[英]PyQt5 QDialog in subsequent threads
我在Python 3.3中有一個PyQt5程序,該程序在每次按下按鈕時都會啟動一個新線程。 該線程將使用彈出對話框。 第一次按下該按鈕時,它將起作用,但是,第二次(第一次完成后)將使程序崩潰。 我可以在線程中多次調用該對話框,但是第二次運行該線程時,程序將凍結。 此代碼將重現該問題。
import sys
from threading import Thread
from PyQt5 import QtWidgets, QtCore
class Ui_Dialog(object):
def setupUi(self, Dialog):
Dialog.setObjectName("Dialog")
self.pushButton = QtWidgets.QPushButton(Dialog)
self.pushButton.setGeometry(QtCore.QRect(100, 100, 100, 50))
self.pushButton.setObjectName("pushButton")
self.retranslateUi(Dialog)
QtCore.QMetaObject.connectSlotsByName(Dialog)
def retranslateUi(self, Dialog):
_translate = QtCore.QCoreApplication.translate
Dialog.setWindowTitle(_translate("Dialog", "Test"))
self.pushButton.setText(_translate("Dialog", "OK"))
class Ui_MainWindow(object):
def setupUi(self, mainWindow):
mainWindow.setObjectName("mainWindow")
self.pushButton = QtWidgets.QPushButton(mainWindow)
self.pushButton.setGeometry(QtCore.QRect(30, 20, 100, 60))
self.pushButton.setObjectName("pushButton")
self.retranslateUi(mainWindow)
QtCore.QMetaObject.connectSlotsByName(mainWindow)
def retranslateUi(self, Dialog):
_translate = QtCore.QCoreApplication.translate
Dialog.setWindowTitle(_translate("mainWindow", "Test"))
self.pushButton.setText(_translate("mainWindow", "Push Me!"))
class TestDialog(QtWidgets.QDialog):
def __init__(self, parent=None):
super(TestDialog, self).__init__(parent)
self.ui = Ui_Dialog()
self.ui.setupUi(self)
# This message simply needs to go away when the button is pushed
self.ui.pushButton.clicked.connect(self.close)
def show_message(self):
super(TestDialog, self).exec_()
class Main(QtWidgets.QMainWindow):
def __init__(self):
super(Main, self).__init__()
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
self.dialog = TestDialog()
self.ui.pushButton.clicked.connect(self.start_thread)
def start_thread(self):
t = Thread(target=self.show_dialog)
t.daemon = True
t.start()
def show_dialog(self):
# Do lots of background stuff here
self.dialog.show_message()
# The dialog can be shown multiple times within the same thread
self.dialog.show_message()
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = Main()
window.show()
sys.exit(app.exec_())
刪除對話框消息,它起作用。 那為什么不能從第二個線程調用對話框呢? 我不是試圖同時運行兩個線程,而是一個接一個。
感謝塞巴斯蒂安的幫助,我找到了答案。 我創建了一個信號對象,將其連接到show_message函數。 我還添加了一個信號來告知線程何時接受對話框。 這是工作代碼。
import sys
from threading import Thread
from PyQt5 import QtWidgets, QtCore
class Ui_Dialog(object):
def setupUi(self, Dialog):
Dialog.setObjectName("Dialog")
self.pushButton = QtWidgets.QPushButton(Dialog)
self.pushButton.setGeometry(QtCore.QRect(100, 100, 100, 50))
self.pushButton.setObjectName("pushButton")
self.retranslateUi(Dialog)
QtCore.QMetaObject.connectSlotsByName(Dialog)
def retranslateUi(self, Dialog):
_translate = QtCore.QCoreApplication.translate
Dialog.setWindowTitle(_translate("Dialog", "Test"))
self.pushButton.setText(_translate("Dialog", "OK"))
class Ui_MainWindow(object):
def setupUi(self, mainWindow):
mainWindow.setObjectName("mainWindow")
self.pushButton = QtWidgets.QPushButton(mainWindow)
self.pushButton.setGeometry(QtCore.QRect(30, 20, 100, 60))
self.pushButton.setObjectName("pushButton")
self.retranslateUi(mainWindow)
QtCore.QMetaObject.connectSlotsByName(mainWindow)
def retranslateUi(self, Dialog):
_translate = QtCore.QCoreApplication.translate
Dialog.setWindowTitle(_translate("mainWindow", "Test"))
self.pushButton.setText(_translate("mainWindow", "Push Me!"))
