window 当我尝试使用 python 的线程在 Qt window 中显示不断变化的值时冻结

Question

我的主要objective是在Qt-windowtextEdit上展示一个不断发展的价值。（这个 window 只包含一个 checkBox 和一个 textEdit）。遗憾的是，我无法单击该复选框并且 window 被冻结，直到我关闭终端。

import sys
from threading import Thread
from random import randint
import time
from PyQt4 import QtGui,uic


class MyThread(Thread):

    def __init__(self):
        Thread.__init__(self)

    #function to continually change the targeted value
    def run(self):
        for i in range(1, 20):
            self.a = randint (1, 10)
            secondsToSleep = 1
            time.sleep(secondsToSleep)


class MyWindow(QtGui.QMainWindow,Thread):
    def __init__(self):
        Thread.__init__(self)
        super(MyWindow,self).__init__()
        uic.loadUi('mywindow.ui',self)
        self.checkBox.stateChanged.connect(self.checkeven)
        self.show()

    #i show the value only if the checkbox is checked
    def checkeven(self):
        while self.checkBox.isChecked():
            self.textEdit.setText(str(myThreadOb1.a)) 



# Run following code when the program starts
if __name__ == '__main__':
   app = QtGui.QApplication(sys.argv)


   # Declare objects of MyThread class
   myThreadOb1 = MyThread()
   myThreadOb2 = MyWindow()

   # Start running the threads!
   myThreadOb1.start()
   myThreadOb2.start()

   sys.exit(app.exec_())

目前我正在使用线程将随机值设置为 a，但最后应该会更复杂一些，因为我必须从自动化中获取值.

你知道为什么我的代码会这样吗？非常感谢您的帮助。

Answer 1

问题是 while self.checkBox.isChecked() 正在阻塞，阻止 GUI 处理其他事件。

此外，您不应该运行在主线程之外的另一个线程上运行 PyQt GUI。

如果您想将数据从一个线程发送到另一个线程，一个不错的选择是使用信号。

考虑到所有这些因素，我们有以下几点：

import sys
from threading import Thread
from random import randint
import time
from PyQt4 import QtGui, uic, QtCore


class MyThread(Thread, QtCore.QObject):
    aChanged = QtCore.pyqtSignal(int)

    def __init__(self):
        Thread.__init__(self)
        QtCore.QObject.__init__(self)
    #function to continually change the targeted value
    def run(self):
        for i in range(1, 20):
            self.aChanged.emit(randint(1, 10))
            secondsToSleep = 1
            time.sleep(secondsToSleep)


class MyWindow(QtGui.QMainWindow):
    def __init__(self):
        super(MyWindow,self).__init__()
        uic.loadUi('mywindow.ui',self)
        self.thread = MyThread()
        self.thread.aChanged.connect(self.on_a_changed, QtCore.Qt.QueuedConnection)
        self.thread.start()
        self.show()

    #i show the value only if the checkbox is checked
    @QtCore.pyqtSlot(int)
    def on_a_changed(self, a):
        if self.checkBox.isChecked():
            self.textEdit.setText(str(a)) 



# Run following code when the program starts
if __name__ == '__main__':
   app = QtGui.QApplication(sys.argv)


   # Declare objects of MyThread class
   w = MyWindow()

   sys.exit(app.exec_())

更符合 Qt 风格的选项是使用 QThread，因为它是一个 class 处理线程并且是一个 QObject，因此它可以轻松处理信号。

import sys
from random import randint
from PyQt4 import QtGui, uic, QtCore


class MyThread(QtCore.QThread):
    aChanged = QtCore.pyqtSignal(int)
    def run(self):
        for i in range(1, 20):
            self.aChanged.emit(randint(1, 10))
            secondsToSleep = 1
            QtCore.QThread.sleep(secondsToSleep)


class MyWindow(QtGui.QMainWindow):
    def __init__(self):
        super(MyWindow,self).__init__()
        uic.loadUi('mywindow.ui',self)
        self.thread = MyThread()
        self.thread.aChanged.connect(self.on_a_changed, QtCore.Qt.QueuedConnection)
        self.thread.start()
        self.show()

    #i show the value only if the checkbox is checked
    @QtCore.pyqtSlot(int)
    def on_a_changed(self, a):
        if self.checkBox.isChecked():
            self.textEdit.setText(str(a)) 



# Run following code when the program starts
if __name__ == '__main__':
   app = QtGui.QApplication(sys.argv)


   # Declare objects of MyThread class
   w = MyWindow()

   sys.exit(app.exec_())

window 当我尝试使用 python 的线程在 Qt window 中显示不断变化的值时冻结

window freezes when I try to show an evolving value in a Qt window using threads with python

python

multithreading

pyqt

pyqt4