Question

我目前正在寻找一种方法，可以将TCP / IP通信添加到以pyQt编写的应用程序中。尽管我未测试的代码似乎都无法正常工作（在大多数情况下，它只是冻结了GUI），但我还是开始研究线程。

在堆栈上找到了一些代码，并增加了睡眠延迟。我已经阅读了一些有关按顺序运行的线程的信息，并且我认为我对线程的工作原理有所了解，但是得到的结果确实不是我期望的。

import sys
from PyQt5 import QtWidgets, QtCore, QtGui
import time
import threading

class Example(QtWidgets.QWidget):

    def __init__(self):
        super(Example, self).__init__()
        QtWidgets.QToolTip.setFont(QtGui.QFont('SansSerif', 10))
        btn = QtWidgets.QPushButton('Button', self)
        self.show()
        self.background = MyThread(self)
        t = threading.Thread(target=self.background.process)
        t2 = threading.Thread(target=self.background.process2)
        t.start()
        t2.start()
        self.background.notify.connect(self.notify)
        self.background.notify2.connect(self.notify2)

    @QtCore.pyqtSlot()
    def notify(self):
        print("I have been notified")

    @QtCore.pyqtSlot()
    def notify2(self):
        print("I have been notified two")

class MyThread(QtCore.QObject):

    notify = QtCore.pyqtSignal()
    notify2 = QtCore.pyqtSignal()

    def __init__(self, parent):
        super(MyThread, self).__init__(parent)
        self.should_continue = True

    def process(self):
        while self.should_continue:
            # Here, do your server stuff.
            time.sleep(0.001)
            self.notify.emit()

    def process2(self):
        while self.should_continue:
            # Here, do your server stuff.
            time.sleep(0.1)
            self.notify2.emit()

def main():

    app = QtWidgets.QApplication(sys.argv)
    ex = Example()
    sys.exit(app.exec_())

if __name__ == '__main__':
    main()

我为什么得到：

I have been notified
I have been notified two
I have been notified two
I have been notified
I have been notified
I have been notified two
I have been notified
I have been notified two

代替

I have been notified
I have been notified two
I have been notified
I have been notified
I have been notified
I have been notified
I have been notified

notify与notify2打印数量之比不等于时间延迟率吗？

Answer 1

您将错误的方法连接到错误的信号。 process2发出notify2而不是notify，因此您必须进行以下更改：

self.background.notify.connect(self.notify)
self.background.notify.connect(self.notify2)

到

self.background.notify.connect(self.notify)
self.background.notify2.connect(self.notify2)

它将正常工作。

具有巨大时间延迟差异的线程按顺序运行

1 个答案: