I would like to execute a method which can only be called once my QApplication is displayed, i.e. when it has entered its main event loop exec_(). I'm new to Qt4 (using PyQt4): i was hoping to have a on_start()-like callback, but didn't find one.
Do i need to create a thread or a timer? Or is there some callback included in the API already?
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3 Answers
You can use a single-shot timer for this, as in the following simple script:
import sys
from PyQt4 import QtGui, QtCore
app = QtGui.QApplication(sys.argv)
def on_start():
print(' in event loop!')
print(' telling app to exit ...')
app.exit(123)
QtCore.QTimer.singleShot(0, on_start)
print('About to enter event loop')
rc = app.exec_()
print('All done - returned %d' % rc)
when you run this, you should see
About to enter event loop
in event loop!
telling app to exit ...
All done - returned 123
1 Comment
neydroydrec
ha thanks, i thought of using QTimer too. I chose for QThread eventually, which i thought was more simple to use. But your example makes it more clear than the documentation :)
I haven't used Qt before, so I might be wrong on this, but I imagine you could bind your on_start() method to the ApplicationActivate event, and set a flag from inside your on_start() method so that the code would only be run that very first time and not any other times the ApplicationActivate event is triggered during execution of your program.
On the other hand if the on_start() method would be called by multiple threads, you might want to use thread-specific flags. Haven't done much multithreading myself, so I'm not sure of what the specifics for that would be, or how simple/complicated it would be.
7 Comments
neydroydrec
Hey thanks for that. I've seen the
Qt.ApplicationActivate but i remain confused because the QApplication does not have that signal implemented. If i have to bind this signal to QApplication, it would have to take place after the event loop has started too, which brings me back to my initial problem. But perhaps i am just not understanding how signals work in Qt :)
JAB
"but i remain confused because the QApplication does not have that signal implemented." That's what subclassing is for.
neydroydrec
@JAB: Yep i thought of that in the time between my comment and yours. But then can the event of
QApplication be passed to QWidget even as it the latter is not the child of the former? I've seen event-handling between children of or with the topmost widget only.JAB
I think it should be possible, but as I said I don't have any experience actually using Qt. riverbankcomputing.co.uk/static/Docs/PyQt4/html/…
neydroydrec
@JAB: ha! perhaps i can connect the
Widget() instance to the QApplication outside the Widget() class definition, i.e. in the main() function. I'll try and let you know how it worked. Thanks. |
For now i have chosen to use QThread this way:
class MyThread(QtCore.QThread):
def run(self):
''' reinplemented from parent '''
# make thread sleep to make sure
# QApplication is running before doing something
self.sleep(2)
do_something()
class MyWidget(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.attr = 'foo'
self.thread = MyThread(self)
self.thread.start()
def main():
app = QtGui.QApplication(sys.argv)
w = MyWidget()
w.show()
sys.exit(app.exec_())
It works for my purpose because my thread in my (real) code actually looks for the application window in X display and will try doing so until it finds it. But otherwise that's not an elegant way to solve the problem.
It would be nicer if there was a signal emitted by QApplication when entering the event-loop. JAB proposed the Qt.ApplicationActivate but it doesn't seem to be emitted by QApplication, and even if it was, because the MyWidget() is not instantiated as a child of QApplication, i wouldn't know how to pass the signal from app to w
I'll wait for a better answer, if any, before accepting my answer as the chosen solution.