class TestDialog(QtWidgets.QDialog):
signal = QtCore.pyqtSignal()
def __init__(self, parent=None):
super(TestDialog, self).__init__(parent)
self.ui = Ui_Dialog()
self.ui.setupUi(self)
# This message simply needs to go away
self.ui.pushButton.clicked.connect(self.close)
def show_message(self):
# Use this to display the pop-up so the text can be altered
super(TestDialog, self).exec_()
self.signal.emit()
class Main(QtWidgets.QMainWindow):
signal = QtCore.pyqtSignal()
def __init__(self):
super(Main, self).__init__()
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
self.dialog = TestDialog()
self.dialog_done = False
self.ui.pushButton.clicked.connect(self.start_thread)
def complete_dialog(self):
self.dialog_done = True
def wait_for_dialog(self):
while not self.dialog_done:
pass
self.dialog_done = False
def start_thread(self):
t = Thread(target=self.show_dialog)
t.daemon = True
t.start()
def show_dialog(self):
# Do lots of background stuff here
self.signal.emit()
# Wait for the dialog to get closed
self.wait_for_dialog()
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = Main()
window.show()
dialog = TestDialog()
window.signal.connect(dialog.show_message)
dialog.signal.connect(window.complete_dialog)
sys.exit(app.exec_())
感謝@Brickmastr發布的答案,我可以在稍有不同的情況下重新制作此代碼以供自己使用。 如果有其他人想做我曾經做過的事情,希望這種稍微不同的方法可以有所幫助。 如果您要運行的程序要顯示“正在運行”對話框,然后在過程完成后更新該對話框,則這是一種實現方法。 此外,這是將相同的彈出消息用於多個耗時功能的方法。
import PyQt5
#This below import is the python file that gets created from using QtDesigner
#And running pyuic5 to create a .py file from your .ui file - hopefully
#whomever reads this is familiar with using QtDesigner
import dialogBox as fxRun
#This is the file that would contain your primary UI, also created using QtDesigner
import mainUI
import threading
class MAIN_UI(PyQt5.QtWidgets.QMainWindow, mainUI.Ui_interface):
startSignal = PyQt5.QtCore.pyqtSignal()
endSignal = PyQt5.QtCore.pyqtSignal()
def __init__(self,parent=None):
super(MAIN_UI, self).__init__(parent)
self.setupUi(self)
self.buttonStartFunction1.clicked.connect(self.startFunction1)
self.buttonStartFunction2.clicked.connect(self.startFunction2)
def startFunction1(self):
self.startThread(self.exampleMethod1)
def startFunction2(self):
self.startThread(self.exampleMethod2)
def startThread(self,functionName):
t = threading.Thread(target=functionName)
t.daemon = True
t.start()
def exampleMethod1(self):
#This function will show the dialog box at the beginning of the process
# and will update the text and button once the process is complete
FULLPROGRAM.mainUI.startSignal.emit()
#Do lots of things here that take a long time
FULLPROGRAM.mainUI.endSignal.emit()
def exampleMethod2(self):
#This can be a different function, just showing that you can send
#whatever function into the startThread() method and it will work
#the same way
FULLPROGRAM.mainUI.startSignal.emit()
#Do lots of things here that take a long time
FULLPROGRAM.mainUI.endSignal.emit()
class PROCESS_BOX(PyQt5.QtWidgets.QDialog, fxRun.Ui_dialogBox):
def __init__(self,parent=None):
super(PROCESS_BOX,self).__init__(parent)
self.setupUi(self)
self.buttonProcessCompleted.clicked.connect(self.close)
def show_dialogbox(self):
self.setWindowTitle("RUNNING")
self.labelProcessStatus.setText("PROCESSING REQUEST... \n PLEASE WAIT...")
self.buttonProcessCompleted.setEnabled(False)
super(PROCESS_BOX,self).exec_()
def processComplete(self):
self.setWindowTitle("FINISHED")
self.labelProcessStatus.setText("PROCESS COMPLETE! \n CLICK OK")
self.buttonProcessCompleted.setEnabled(True)
class FULLPROGRAM:
def __init__(self):
app = PyQt5.QtWidgets.QApplication(sys.argv)
FULLPROGRAM.fxRun = PROCESS_BOX()
FULLPROGRAM.mainUI = MAIN_UI()
FULLPROGRAM.mainUI.startSignal.connect(FULLPROGRAM.fxRun.show_dialogbox)
FULLPROGRAM.mainUI.endSignal.connect(FULLPROGRAM.fxRun.processComplete)
FULLPROGRAM.mainUI.show()
app.exec_()
def main():
program = FULLPROGRAM()
if __name__ == '__main__':
main()
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