PyQt4:如何在发出信号之前暂停线程？

Question

我有以下pyqtmain.py:

#!/usr/bin/python3
import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import *
from pyqtMeasThread import *


class MainWindow(QMainWindow):
    def __init__(self, parent=None):
        self.qt_app = QApplication(sys.argv)
        QMainWindow.__init__(self, parent)

        buttonWidget = QWidget()
        rsltLabel = QLabel("Result:")
        self.rsltFiled = QLineEdit()
        self.buttonStart = QPushButton("Start")

        verticalLayout = QVBoxLayout(buttonWidget)
        verticalLayout.addWidget(rsltLabel)
        verticalLayout.addWidget(self.rsltFiled)
        verticalLayout.addWidget(self.buttonStart)

        butDW = QDockWidget("Control", self)
        butDW.setWidget(buttonWidget)
        self.addDockWidget(Qt.LeftDockWidgetArea, butDW)

        self.mthread = QThread()  # New thread to run the Measurement Engine
        self.worker = MeasurementEngine()  # Measurement Engine Object

        self.worker.moveToThread(self.mthread)
        self.mthread.finished.connect(self.worker.deleteLater)  # Cleanup after thread finished

        self.worker.measure_msg.connect(self.showRslt)

        self.buttonStart.clicked.connect(self.worker.run)

        # Everything configured, start the worker thread.
        self.mthread.start()

    def run(self):
        """ Show the window and start the event loop """
        self.show()
        self.qt_app.exec_()  # Start event loop

    @pyqtSlot(str)
    def showRslt(self, mystr):
        self.rsltFiled.setText(mystr)


def main():
    win = MainWindow()
    win.run()


if __name__ == '__main__':
    main()

另一个执行实际测量的线程脚本:

from PyQt4.QtCore import *
import time

class MeasurementEngine(QObject):
    measure_msg = pyqtSignal(str)
    def __init__(self):
        QObject.__init__(self)  # Don't forget to call base class constructor

    @pyqtSlot()
    def run(self):
        self.measure_msg.emit('phase1')
        time.sleep(2) # here I would like to make it as an interrupt
        self.measure_msg.emit('phase2')

这段代码现在的作用是按下Start按钮后,将执行线程中运行的函数.但是,实际上在功能运行中,测量有两个阶段.现在我用了一段时间.

但我想要实现的是在"阶段1"测量完成之后.将弹出一个消息框,同时线程将被暂停/保持.在用户关闭消息框之前,线程功能将恢复.

Answer 1

使用QWaitCondition从QtCore模块.使用互斥锁,将后台线程设置为等待/休眠,直到前台线程将其唤醒.然后它将继续从那里开始工作.

#!/usr/bin/python3
import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import *
from pyqtMeasThread import *


class MainWindow(QMainWindow):
    def __init__(self, parent=None):
        self.qt_app = QApplication(sys.argv)
        QMainWindow.__init__(self, parent)

        buttonWidget = QWidget()
        rsltLabel = QLabel("Result:")
        self.rsltFiled = QLineEdit()
        self.buttonStart = QPushButton("Start")

        verticalLayout = QVBoxLayout(buttonWidget)
        verticalLayout.addWidget(rsltLabel)
        verticalLayout.addWidget(self.rsltFiled)
        verticalLayout.addWidget(self.buttonStart)

        butDW = QDockWidget("Control", self)
        butDW.setWidget(buttonWidget)
        self.addDockWidget(Qt.LeftDockWidgetArea, butDW)

        self.mutex = QMutex()
        self.cond = QWaitCondition()
        self.mthread = QThread()  # New thread to run the Measurement Engine
        self.worker = MeasurementEngine(self.mutex, self.cond)  # Measurement Engine Object

        self.worker.moveToThread(self.mthread)
        self.mthread.finished.connect(self.worker.deleteLater)  # Cleanup after thread finished

        self.worker.measure_msg.connect(self.showRslt)

        self.buttonStart.clicked.connect(self.worker.run)

        # Everything configured, start the worker thread.
        self.mthread.start()

    def run(self):
        """ Show the window and start the event loop """
        self.show()
        self.qt_app.exec_()  # Start event loop

    # since this is a slot, it will always get run in the event loop in the main thread
    @pyqtSlot(str)
    def showRslt(self, mystr):
        self.rsltFiled.setText(mystr)
        msgBox = QMessageBox(parent=self)
        msgBox.setText("Close this dialog to continue to Phase 2.")
        msgBox.exec_()
        self.cond.wakeAll()


def main():
    win = MainWindow()
    win.run()


if __name__ == '__main__':
    main()

和:

from PyQt4.QtCore import *
import time

class MeasurementEngine(QObject):
    measure_msg = pyqtSignal(str)
    def __init__(self, mutex, cond):
        QObject.__init__(self)  # Don't forget to call base class constructor
        self.mtx = mutex
        self.cond = cond

    @pyqtSlot()
    def run(self):
        # NOTE: do work for phase 1 here
        self.measure_msg.emit('phase1')
        self.mtx.lock()
        try:
            self.cond.wait(self.mtx)
            # NOTE: do work for phase 2 here
            self.measure_msg.emit('phase2')
        finally:
            self.mtx.unlock()

尽管如此,你的时间有点偏差.您甚至可以在显示窗口之前创建应用程序并启动线程.因此,在主窗口弹出之前将弹出消息框.要获得正确的事件序列,您应该在已经使主窗口可见之后将线程作为runMainWindow方法的一部分启动.如果您希望等待条件与消息设置分开,您可能需要一个单独的信号和插槽来处理它.

Answer 2

您无法显示QDialoga 中的 a QThread。所有与 GUI 相关的事情都必须在 GUI 线程（创建对象的线程）中完成QApplication。你可以做的是使用 2 QThread：

• 第一：执行第一阶段。您可以将finished其信号连接到将显示弹出窗口的QThread插槽（使用时它将是模态的）。QMainWindowQDialog.exec_()
• 第二：执行phase2。QThread您可以在上面显示的弹出窗口关闭后创建。