能够实现德摩斯梯尼的答案的退出动作）。 从PyQt4中移植到PyQt5 源

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

通过一个pyqt5连接事件：

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 6

我无法得到任何上述答案的PyQt5工作（退出在系统托盘菜单，实际上不会退出），但我设法把它们结合在一起的，做工作的解决方案。我还在努力，以确定是否exitAction应进一步以某种方式使用。

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)