PyQt - How to simulate signal? Or some other solution?

Question

In have the modified the code from This Answer

from PyQt5.QtWidgets import *
from PyQt5.QtCore import *
from PyQt5.QtGui import *
import time

class MyWorker(QObject):

    @pyqtSlot()
    def firstWork(self):
        print ('doing first work')
        time.sleep(2)
        print ('first work done')

    @pyqtSlot()
    def secondWork(self):
        print ('doing second work')
        time.sleep(2)
        print ('second work done')

class Window(QWidget):
    def __init__(self, parent = None):
        super(Window, self).__init__()

        self.initUi()
        self.setupThread()

    def initUi(self):
        layout = QVBoxLayout()
        self.button1 = QPushButton('1')
        self.button2 = QPushButton('2')
        layout.addWidget(self.button1)
        layout.addWidget(self.button2)
        self.setLayout(layout)
        self.show()

    def setupThread(self):
        self.thread = QThread()
        self.worker = MyWorker()

        self.worker.moveToThread(self.thread)

        self.button2.clicked.connect(self.worker.secondWork)
        self.button1.clicked.connect(self.worker.firstWork)

        # Start thread
        self.thread.start()    

if __name__ == "__main__":
    app = QApplication([])
    w = Window()
    app.exec_()

Everything works perfectly but i want to start methods in separate threads by simple line of code and not by event signals.

Is there was how to perform this?

Something like self.worker.secondWork() instead of event from button self.button2.clicked.connect(self.worker.secondWork) (This of course does not works)

The only way i have found with my skill is create invisible button and simulate his click event for this but it is not good solution of course!

The question you linked to already gave you the code to solve this, so I'm not sure why you removed that from your example. All you need to do is connect the `thread.started` signal to the two worker slots. There's no need to connect up the buttons as well, since you're not sending anything back to the worker. — ekhumoro, Feb 03 '17 at 18:51
I did not know that signal can easily work in opposite direction — Erik Šťastný, Feb 06 '17 at 07:30

score 0 · Accepted Answer · answered Feb 03 '17 at 13:55

I have solved this by adding custom PyQt signal.

from PyQt5.QtWidgets import *
from PyQt5.QtCore import *
from PyQt5.QtGui import *
import time

class MyWorker(QObject):

    @pyqtSlot()
    def firstWork(self):
        print ('doing first work')
        time.sleep(2)
        print ('first work done')

    @pyqtSlot()
    def secondWork(self):
        print ('doing second work')
        time.sleep(2)
        print ('second work done')

class Window(QWidget):

    Signal = pyqtSignal()

    def __init__(self, parent = None):
        super(Window, self).__init__()

        self.initUi()
        self.setupThread()

    def initUi(self):
        layout = QVBoxLayout()
        self.button1 = QPushButton('1')
        self.button2 = QPushButton('2')
        layout.addWidget(self.button1)
        layout.addWidget(self.button2)
        self.setLayout(layout)
        self.show()

    def setupThread(self):
        self.thread = QThread()
        self.worker = MyWorker()

        self.worker.moveToThread(self.thread)

        self.Signal.connect(self.worker.secondWork)
        self.Signal.connect(self.worker.firstWork)

        self.thread.start()
        self.Signal.emit()

if __name__ == "__main__":
    app = QApplication([])
    w = Window()
    app.exec_()

PyQt - How to simulate signal? Or some other solution?

1 Answers1