I have a button connected to a function called OpenSupplyWidget() which is supposed to start a QWidget class I have in another file (the file is SupplyWidget.py and is already imported).
def OpenSupplyWidget(self):
sw = SupplyWidget()
sw.show()
The function only opens the window for a split second. Using sys.exit(app.exec_()) only returns an error saying the QApplication event loop is already running.
What method do I use to get what I want (opening a widget)?
Thanks!
Try having the SupplyWidget outside of the OpenSupplyWidget-function. The sw probably gets destroyed because the function terminates.
...
def __init__(self):
self.sw = SupplyWidget()
def OpenSupplyWidget(self):
self.sw.show()
..
Related
Basically I have created a Dialog without buttons window as an alert box using Qt5 designer and compiled to Python using Pyuic5 which resulted into partial code as below:
class Ui_alertwindow(object):
def setupUi(self, alertwindow):
alertwindow.setObjectName("alertwindow")
alertwindow.resize(400, 300)
alertwindow.setStyleSheet("background:#222;")
# much more code
Later on, I will have to modify parts of this window which I could not using Qt5 designer, hence I created a separate class to make the changes, as below:
class alertWindowCustom(QDialog, Ui_alertwindow):
def __init__(self):
QDialog.__init__(self)
self.setupUi(self)
self.run()
def run(self):
print("Changes will be made here")
And then finally call this function in QMainWindow when a button is pressed:
def CreateAlertWindow(self):
alertWindow = alertWindowCustom()
alertWindowInput = alertWindow.exec_()
This creates a white dialog box while not responding and in the console I am getting below errors:
Changes will be made here
Unknown property justification
Unknown property justification
QBasicTimer::stop: Failed. Possibly trying to stop from a different thread
QBasicTimer::stop: Failed. Possibly trying to stop from a different thread
Can anyone tell me what is wrong here?
The following is a loop that I created:
import mainui
import loginui
from PyQt5 import QtWidgets
import sys
while True:
print('test')
app = QtWidgets.QApplication(sys.argv)
ui = loginui.Ui_MainWindow()
ui.setupUi()
ui.MainWindow.show()
app.exec_()
username=ui.username
app2 = QtWidgets.QApplication(sys.argv)
ui2 = mainui.Ui_MainWindow(username)
ui2.setupUi()
ui2.MainWindow.show()
app2.exec_()
if ui2.exitFlag=='repeat':#Repeat Condition
continue
else: #Exit Condition
sys.exit()
This is a loop containing a couple of PyQt5 windows, which are displayed in order. The windows run normally when they are not contained within a loop, and they also run pretty well in the first iteration of the loop.
But, when the repeat condition is satisfied, even though the loop does iterate (prints 'test' again) - the ui and ui2 windows do not get displayed again, and subsequently the program hits the exit condition and stops.
Any suggestions about why the windows do not get displayed, and how I can get them displayed would be very much appreciated.
An important premise: usually you need only one QApplication instance.
Proposed solutions
In the following examples I'm using a single QApplication instance, and switch between windows using signals.
Since you probably need to wait for the window to be closed in some way, you might prefer to use a QDialog instead of a QMainWindow, but if for some reason you need the features provided by QMainWindow (menus, dockbars, etc) this is a possible solution:
class First(QtWidgets.QMainWindow):
closed = QtCore.pyqtSignal()
def __init__(self):
super().__init__()
central = QtWidgets.QWidget()
self.setCentralWidget(central)
layout = QtWidgets.QHBoxLayout(central)
button = QtWidgets.QPushButton('Continue')
layout.addWidget(button)
button.clicked.connect(self.close)
def closeEvent(self, event):
self.closed.emit()
class Last(QtWidgets.QMainWindow):
shouldRestart = QtCore.pyqtSignal()
def __init__(self):
super().__init__()
central = QtWidgets.QWidget()
self.setCentralWidget(central)
layout = QtWidgets.QHBoxLayout(central)
restartButton = QtWidgets.QPushButton('Restart')
layout.addWidget(restartButton)
closeButton = QtWidgets.QPushButton('Quit')
layout.addWidget(closeButton)
restartButton.clicked.connect(self.restart)
closeButton.clicked.connect(self.close)
def restart(self):
self.exitFlag = True
self.close()
def showEvent(self, event):
# ensure that the flag is always false as soon as the window is shown
self.exitFlag = False
def closeEvent(self, event):
if self.exitFlag:
self.shouldRestart.emit()
app = QtWidgets.QApplication(sys.argv)
first = First()
last = Last()
first.closed.connect(last.show)
last.shouldRestart.connect(first.show)
first.show()
sys.exit(app.exec_())
Note that you can add menubars to a QWidget too, by using setMenuBar(menuBar) on their layout.
