I'm currently trying to make an API call when my button is clicked without pausing my GUI for the duration of the call. I was using threads to set the text of my Qlable to the response of the API call. This worked however it was unsafe as I was accessing GUI elements from a separate thread.
Currently, I'm attempting to use QThreads to make the API call and then emit the response to the GUI thread however, when I create the Qthread object my program finishes with exit code 3. I've simplified the problem for more clarity.
class MainWindow(QtWidgets.QWidget):
def __init__(self):
super(MainWindow, self).__init__()
self.setWindowTitle("TestWindow")
self.setFixedSize(300,75)
self.main_layout = QtWidgets.QGridLayout()
self.setLayout(self.main_layout)
self.txt_0 = QtWidgets.QLabel()
self.btn_0 = QtWidgets.QPushButton('Press Me')
self.btn_0.clicked.connect(self.btn_0_clicked)
self.main_layout.addWidget(self.txt_0, 0, 0)
self.main_layout.addWidget(self.btn_0, 1, 0)
self.show()
def btn_0_clicked(self):
temp_thread = StringThread("name")
temp_thread.start()
class StringThread(QtCore.QThread):
str_signal = QtCore.pyqtSignal(str)
_name = ''
def __init__(self, name):
QtCore.QThread.__init__(self)
self._name = name
print("Thread Created")
def run(self):
self.str_signal.emit('Emitted message from StringThread. Name = ' + self._name)
print("Done run")
My intention is to set the text of my Qlable to the message emitted from the pyqtSignal in the StringThread class however, as soon as I click the button my program finishes with exit code 3.
edit:
I made the following changes to the btn_0_clicked method
def btn_0_clicked(self):
self.temp_thread = StringThread("hello")
self.temp_thread.str_signal.connect(self.txt_0.setText)
self.temp_thread.start()
It's working now.