I'm trying (and researching) with little success to emit a signal from a working Qthread to the main window. I don't seem to understand how I should go about this in the new syntax.
Here's a simple example.
from PySide.QtCore import *
from PySide.QtGui import *
import sys
import time
class Dialog(QDialog):
def __init__(self, parent=None):
super(Dialog, self).__init__(parent)
button = QPushButton("Test me!")
layout = QVBoxLayout()
layout.addWidget(button)
self.setLayout(layout)
#self.button.clicked.connect(self.test) ----> 'Dialog' object has no attribute 'button'
self.connect(button, SIGNAL('clicked()'), self.test)
self.workerThread = WorkerThread()
def test(self):
self.workerThread.start()
QMessageBox.information(self, 'Done!', 'Done.')
class WorkerThread(QThread):
def __init__(self, parent=None):
super(WorkerThread, self).__init__(parent)
def run(self):
time.sleep(5)
print "Thread done!"
app = QApplication(sys.argv)
dialog = Dialog()
dialog.show()
app.exec_()
I understand that if I didn't have another thread I'd create the signal inside the Dialog class and connect it in the __init__
but how can I create a custom signal that can be emitted from WorkerThread and be used test()?
As a side question. You can see it commented out of the code that the new syntax for connecting the signal errors out. Is it something in my configurations?
I'm on OsX El Capitan, Python 2.7
Any help is highly appreciated! Thanks a lot
TL:DR: I'd like to emmit a signal from the WorkerThread after 5 seconds so that the test function displays the QMessageBox only after WorkingThread is done using the new syntax.