On the other hand, QDialogs are more indicated for these cases, as they provide their exec_() method which has its own event loop and blocks everything else until the dialog is closed.
class First(QtWidgets.QDialog):
def __init__(self):
super().__init__()
layout = QtWidgets.QHBoxLayout(self)
button = QtWidgets.QPushButton('Continue')
layout.addWidget(button)
button.clicked.connect(self.accept)
class Last(QtWidgets.QDialog):
def __init__(self):
super().__init__()
layout = QtWidgets.QHBoxLayout(self)
restartButton = QtWidgets.QPushButton('Restart')
layout.addWidget(restartButton)
closeButton = QtWidgets.QPushButton('Quit')
layout.addWidget(closeButton)
restartButton.clicked.connect(self.accept)
closeButton.clicked.connect(self.reject)
def start():
QtCore.QTimer.singleShot(0, first.exec_)
app = QtWidgets.QApplication(sys.argv)
app.setQuitOnLastWindowClosed(False)
first = First()
last = Last()
first.finished.connect(last.exec_)
last.accepted.connect(start)
last.rejected.connect(app.quit)
start()
sys.exit(app.exec_())
Note that in this case I had to use a QTimer to launch the first dialog. This is due to the fact that in normal conditions signals wait for theirs slot to be completed before returning control to the emitter (the dialog). Since we're constantly recalling the same dialog, this leads to recursion:
First is executed
First is closed, emitting the finished signal, which causes the following:
Second is executed
at this point the finished signal has not returned yet
Second is accepted, emitting the accepted signal, which causes:
First hasn't returned its exec_() yet, but we're trying to exec it again
Qt crashes showing the error StdErr: QDialog::exec: Recursive call detected
Using a QTimer.singleShot ensures that the signal returns instantly, avoiding any recursion for exec_().
Ok, but why doesn't it work?
As said, only one Q[*]Application instance should usually exists for each process. This doesn't actually prevent to create more instances subsequently: in fact, your code works while it's in the first cycle of the loop.
The problem is related to python garbage collection and how PyQt and Qt deals with memory access to the C++ Qt objects, most importantly the application instance.
When you create the second QApplication, you're assigning it to a new variable (app2). At that point, the first one still exists, and will be finally deleted (by Qt) as soon as the process is completed with sys.exit.
When the cycle restarts, instead, you're overwriting app, which would normally cause python to garbage collect the previous object as soon as possible.
This represents a problem, as Python and Qt need to do "their stuff" to correctly delete an existing QApplication object and the python reference.
If you put the following line at the beginning, you'll see that the first time the instance is returned correctly, while the second returns None:
app = QtWidgets.QApplication(sys.argv)
print('Instance: ', QtWidgets.QApplication.instance())
There's a related question here on StackOverflow, and an important comment to its answer:
In principle, I don't see any reason why multiple instances of QApplication cannot be created, so long as no more than one exists at the same time. In fact, it may often be a requirement in unit-testing that a new application instance is created for each test. The important thing is to ensure that each instance gets deleted properly, and, perhaps more importantly, that it gets deleted at the right time.
A workaround to avoid the garbage collection is to add a persistent reference to the app:
apps = []
while True:
print('test')
app = QtWidgets.QApplication(sys.argv)
apps.append(app)
# ...
app2 = QtWidgets.QApplication(sys.argv)
apps.append(app2)
But, as said, you should not create a new QApplication instance if you don't really need that (which is almost never the case).
As already noted in the comments to the question, you should never modify the files generated with pyuic (nor try to mimic their behavior). Read more about using Designer.
