PyQt - How can I re-initialise the whole code without restarting it? - python

I wrote a python GUI program consisting of two separate files; One is for logic code and the other for GUI using PyQt4. The behaviour of some objects (buttons, text fields ...) changes throughout the code and I need to reset everything to its original status by clicking on a QAction class menu item. How can I do that?
EDIT: the function that supposed to reset the GUI to the original status:
def newSession(self):
self.ui.setupUi(self)
self.filename = ""
self.paramsSplitted = []
self.timestep = None
self.index = None
self.selectedParam = None
self.selectedMethod = None
--Snip--

What you could do:
Define a ResetHandler(QtCore.QObject) object with a reset_everything signal
During startup create an instance and set it on the globally available QApplication like qapplication.reset_handler = ResetHandler()
Every UI element that needs to update itself defines a on_reset_everything_triggered() slot. (Optional: You could also just use update for example).
When You create UI elements that are supposed to update, connect them to the globally available reset_everything signal from the handler on the QApplication.
Connect your QAction.triggered with the ResetHandler.reset_everything signal.
Now every time you press the QAction the reset_everything signal is invoked, and all UI elements that you connected will update themself.

Like you requested in your comment here is a schematic way of utilizing a function to connect all signals and the method setupUi.
class MainWindow(QtGui.QMainWindow) :
def __init__(self) :
QtGui.QMainWindow.__init__(self)
self.ui.setupUi(self)
# Some code
self.connectAllSignals()
def connectAllSignals(self) :
self.someWidget.clicked.connect(self.someFunction)
self.someAction.triggered.connect(self.otherFunction)
# All the other signals
def disconnectAllSignals(self) :
try :
self.someWidget.clicked.disconnect()
self.someAction.triggered.disconnect()
# All the other signals
except :
print("Something went wrong. Check your code.")
pass
def newSession(self) :
self.ui.setupUi(self)
self.disconnectAllSignals()
self.connectAllSignals()
# Do whatever it takes
By this you ensure you have only the initial settings for your signals and all dynamically added connections are broken. In the method disconnectAllSignals be sure all widgets exist and all signals have at least one connection by the time you call it. If you have new widgets invoked dynamically you should consider deleting them in th method newSession after calling connectAllSignals.

Related

Why is QWidget being destroyed? (PyQt)

