后续线程中的PyQt5 QDialog
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我在Python 3.3中有一个PyQt5程序,每次按下按钮时都会启动一个新线程。该线程将使用弹出对话框。它在第一次按下按钮时起作用,但是,第二次(在第一次完成之后)将使程序崩溃。我可以在线程中多次调用对话框,但第二次运行该程序时程序会冻结。此代码将重现该问题。
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_())
删除对话框消息,它的工作原理。那么为什么我不能从第二个线程调用对话框呢?我不是试图同时运行两个线程,而是一个接一个地运行。
答案
我想通了,谢谢塞巴斯蒂安的帮助。我创建了一个信号对象,将其连接到show_message函数。我还添加了一个信号,告诉线程何时接受了对话框。这是工作代码。
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_())
另一答案
感谢@Brickmastr发布的答案,我能够在稍微不同的场景中为自己的用途重做这个。如果有其他人想要做我想做的事情,希望这种稍微变化的方法可以提供帮助。如果您正在运行要显示“正在运行”对话框的程序,然后在该过程完成后更新该框,则这是实现此目的的一种方法。此外,这是您可以将相同的弹出消息用于多个耗时的功能。
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...
PLEASE WAIT...")
self.buttonProcessCompleted.setEnabled(False)
super(PROCESS_BOX,self).exec_()
def processComplete(self):
self.setWindowTitle("FINISHED")
self.labelProcessStatus.setText("PROCESS COMPLETE!
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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