The following is a loop that I created:
import mainui
import loginui
from PyQt5 import QtWidgets
import sys
while True:
print('test')
app = QtWidgets.QApplication(sys.argv)
ui = loginui.Ui_MainWindow()
ui.setupUi()
ui.MainWindow.show()
app.exec_()
username=ui.username
app2 = QtWidgets.QApplication(sys.argv)
ui2 = mainui.Ui_MainWindow(username)
ui2.setupUi()
ui2.MainWindow.show()
app2.exec_()
if ui2.exitFlag=='repeat':#Repeat Condition
continue
else: #Exit Condition
sys.exit()
This is a loop containing a couple of PyQt5 windows, which are displayed in order. The windows run normally when they are not contained within a loop, and they also run pretty well in the first iteration of the loop.
But, when the repeat condition is satisfied, even though the loop does iterate (prints 'test' again) - the ui and ui2 windows do not get displayed again, and subsequently the program hits the exit condition and stops.
Any suggestions about why the windows do not get displayed, and how I can get them displayed would be very much appreciated.
An important premise: usually you need only one QApplication instance.
Proposed solutions
In the following examples I'm using a single QApplication instance, and switch between windows using signals.
Since you probably need to wait for the window to be closed in some way, you might prefer to use a QDialog instead of a QMainWindow, but if for some reason you need the features provided by QMainWindow (menus, dockbars, etc) this is a possible solution:
class First(QtWidgets.QMainWindow):
closed = QtCore.pyqtSignal()
def __init__(self):
super().__init__()
central = QtWidgets.QWidget()
self.setCentralWidget(central)
layout = QtWidgets.QHBoxLayout(central)
button = QtWidgets.QPushButton('Continue')
layout.addWidget(button)
button.clicked.connect(self.close)
def closeEvent(self, event):
self.closed.emit()
class Last(QtWidgets.QMainWindow):
shouldRestart = QtCore.pyqtSignal()
def __init__(self):
super().__init__()
central = QtWidgets.QWidget()
self.setCentralWidget(central)
layout = QtWidgets.QHBoxLayout(central)
restartButton = QtWidgets.QPushButton('Restart')
layout.addWidget(restartButton)
closeButton = QtWidgets.QPushButton('Quit')
layout.addWidget(closeButton)
restartButton.clicked.connect(self.restart)
closeButton.clicked.connect(self.close)
def restart(self):
self.exitFlag = True
self.close()
def showEvent(self, event):
# ensure that the flag is always false as soon as the window is shown
self.exitFlag = False
def closeEvent(self, event):
if self.exitFlag:
self.shouldRestart.emit()
app = QtWidgets.QApplication(sys.argv)
first = First()
last = Last()
first.closed.connect(last.show)
last.shouldRestart.connect(first.show)
first.show()
sys.exit(app.exec_())
Note that you can add menubars to a QWidget too, by using setMenuBar(menuBar) on their layout.
On the other hand, QDialogs are more indicated for these cases, as they provide their exec_() method which has its own event loop and blocks everything else until the dialog is closed.
class First(QtWidgets.QDialog):
def __init__(self):
super().__init__()
layout = QtWidgets.QHBoxLayout(self)
button = QtWidgets.QPushButton('Continue')
layout.addWidget(button)
button.clicked.connect(self.accept)
class Last(QtWidgets.QDialog):
def __init__(self):
super().__init__()
layout = QtWidgets.QHBoxLayout(self)
restartButton = QtWidgets.QPushButton('Restart')
layout.addWidget(restartButton)
closeButton = QtWidgets.QPushButton('Quit')
layout.addWidget(closeButton)
restartButton.clicked.connect(self.accept)
closeButton.clicked.connect(self.reject)
def start():
QtCore.QTimer.singleShot(0, first.exec_)
app = QtWidgets.QApplication(sys.argv)
app.setQuitOnLastWindowClosed(False)
first = First()
last = Last()
first.finished.connect(last.exec_)
last.accepted.connect(start)
last.rejected.connect(app.quit)
start()
sys.exit(app.exec_())
Note that in this case I had to use a QTimer to launch the first dialog. This is due to the fact that in normal conditions signals wait for theirs slot to be completed before returning control to the emitter (the dialog). Since we're constantly recalling the same dialog, this leads to recursion:
First is executed
First is closed, emitting the finished signal, which causes the following:
Second is executed
at this point the finished signal has not returned yet
Second is accepted, emitting the accepted signal, which causes:
First hasn't returned its exec_() yet, but we're trying to exec it again
Qt crashes showing the error StdErr: QDialog::exec: Recursive call detected
Using a QTimer.singleShot ensures that the signal returns instantly, avoiding any recursion for exec_().
