Question

我在使用PyQt5的菜单系统托盘中工作。我是PyQt5的新手，我想做的是在没有菜单被阻止的情况下触发一个动作（多线程）。在很多地方读完之后，我得出的结论是使用Qthread应该是要走的路（但是如果我能理解那个班的工作方式......）。但是，考虑到我的应用程序非常简单，使用threading也不会那么糟糕。所以，我使用import threading尝试了以下代码：

from PyQt5 import QtCore, QtGui, QtWidgets
import threading

class menubar(object):
    def __init__(self):
    signal.signal(signal.SIGINT, signal.SIG_DFL)
    self.systray = True
    self.stopped = False

    def search_menu(self):
        self.SearchAction = menu.addAction("Search")
        self.SearchAction.triggered.connect(self.search_cast)

    def _search_cast_(self):
        args.select_cc = True
        self.cc.initialize_cast()
        self.cast_list()

    def search_cast(self):
        threading.Thread(target=self._search_cast_).start()

#some more methods here...

def main():

    menubar()
    app = QtWidgets.QApplication(sys.argv)
    tray = QtWidgets.QSystemTrayIcon(icon)

    menu = QtWidgets.QMenu()
    start = menubar()
    start.search_menu()
    start.separator_menu()
    start.populating_menu()
    start.separator_menu()
    start.stop_menu()
    start.resetaudio_menu()
    start.about_menu()
    start.exit_menu()

    tray.setContextMenu(menu)
    tray.show()
    app.exec_()

if __name__ == '__main__':
     main()

当我开始我的菜单时，一切都按照我的预期到位。然后，当我点击菜单Search时，操作会触发self.search_cast方法，并且我的菜单会填充它找到的列表。我还可以看到我的应用程序正在进行搜索而不会被阻止但是当它完成时我会收到以下错误：

QObject: Cannot create children for a parent that is in a different thread.
(Parent is QMenu(0x7fcef497c160), parent's thread is     QThread(0x7fcef2603d10), current thread is QThread(0x7fcef4a89360)
QObject: Cannot create children for a parent that is in a different thread.
(Parent is QMenu(0x7fcef497c160), parent's thread is  QThread(0x7fcef2603d10), current thread is QThread(0x7fcef4a89360)
QObject: Cannot create children for a parent that is in a different thread.

在此之后，菜单仍然是＆＃34;功能性＆＃34;从某种意义上说，它是响应式的，但不能再触发任何动作。此外，似乎不再创建线程。如果有人能解释我为什么会这样，我会很高兴的。我没有看到光......

更新：

我现在创建了一个worker.py，其中包含：

from PyQt5.QtCore import QThread, QObject, pyqtSignal, pyqtSlot
#some other imports


class Worker(QObject):
    finished = pyqtSignal()


@pyqtSlot()
def _search_cast_(self):
    self.cc = casting()
    self.cc.initialize_cast()
    self.finished.emit()

然后我在class menubar添加了以下内容：

class menubar(object):
    def __init__(self):
        self.cc = casting()
        signal.signal(signal.SIGINT, signal.SIG_DFL)
        self.cc.cast = None
        self.systray = True
        self.stopped = False

        self.obj = worker.Worker()  # no parent!
        self.thread = QThread()  # no parent!
        self.obj.moveToThread(self.thread)
        self.obj.finished.connect(self.thread.quit)
        self.thread.started.connect(self.obj._search_cast_)

  def search_menu(self):
        self.SearchAction = menu.addAction("Search")
        self.SearchAction.triggered.connect(self.search_cast)

  def search_cast(self):
    self.thread.start()
    self.cast_list()

  def cast_list(self):
     if len(self.cc.availablecc) == 0:
     # some actions here.

现在我收到以下错误：

 AttributeError: 'casting' object has no attribute 'availablecc'

我确保worker实际上正在从我调用availablecc的外部类中恢复cc。但由于某种原因，menubar班没有收到。我的工作基于https://stackoverflow.com/a/33453124/1995261

Answer 1

我会继续回答自己。受https://stackoverflow.com/a/33453124/1995261的启发，我通过实施以下内容解决了这个问题：

1）我创建了一个worker.py来执行阻止菜单的方法_search_cast_。当此方法完成搜索时，它会发出两个信号：a）一个通知他已恢复list，b）该方法已完成。

#worker.py
from PyQt5.QtCore import QThread, QObject, pyqtSignal, pyqtSlot


class Worker(QObject):
    finished = pyqtSignal()
    intReady = pyqtSignal(list)
    def __init__(self):
        QObject.__init__(self)

    @pyqtSlot()
    def _search_cast_(self):
        self.cc = casting()
        self.cc.initialize_cast()
        availablecc = self.cc.availablecc
        self.intReady.emit(availablecc)
        self.finished.emit()

