如何从正在运行的QThread发信号回到启动它的PyQt Gui？

Question

我试图了解如何使用从Qthread返回到启动的Gui接口的信令.

设置:我有一个需要几乎无限期运行的进程(模拟)(或者至少在很长一段时间内运行).当它运行时,它执行各种计算,并且一些结果必须被发送回GUI,将实时适当地显示它们.我正在使用PyQt进行GUI.我最初尝试使用python的线程模块,然后在SO和其他地方阅读了几个帖子后切换到QThreads.

根据Qt博客上的这篇文章你做错了,使用QThread的首选方法是创建一个QObject,然后将其移动到Qthread.所以我在PyQt中跟随了关于QThread的onBackground线程的建议">这个问题并尝试了一个简单的测试应用程序(下面的代码):它打开了一个简单的GUI,让你启动后台进程,并且它发布了更新步骤值一个旋转盒.

但它不起作用.GUI永远不会更新.我究竟做错了什么？

import time, sys
from PyQt4.QtCore  import *
from PyQt4.QtGui import * 

class SimulRunner(QObject):
    'Object managing the simulation'

    stepIncreased = pyqtSignal(int, name = 'stepIncreased')
    def __init__(self):
        super(SimulRunner, self).__init__()
        self._step = 0
        self._isRunning = True
        self._maxSteps = 20

    def longRunning(self):
        while self._step  < self._maxSteps  and self._isRunning == True:
            self._step += 1
            self.stepIncreased.emit(self._step)
            time.sleep(0.1)

    def stop(self):
        self._isRunning = False

class SimulationUi(QDialog):
    'PyQt interface'

    def __init__(self):
        super(SimulationUi, self).__init__()

        self.goButton = QPushButton('Go')
        self.stopButton = QPushButton('Stop')
        self.currentStep = QSpinBox()

        self.layout = QHBoxLayout()
        self.layout.addWidget(self.goButton)
        self.layout.addWidget(self.stopButton)
        self.layout.addWidget(self.currentStep)
        self.setLayout(self.layout)

        self.simulRunner = SimulRunner()
        self.simulThread = QThread()
        self.simulRunner.moveToThread(self.simulThread)
        self.simulRunner.stepIncreased.connect(self.currentStep.setValue)


        self.connect(self.stopButton, SIGNAL('clicked()'), self.simulRunner.stop)
        self.connect(self.goButton, SIGNAL('clicked()'), self.simulThread.start)
        self.connect(self.simulRunner,SIGNAL('stepIncreased'), self.currentStep.setValue)


if __name__ == '__main__':
    app = QApplication(sys.argv)
    simul = SimulationUi()
    simul.show()
    sys.exit(app.exec_())

Answer 1

这里的问题很简单:你SimulRunner永远不会收到一个信号,导致它开始工作.这样做的一种方法是将它连接到startedThread 的信号.

另外,在python中你应该使用新式的连接信号的方式:

...
self.simulRunner = SimulRunner()
self.simulThread = QThread()
self.simulRunner.moveToThread(self.simulThread)
self.simulRunner.stepIncreased.connect(self.currentStep.setValue)
self.stopButton.clicked.connect(self.simulRunner.stop)
self.goButton.clicked.connect(self.simulThread.start)
# start the execution loop with the thread:
self.simulThread.started.connect(self.simulRunner.longRunning)
...