QDialog switch to non-modal after accept not redrawing - python

So, I'm not sure if the title is the best description, but it's what I came up with.
Here's the deal. I'm working on a PyQt app that has a sort of plugin system where you can just add some sub classes to a folder and the app finds them. These commands have the option of being able to create little uis for themselves. Basically, they look like this:
class Command(object):
def do(self):
self.setupUi()
self.pre()
self.run()
self.post()
def pre(self):
# do setup stuff for run method
def run(self):
# do actual work
def post(self):
# clean up after run
def setupUi(self):
# create a ui for this command
diag = QDialog()
diag.exec_()
Now, the issue I'm running into is, I have one Command that creates a dialog, and waits for the user to accept it. Then, I need to switch the dialog to non-modal while the command is running, and up date the dialog. This all seems to work fine. But, the problem is I can't get the dialog to redraw until after the pre, run, and post methods have finished. So, if I have the setupUi like this:
def setupUi(self):
# create a ui for this command
diag = QDialog()
if diag.exec_():
diag.setModal(False)
diag.show()
I tried processEvents but that didn't seem to do it. Has anyone else run into this issue, or know of any work arounds?
Thanks

Using diag.exec_() will block until the dialog returns (is closed). So, if you will need to call show() on it's own. There are a few ways to proceed from here.
You can have the dialog accept slot run a reference to the rest of the commands
You can poll the dialog to see if the user has accepted
You can move the pre, run, and post commands to the dialog
Assuming you want to keep the meat of the code out of the dialog class, and since periodically polling is best to avoid if possible, here is an example of the first strategy:
import sys, time
from PyQt4 import QtCore, QtGui
class MyDialog(QtGui.QDialog):
def __init__(self,parent=None):
QtGui.QDialog.__init__(self,parent)
layout = QtGui.QVBoxLayout()
self.msg = QtGui.QLabel('some sort of status')
self.buttonbox = QtGui.QDialogButtonBox(QtGui.QDialogButtonBox.Ok, QtCore.Qt.Horizontal, self)
self.buttonbox.accepted.connect(self.accept)
layout.addWidget(self.msg)
layout.addWidget(self.buttonbox)
self.setLayout(layout)
def set_msg(self, new_msg):
self.msg.setText(new_msg)
def set_function_on_accept(self,fcn):
self.function = fcn
def accept(self):
self.function()
class Command(object):
def do(self):
self.setupUi()
def do_work(self):
self.pre()
self.run()
self.post()
def pre(self):
# do setup stuff for run method
time.sleep(1)
self.diag.set_msg("stuff setup")
QtGui.QApplication.processEvents()
def run(self):
# do actual work
time.sleep(1)
self.diag.set_msg("work done")
QtGui.QApplication.processEvents()
def post(self):
# clean up after run
time.sleep(1)
self.diag.set_msg("cleaned up")
QtGui.QApplication.processEvents()
def setupUi(self):
# create a ui for this command
diag = MyDialog()
self.diag = diag
diag.set_function_on_accept(self.do_work)
diag.show()
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
command = Command()
command.do()
sys.exit(app.exec_())

Related

When changing screens with QStackedWidget, how to call a function in the changed screen?