So I have the main window. When I click a button in the main window, a new widget is created (in a new window):
self.doorButton.clicked.connect(self.open_door)
def open_door(self):
self.doorwin = QtWidgets.QWidget()
self.doorui = doorwinActions(self.doors)
self.doorui.setupUi(self.doorwin)
self.doorwin.show()
The new QWidget or doorwin has only one widget - tableWidget
I use the object self.doors to populate the table. Now since I have a worker thread (QThread) updating the said object (self.doors), I use QTimer to repopulate the table every 1 second.
class doorwinActions(Ui_doorWin):
def __init__(self,doors):
self.doors = doors
# update setupUi
def setupUi(self, Widget):
super().setupUi(Widget)
Widget.move(self.left, self.top) # set location for window
Widget.setWindowTitle(self.title) # change title
self.timer = QTimer()
self.timer.timeout.connect(lambda:self.popTable())
self.timer.start(1000)
def popTable(self):
mutex.lock()
entries = len(self.doors)
self.tableWidget.setRowCount(entries)
for i in range(entries):
self.tableWidget.setItem(i,0,QtWidgets.QTableWidgetItem(str(self.doors[i][0])))
self.tableWidget.setItem(i,1,QtWidgets.QTableWidgetItem(self.doors[i][1].get_id()))
self.tableWidget.setItem(i,2,QtWidgets.QTableWidgetItem(self.doors[i][1].get_name()))
self.tableWidget.setItem(i,3,QtWidgets.QTableWidgetItem(str(self.doors[i][2])))
mutex.unlock()
When I run the program, it runs smoothly. It opens a new window. If the self.doors object is updated while the window is open, the GUI reflects the change.
BUT, the problem occurs if I reopen the window. If I close the window and then click on the button again, the program crashes with the error:
RuntimeError: wrapped C/C++ object of type QTableWidget has been deleted
From what I understand about my code, when I close the window, the whole widget window (and the table) is deleted. And when I click on the doorButton, new Qwidget/table are created. So, the question is, why would it delete something it just created?
What (sort of) works? - If I move the setup of the door window to the main window's setup, it works. So the open_door function would just be:
def open_door(self):
self.doorwin.show()
The rest would be in the main window setup. But the problem is, then even when I close the window, the QTimer is still going in the background, just eating up processing power.
So, either,
How do I stop the event when the window is closed OR
How do I stop the tableWidget from being deleted?
Your main problem is garbage collection.
When you do this:
def open_door(self):
self.doorwin = QtWidgets.QWidget()
self.doorui = doorwinActions(self.doors)
self.doorui.setupUi(self.doorwin)
self.doorwin.show()
You are creating a new QWidget. It has absolutely no other reference but self.doorwin. This means that if you call open_door again, you will be overwriting self.doorwin, and since the previous widget was not referenced anywhere else, it will get deleted along with all its contents.
Now, QTimers are tricky. You created a QTimer in doorwinActions and QTimers can be persistent even if they have no parent: they keep going on until they're stopped or deleted, and they can only be deleted explicitly or when their parent is deleted (with the exception of timers created with the static QTimer.singleShot() function).
Finally, you must remember that PyQt (like PySide) is a binding. It creates "connections" with the objects created in Qt (let's call them "C++ objects"), and through those bindings we can access those objects, their functions and so on, through python references.
But, and this is of foremost importance, both objects can have a different lifespan:
the python reference can be deleted, and the Qt object can still exist;
the Qt object can be destroyed, yet we still have the python object that referenced (past tense) it;
This is exactly what happens in your case: the self.doorui object is overwritten, but it has an object (the QTimer, self.timer) that is still alive, so the Python garbage collector will not delete it, and the timer is still able to call popTable. But, at that point, the widget (self.doorwin) and its contents have been destroyed on the "C++ side", which causes your crash: while self.tableWidget still exists as a python reference, the actual widget has been destroyed along with its parent widget, and calling its functions causes a fatal error as the binding cannot find the actual object.
Now, how to solve that?
There are various options, and it depends on what you need to do with that window. But, before that, there is something much more important.
You have been manually editing a file generated by pyuic, but those files are not intended for that. They are to be considered like "resource" files, used only for their purpose (the setupUi method), and never, EVER be manually edited. Doing that is considered bad practice and is conceptually wrong for many reasons - and their header clearly warns about that. To read more about the commonly accepted approaches for those files, read the official guidelines about using Designer.
One of those reasons is exactly related to the garbage collection issue explained above.
Note that subclassing the pyuic form class alone is also discouraged (and pointless if you want to extend the widget behavior); the most common, accepted and suggested practice is to create a class that inherits both from the Qt widget class and the UI class. The following code assumes that you have recreated the file with pyuic and named ui_doorwin.py:
# ...
from ui_doorwin import Ui_DoorWin
# ...
class DoorWin(QtWidgets.QWidget, Ui_DoorWin):
def __init__(self, doors):
super().__init__()
self.doors = doors
self.setupUi(self)
self.timer = QTimer(self) # <-- IMPORTANT! Note the "self" argument
self.timer.timeout.connect(self.popTable)
self.timer.start(1000)
def popTable(self):
# ...
With the above code you can be sure that whenever the widget gets deleted for any reason, the timer will be destroyed along with it, so the function will not be called trying to access objects that don't exist anymore.
If you need to keep using an existing instance of the window, the solution is pretty simple: create a None instance (or class) attribute and check if it already exists before creating a new one:
class SomeParent(QtWidgets.QWidget):
doorwin = None
# ...
def open_door(self):
if not self.doorwin:
self.doorwin = DoorWin()
self.doorwin.show()
The above code will not stop the table from updating, which is something you might not want, so you might choose to start and stop the timer depending on when the window is actually shown:
class DoorWin(QtWidgets.QWidget, Ui_DoorWin):
def __init__(self, doors):
super().__init__()
self.doors = doors
self.setupUi(self)
self.timer = QTimer(self)
self.timer.timeout.connect(self.popTable)
def showEvent(self, event):
if not event.spontaneous():
self.timer.start()
def hideEvent(self, event):
if not event.spontaneous():
self.timer.stop()
The event.spontaneous() check above is to prevent stopping the timer if the show/hide event is caused by system calls, like minimizing the window or changing desktop. It's up to you to decide if you want to keep the timer going on and process all data, even if the window is not shown.
Then, if you want to completely destroy the window when it's closed and when a new one is opened, do the following:
class DoorWin(QtWidgets.QWidget, Ui_DoorWin):
def __init__(self, doors):
# ... (as above)
self.setAttribute(QtCore.Qt.WA_DeleteOnClose)
and then ensure that the widget exists (note that if it's closed by the user the reference still exists):
class SomeParent(QtWidgets.QWidget):
doorwin = None
# ...
def open_door(self):
if self.doorwin:
try:
self.doorwin.close()
except RuntimeError:
pass
self.doorwin = DoorWin()
self.doorwin.show()
I found the solution. My two proposed solutions are the same. If I stop the QTimer on window closing, it no longer gives me the error.
self.exitButton.clicked.connect(self.close_win)
def close_win(self):
self.timer.stop()
self.Widget.close()

