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[英]QObject: Cannot create children for a parent that is in a different thread. PyQt5
[英]PyQt5 QObject: Cannot create children for a parent that is in a different thread
我正在使用PyQt5在菜单系统托盘中工作。 我对PyQt5非常陌生,我想做的是在不阻止菜单(多线程)的情况下触发操作。 在许多地方阅读完之后,我得出的结论是使用Qthread
应该是Qthread
的方法(但是,如果我能理解该类的工作原理……)。 但是,考虑到我的应用程序非常简单,使用threading
也不会很糟糕。 因此,我尝试使用import threading
执行以下代码:
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()
当我开始菜单时,一切都已准备就绪。 然后,当我单击菜单Search
该操作将触发self.search_cast
方法,并且我的菜单中会填充找到的列表。 我还可以看到我的应用程序在进行搜索时不会被阻塞,但是完成时会出现以下错误:
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.
此后,菜单仍然具有“响应性”,但不能触发更多操作,因此它仍然是“功能性”的。 此外,似乎没有更多的线程被创建。 如果有人可以解释我为什么会发生这种情况,我将感到非常高兴。 我看不到光...
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.
现在我得到以下错误:
AttributeError: 'casting' object has no attribute 'availablecc'
我确保实际上该worker
程序正在从我称为cc
的外部类中恢复availablecc
。 但是由于某些原因, menubar
类未收到该消息。 我正在基于此https://stackoverflow.com/a/33453124/1995261
我将继续回答自己。 受https://stackoverflow.com/a/33453124/1995261的启发,我通过实现以下方法解决了这一问题:
1)我创建了一个worker.py
,它执行阻止菜单的_search_cast_
方法。 当此方法完成搜索时,它将发出两个信号:a)一个通知他已恢复list
,以及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
我转储了以下内容,并尝试在代码内用注释进行解释:
#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()
这样,在搜索list
,菜单不会被阻止,屏幕上不会显示任何错误,并且在activity monitor
threads
时, threads
数保持正确。
希望对大家有帮助。 有关更精确的信息(我仍在学习PyQt,并且措辞可能不太好),建议您检查上面发布的链接。
由于这是该错误的Google最佳答案,并且花费了我比预期的时间才能正确解决此问题,因此,我将分享我针对Python 3和PyQt 5的非常简单的解决方案(如果您更改某些导入,它也应该在PyQt4中工作) )。
我遇到的情况是带有右键单击菜单的系统托盘图标,当其他线程请求它时应重新构建该图标。 您当然可以将其应用于要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,您必须创建一个新class MySignal(QtCore.QObject)
why 。 我创建了一个包含两个示例的类-一个示例不发送任何参数,而另一个示例则可以传递字符串。 您当然可以定义其他参数 。 然后,在目标线程中,创建该类的新实例,并将该类中的函数连接到目标内部的函数(本例中为systray图标)。 在那之后,您现在可以像在while循环中一样调用emit(...)
函数。
现在,Qt很高兴,因为与直接从其他线程调用trayIcon.build_menu()
相比,您只发出一个信号即可。
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