I'm trying to call the init function of the screen I'm changing my screen index to
For an example, i have this code:
from PyQt5 import QtWidgets as qtw
from PyQt5 import QtGui as qtg
from sys import argv as sysArgv
from sys import exit as sysExit
arialLarge = qtg.QFont("Arial", 18)
class MainWindow(qtw.QWidget):
def __init__(self):
super().__init__()
# Current screen label;
mainWindowLabel = qtw.QLabel("This is the main window", self)
mainWindowLabel.setFont(arialLarge)
mainWindowLabel.move(20, 40)
# Button for going to the HelloWindow screen;
gotoHelloWindowButton = qtw.QPushButton("Go to hello window", self, clicked=lambda: appStack.setCurrentIndex(appStack.currentIndex()+1))
gotoHelloWindowButton.move(100, 100)
class HelloWindow(qtw.QWidget):
def __init__(self):
super().__init__()
# EG: print hello world when I visit this page
print("hello world")
# Current screen label;
helloWindowLabel = qtw.QLabel("This is the hello window", self)
helloWindowLabel.setFont(arialLarge)
helloWindowLabel.move(20, 40)
# Button for going to the MainWindow screen;
gotoMainWindowButton = qtw.QPushButton("Go to main window", self, clicked=lambda: appStack.setCurrentIndex(appStack.currentIndex()-1))
gotoMainWindowButton.move(100, 100)
if __name__ == "__main__":
app = qtw.QApplication(sysArgv)
appStack = qtw.QStackedWidget()
appStack.addWidget(MainWindow())
appStack.setFixedSize(300, 300)
appStack.show()
appStack.addWidget(HelloWindow())
sysExit(app.exec())
If im visiting the HelloWindow from the MainWindow, how can i run the init function of the HelloWindow screen so I can run whatever code I want in there?
I need to be able to do this as on the app im working on as on the mainpage i have dynamically created buttons that all have functions parameters with different indexes to my server, and i need to be able to fetch the data from server based off the clicked button's data index so on the other page I can view the desired data.
The __init__ of a python class is what is called when an instance is created (using SomeClass()), so you should not try (or even think) to call it again, as it could create serious problems and bugs that are hard to track.
I strongly suggest you to read the documentation about classes in Python, as you cannot ignore that aspect in object oriented programming.
If you need to call something everytime the index is changed, then you should better subclass QStackedWidget and control everything from there.
A good solution is to create a standardized function that will be called everytime the page is presented, and ensure that the stack widget correctly calls it.
class FirstPage(QtWidgets.QWidget):
def __init__(self):
super().__init__(self)
# ...
self.nextButton = QtWidgets.QPushButton('Next')
self.doSomething()
def doSomething(self):
...
class SecondPage(QtWidgets.QWidget):
def __init__(self):
super().__init__(self)
# ...
self.prevButton = QtWidgets.QPushButton('Previous')
self.doSomething()
def doSomething(self):
...
class Stack(QtWidgets.QStackedWidget):
def __init__(self):
super().__init__(self)
self.first = FirstPage()
self.first.nextButton.clicked.connect(self.goNext)
self.addWidget(self.first)
self.second = SecondPage()
self.second.prevButton.clicked.connect(self.goPrev)
self.currentChanged.connect(self.initCurrent)
def goNext(self):
self.setCurrentIndex(1)
def goPrev(self):
self.setCurrentIndex(0)
def initCurrent()
if self.currentWidget():
self.currentWidget().doSomething()
if __name__ == "__main__":
app = qtw.QApplication(sysArgv)
appStack = Stack()
appStack.setFixedSize(300, 300)
appStack.show()
sysExit(app.exec())
Note that adding a QMainWindow to a parent is not a good idea, as Qt main windows are intended to be used as top level windows; also note that using fixed geometries (positions and sizes) is often considered bad practice, and you should use layout managers instead.

Python GUI with PyQt5 Multi Threading

I'm trying to make my app support multi threading in connection with GUI , I'm trying to connect to method inside GUI from threading outside GUI, I have inspired this idea from Simplest way for PyQT Threading and it was marked as working solution , where is my fault
Below is the error.
class Communicate(QtCore.QObject):
myGUI_signal = QtCore.pyqtSignal(str)
def myThread(callbackFunc):
# Setup the signal-slot mechanism.
mySrc = Communicate()
mySrc.myGUI_signal.connect(callbackFunc)
# Endless loop. You typically want the thread
# to run forever.
while(True):
# Do something useful here.
msgForGui = 'This is a message to send to the GUI'
mySrc.myGUI_signal.emit(msgForGui)
FORM_CLASS, _ = loadUiType(os.path.join(os.path.dirname('__file__'), "main.ui"))
class MainApp(QMainWindow, FORM_CLASS): # QMainWindow refere to window type used in ui file
# this is constructor
def __init__(self, parent=None):
super(MainApp, self).__init__(parent)
QMainWindow.__init__(self)
self.setupUi(self)
self.ui()
self.actions()
def ui(self):
self.setFixedSize(848, 663)
def actions(self):
self.pushButton.clicked.connect(self.startTheThread)
def theCallbackFunc(self, msg):
print('the thread has sent this message to the GUI:')
print(msg)
print('---------')
def startTheThread(self):
# Create the new thread. The target function is 'myThread'. The
# function we created in the beginning.
t = threading.Thread(name = 'myThread', target = myThread, args =(self.theCallbackFunc))
t.start()
def main():
app = QApplication(sys.argv)
window = MainApp() # calling class of main window (first window)
window.show() # to show window
app.exec_() # infinit loop to make continous show for window
if __name__ == '__main__':
main()
When instantiating the Python thread, args must be a tuple or list (or other iterable). args =(self.theCallbackFunc) is not a tuple. args =(self.theCallbackFunc,) is a tuple (note the extra comma needed for tuples containing a single value).

Change Button Color in Qt Thread Python

I need to change the color of QPushButton, but an error occurred: "AttributeError: type object 'ProyectoTFM' has no attribute 'ui'".
I don't know hoy to acced to a ui variable from my thread.
This is my code:
import sys
import OpenOPC
import time
import threading
from proyectoQt import *
def actualizarDatosOPC():
while 1:
time.sleep(5)
if(itemsOPC[15])[1]!=0:
#Error on next line
ProyectoTFM.ui.AP08Button.setStyleSheet("background-color: red")
return
class ProyectoTFM(QtGui.QMainWindow):
def __init__(self,parent=None):
QtGui.QMainWindow.__init__(self,parent)
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
self.startTheThread()
print('Init')
def startTheThread(self):
threadQt = threading.Thread(target = actualizarDatosOPC)
threadQt.start()
def clienteOPC():
opc=OpenOPC.client()
opc.connect('Kepware.KEPServerEX.V6')
global itemsOPC
while 1:
itemsOPC = opc.read(opc.list('PLC.PLC.TAGS'))
time.sleep(5)
return
threads = list()
threadOPC = threading.Thread(target=clienteOPC)
threads.append(threadOPC)
threadOPC.start()
time.sleep(5)
if __name__== "__main__":
app=QtGui.QApplication(sys.argv)
myapp = ProyectoTFM()
myapp.show()
sys.exit(app.exec_())
threadOPC.__delete()
Sorry for my English and thanks.
It is not correct to modify the view from a different thread to the main one, a way to solve the problem without using QThread is to create a signal that connects to some slot that changes the color of the button. To be able to emit the signal from the new thread we must pass the object to him through the parameter args.
def actualizarDatosOPC(obj):
while 1:
time.sleep(5)
if(itemsOPC[15])[1]!=0:
#Error on next line
obj.sendChangeColor.emit()
return
class ProyectoTFM(QtGui.QMainWindow):
sendChangeColor = QtCore.pyqtSignal()
def __init__(self,parent=None):
QtGui.QMainWindow.__init__(self,parent)
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
self.startTheThread()
print('Init')
self.sendChangeColor.connect(lambda: self.ui.AP08Button.setStyleSheet("background-color: red"))
def startTheThread(self):
threadQt = threading.Thread(target = actualizarDatosOPC, args=(self,))
threadQt.start()
Even if you got this to work, you can't modify the UI from a thread directly.
A few things:
You never actually pass the UI to the function actualizarDatosOPC
so it doesn't know it exists.
Is there any reason you can't use PyQt's built in threading tools? If you are going to use PyQt it might make sense to buy into the whole framework.
def startTheThread(self):
self.threadQt = QThread()
d = actualizarDatosOPC(self)
d.moveToThread(self.threadQt)
self.threadQt.start()
def actualizarDatosOPC(widget):
.... widget.AP08Button.setStyleSheet("background-color: red")
If you do choose to go this route, I'd take a look at this thread which has a good example:
How to use QThread correctly in pyqt with moveToThread()?
Additionally, while the way you initialize your Window works, this is the more standard way to do it:
class ProyectoTFM(QMainWindow, Ui_MainWindow):
def __init__(self, parent):
# General Init Stuff
super(Login, self).__init__(parent)
self.setupUi(self)
After that, whenever you want to refer to something in the UI all you need to do is refer to self._____. For example, if you have a button named buttonA, self.buttonA would be the appropriate reference.
Edit:
As mentioned in another answer, the proper way to actually change the button color would be to emit a trigger that to the main thread which could then respond by changing the button color.