PyQt5: Closing / relaunching application causes seg fault [duplicate]

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.

How do I open (and close) a PyQt5 application inside a loop, and get that loop running multiple times

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.

Multiple QRadioButton error

I'm using PyQt5 to create a program. I created 3 Radio Buttons, but when I check the first button and check the second button after that. The program will run both of the functions which are connected to these buttons. How I can make it only run the function which is connected to that button. Thanks.
def __init__(self):
super(Program, self).__init__()
self.ui = Ui_APIManager()
self.ui.setupUi(self)
self.show()
self.ui.add_btn.toggled.connect(self.start)
self.ui.check_btn.toggled.connect(self.start)
self.ui.delete_btn.toggled.connect(self.start)
def start(self):
if self.ui.add_btn.isChecked():
self.ui.third_lbl.setEnabled(True)
self.ui.first_lbl.setText('Tool name')
self.ui.second_lbl.setText('ID')
self.ui.third_lbl.setText('Username')
self.ui.action_btn.clicked.connect(self.add_user)
elif self.ui.check_btn.isChecked():
self.ui.first_lbl.setText('Type of search')
self.ui.second_lbl.setText('Keyword')
self.ui.third_lbl.setEnabled(False)
self.ui.action_btn.clicked.connect(self.check_user)
elif self.ui.delete_btn.isChecked():
self.ui.first_lbl.setText('Type of search')
self.ui.second_lbl.setText('Keyword')
self.ui.third_lbl.setEnabled(False)
self.ui.action_btn.clicked.connect(self.delete_user)
Qt signals can have multiple slots attached to them. Every time you click a button the start function is adding another connection to the action_button.clicked signal.
You will need to disconnect any existing slots from the signal first to achieve the desired behaviour. You can disconnect everything from self.ui.action_btn all at once by calling its disconnect() function.
Rather than trying to reassign the roles of the GUI elements you have created, you would be better off creating separate widgets containing the elements for each checkbox state and switching between them. You might find QStackedWidget useful.

update view immediately on datachanged

I have a table with cells like Excel. I know I need to emit dataChanged signal from my Model in order to update my View, it works perfectly. However, when the update comes from another (python) thread, I need to set my mouse (click once) on the updated cell in order the new value to be shown. I know the new value is correctly set in the Model (i can see the log), but the View is lagged behind..
If the update relates to the cell which is already active, the update is reflected immediatly.
What signal should I emit in order to force View to be updated without me clicking on the right cell?
EDIT It works now. I need to implement QThread rather than plain pythhon thread and to connect the signal to a handler from data model. Solved
The warning QObject::connect: Cannot queue arguments of type 'QVector<int>' was preventing dataChanged signal from being captured. This happened because this signal was emitted from another (pythonic) thread. In order to make it work, I needed to subclass QThead like:
class MyThread(QThread):
updated = QtCore.pyqtSignal(str)
def run(self):
while 1:
result= do stuff...
self.updated.emit(result)
And in my QAbstractTableModel, connect the signal to my model:
class TableModel(QAbstractTableModel):
def __init__(self, rowCount: int, colCount: int, parent=None):
super(QAbstractTableModel, self).__init__(parent)
self._listenerth = MyThread()
self._listenerth.updated.connect(self.remote_update_handler)
self._listenerth.start(QThread.LowPriority)
def remote_update_handler(self,result):
self.setData(QModelIndex,result, QtCore.Qt.EditRole)
This made the warning gone, and update worked flawlessly

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