Ok, but why doesn't it work?
As said, only one Q[*]Application instance should usually exists for each process. This doesn't actually prevent to create more instances subsequently: in fact, your code works while it's in the first cycle of the loop.
The problem is related to python garbage collection and how PyQt and Qt deals with memory access to the C++ Qt objects, most importantly the application instance.
When you create the second QApplication, you're assigning it to a new variable (app2). At that point, the first one still exists, and will be finally deleted (by Qt) as soon as the process is completed with sys.exit.
When the cycle restarts, instead, you're overwriting app, which would normally cause python to garbage collect the previous object as soon as possible.
This represents a problem, as Python and Qt need to do "their stuff" to correctly delete an existing QApplication object and the python reference.
If you put the following line at the beginning, you'll see that the first time the instance is returned correctly, while the second returns None:
app = QtWidgets.QApplication(sys.argv)
print('Instance: ', QtWidgets.QApplication.instance())
There's a related question here on StackOverflow, and an important comment to its answer:
In principle, I don't see any reason why multiple instances of QApplication cannot be created, so long as no more than one exists at the same time. In fact, it may often be a requirement in unit-testing that a new application instance is created for each test. The important thing is to ensure that each instance gets deleted properly, and, perhaps more importantly, that it gets deleted at the right time.
A workaround to avoid the garbage collection is to add a persistent reference to the app:
apps = []
while True:
print('test')
app = QtWidgets.QApplication(sys.argv)
apps.append(app)
# ...
app2 = QtWidgets.QApplication(sys.argv)
apps.append(app2)
But, as said, you should not create a new QApplication instance if you don't really need that (which is almost never the case).
As already noted in the comments to the question, you should never modify the files generated with pyuic (nor try to mimic their behavior). Read more about using Designer.
For instance, I connect the 'clicked' signal of QPushButton to a function named 'func_with_return'. Assumes that there are just three statements in this function: The first one is 'print('start')', the second one is 'return 1' and the last one is 'print('end')'. There is my python code based on PyQt5.
import sys
from PyQt5.QtWidgets import QApplication, QFrame, QPushButton
class MyWindow(QFrame):
def __init__(self):
super(MyWindow, self).__init__()
self.layout_init()
self.layout_manage()
def layout_init(self):
self.setFixedSize(800, 600)
self.button01 = QPushButton('click!', self)
self.button01.setFixedSize(100, 100)
self.button01.clicked.connect(self.func_with_return)
def layout_manage(self):
pass
def func_with_return(self):
print('---------func_with_return starts---------')
return 1
print('---------func_with_return ends---------')
if __name__ == '__main__':
app = QApplication(sys.argv)
mywindow = MyWindow()
mywindow.show()
sys.exit(app.exec_())
Basically, there is no error after clicking on this button. What I am curious about is the interruption caused by 'return' inside a 'slot'. Will this interruption have collision with the signal&slot mechanism?
None. The signals only invoke the function, if the function returns Qt will not use it.
On the other hand in Qt/PyQt it is said that a function is a slot if you use the decorator #QtCore.pyqtSlot(). In your case it is a simple function. Even so for a signal will not serve the data that returns the slot or function invoked.
Will this interruption have collision with the signal&slot mechanism?
No, it does not have a collision. Returning to the beginning, middle or end is irrelevant, remember every function returns something (if you do not use return the function will implicitly return None at the end).
On the other hand in a GUI the tasks of the functions must be light, if they are heavy you must execute it in another thread.
I show Second QMainWindow after click on button in parent QMainWindow
def on_click(self):
window = second_window.MainWindow()
window.show()
Second window not shown (Without any error). But if in Second window I add line:
self.func = functools.partial(self.some_func)
All work correct.
Why it's happens?
I think the problem here is that you are creating the window as a local variable inside the on_click scope. As soon as on_click finishes the window attribute will be destroyed.
Try storing the window in an instance variable:
def on_click(self):
self._window = second_window.MainWindow()
self._window.show()
The functools.partial approach is probably working just because you are already storing it at the instance.