2）在main.py我抛弃了以下内容，我尝试用注释解释代码内部：

#main.py
from PyQt5.QtCore import QThread, QObject, pyqtSignal, pyqtSlot
import worker # This is to import worker.py
class menubar(object):
    def __init__(self):
        signal.signal(signal.SIGINT, signal.SIG_DFL)
        self.cc.cast = None
        self.systray = True
        self.stopped = False

        self.obj = worker.Worker()  # The worker is started with no parent!
        self.thread = QThread()  # We initialise the Qthread class with no parent!
        self.obj.intReady.connect(self.onIntReady) # We receive the signal that the list is ready
        self.obj.moveToThread(self.thread) # Moving the object to the thread
        self.obj.finished.connect(self.thread.quit) # When the method is finished we receive the signal that it is finished
        self.thread.started.connect(self.obj._search_cast_) # We need to connect the above with the desired method inside the work.py

        self.app = QtWidgets.QApplication(sys.argv)

        def search_menu(self):
            self.SearchAction = self.menu.addAction("Search")
            self.SearchAction.triggered.connect(self.search_cast)

        def onIntReady(self, availablecc):     # This method receives the list from the worker
            print ('availablecc', availablecc)  # This is for debugging reasons to verify that I receive the list with the correct content
            self.availablecc = availablecc

        def search_cast(self):   #This method starts the thread when  self.SearchAction is triggered
            args.select_cc = True
            self.thread.start()

通过这种方式，当搜索list时，菜单不会被屏蔽，屏幕上不会显示任何错误，并且threads中activity monitor监控时的数量会保持正确。

我希望这有助于人们。为了获得更准确的信息（我还在学习PyQt，我的措辞可能不是很好），我建议你查看我上面发布的链接。

Answer 2

由于这是google针对此错误的最佳答案，我花了比预期更长的时间才能正常工作，我将分享我对Python 3和PyQt 5的非常简单的解决方案（如果你更改了一些导入它应该在PyQt4中工作我猜也是如此）。

我遇到的情况是带有右键菜单的系统托盘图标，应该在另一个线程请求时重新构建。您当然可以将此应用于您希望通过线程限制进行通信的其他问题。

import time
import sys
import threading
from PyQt5 import QtGui
from PyQt5 import QtWidgets
from PyQt5 import QtCore



class SystemTrayIcon(QtWidgets.QSystemTrayIcon):
    def __init__(self, icon=None, parent=None):
        icon = QtGui.QIcon(QtWidgets.QApplication.style().standardPixmap(QtWidgets.QStyle.SP_MediaPlay))
        QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)

        self.menu = QtWidgets.QMenu(parent)
        self.setContextMenu(self.menu)

        self.build_menu()
        self.show()

        # see http://pyqt.sourceforge.net/Docs/PyQt5/signals_slots.html for more information
        self.signal = MySignal()
        self.signal.sig_no_args.connect(self.build_menu)
        self.signal.sig_with_str.connect(self.print_string)


    def build_menu(self):
        ''' This function should be called in order to rebuild 
        the right-click menu for the systray icon'''
        global list_dict_streams
        self.menu.clear()

        exitAction = self.menu.addAction("Exit")
        exitAction.triggered.connect(self._exit)

        for x in list_dict_streams :
            self.menu.addAction(x)


    def print_string(self, str):
        print(str)


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



class MySignal(QtCore.QObject):
    ''' Why a whole new class? See here: 
    https://stackoverflow.com/a/25930966/2441026 '''
    sig_no_args = QtCore.pyqtSignal()
    sig_with_str = QtCore.pyqtSignal(str)


list_dict_streams = ["1"]
def work_thread(trayIcon):
    ''' Will add one menu item to the systray menu every 5 seconds
    and will send a signal with a string '''
    global list_dict_streams

    while True:
        trayIcon.signal.sig_no_args.emit()
        trayIcon.signal.sig_with_str.emit("String emitted")
        list_dict_streams.append(str(len(list_dict_streams)+1))
        time.sleep(5)


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

    t = threading.Thread(target=work_thread, args=(trayIcon,))
    t.daemon = True     # otherwise the 'Exit' from the systray menu will not work
    t.start()

    sys.exit(app.exec_())


if __name__ == '__main__':
    main()

基本上，您必须创建新的class MySignal(QtCore.QObject) why。我创建了一个包含两个示例的类 - 一个不向另一个发送参数的参数，您可以传递一个字符串。你当然可以define other arguments。然后在目标线程中创建此类的新实例，并将该类中的函数连接到目标内的函数（在我的情况下为系统托盘图标）。之后，您现在可以像在while循环中一样调用emit(...)函数现在Qt很高兴，因为你只是发出一个信号，而不是直接从另一个线程调用trayIcon.build_menu()。

PyQt5 QObject：无法为不同线程中的父级创建子级

2 个答案: