Issue
I have a simple PySide Gui which I want to show, however there currently is a blocking task which is a long process, populating the dropdown list with items. Which means the UI does not show until the dropdown list is finished being populated. I want to know is there a way to force show the UI before it attempts to populate the list. I would prefer the dialog show so users know they opened the tool before assuming it's crashed or something.
I'm using Qt since my application needs to run in both PySide and PySide2. I was initially trying to use the qApp.processEvents() but it doesn't seem to be available in the Qt wrapper, or I may have been missing something. If you know what the equivalent would be, I'm fine with the process events being the solution. Optionally if there is an elegant way to populate the list from a background thread somehow...
from Qt import QtGui, QtCore, QtWidgets
class Form(QtWidgets.QWidget):
def __init__(self, parent=None):
super(Form, self).__init__(parent)
self.resize(200,50)
self.items = QtWidgets.QComboBox()
layout = QtWidgets.QVBoxLayout()
layout.addWidget(self.items)
self.setLayout(layout)
# init
self.long_process()
def long_process(self):
for i in range(30000):
self.items.addItem('Item {}'.format(i))
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
form = Form()
form.show()
sys.exit(app.exec_())
Solution
A good option for these cases is always to use QTimer
:
class Form(QtWidgets.QWidget):
def __init__(self, parent=None):
super(Form, self).__init__(parent)
[...]
# init
timer = QtCore.QTimer(self)
timer.timeout.connect(self.on_timeout)
timer.start(0)
def on_timeout(self):
self.items.addItem('Item {}'.format(self.counter))
self.counter += 1
if self.counter == 30000:
self.sender().stop()
Answered By - eyllanesc
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