How to prevent the GUI from freezing using QThread?

I have a GUI which needs to perform work that takes some time and I want to show the progress of this work, similar to the following:
import sys
import time
from PyQt4 import QtGui, QtCore
class MyProgress(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
# start loop with signal
self.button = QtGui.QPushButton('loop', self)
self.connect(self.button, QtCore.SIGNAL('clicked()'), self.loop)
# test button
self.test_button = QtGui.QPushButton('test')
self.connect(self.test_button, QtCore.SIGNAL('clicked()'), self.test)
self.pbar = QtGui.QProgressBar(self)
self.pbar.setMinimum(0)
self.pbar.setMaximum(100)
# layout
vbox = QtGui.QVBoxLayout()
vbox.addWidget(self.test_button)
vbox.addWidget(self.button)
vbox.addWidget(self.pbar)
self.setLayout(vbox)
self.show()
def update(self):
self.pbar.setValue(self.pbar.value() + 1)
def loop(self):
for step in range(100):
self.update()
print step
time.sleep(1)
def test(self):
if self.test_button.text() == 'test':
self.test_button.setText('ok')
else:
self.test_button.setText('test')
app = QtGui.QApplication(sys.argv)
view = MyProgress()
view.loop() # call loop directly to check whether view is displayed
sys.exit(app.exec_())
When I execute the code the loop method is called and it prints out the values as well as updates the progress bar. However the view widget will be blocked during the execution of loop and although this is fine for my application it doesn't look nice with Ubuntu. So I decided to move the work to a separate thread like this:
import sys
import time
from PyQt4 import QtGui, QtCore
class Worker(QtCore.QObject):
def __init__(self, parent=None):
QtCore.QObject.__init__(self, parent)
def loop(self):
for step in range(10):
print step
time.sleep(1)
class MyProgress(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
# test button
self.test_button = QtGui.QPushButton('test')
self.connect(self.test_button, QtCore.SIGNAL('clicked()'), self.test)
self.pbar = QtGui.QProgressBar(self)
self.pbar.setMinimum(0)
self.pbar.setMaximum(100)
# layout
vbox = QtGui.QVBoxLayout()
vbox.addWidget(self.test_button)
vbox.addWidget(self.pbar)
self.setLayout(vbox)
self.show()
def test(self):
if self.test_button.text() == 'test':
self.test_button.setText('ok')
else:
self.test_button.setText('test')
app = QtGui.QApplication(sys.argv)
view = MyProgress()
work = Worker()
thread = QtCore.QThread()
work.moveToThread(thread)
# app.connect(thread, QtCore.SIGNAL('started()'), work.loop) # alternative
thread.start()
work.loop() # not called if thread started() connected to loop
sys.exit(app.exec_())
When I run this version of the script the loop starts running (the steps are displayed in the terminal) but the view widget is not shown. This is the first thing I can't quite follow. Because the only difference from the previous version here is that the loop runs in a different object however the view widget is created before and therefore should be shown (as it was the case for the previous script).
However when I connected the signal started() from thread to the loop function of worker then loop is never executed although I start the thread (in this case I didn't call loop on worker). On the other hand view is shown which makes me think that it depends whether app.exec_() is called or not. However in the 1st version of the script where loop was called on view it showed the widget although it couldn't reach app.exec_().
Does anyone know what happens here and can explain how to execute loop (in a separate thread) without freezing view?
EDIT: If I add a thread.finished.connect(app.exit) the application exits immediately without executing loop. I checked out the 2nd version of this answer which is basically the same what I do. But in both cases it finishes the job immediately without executing the desired method and I can't really spot why.
The example doesn't work because communication between the worker thread and the GUI thread is all one way.
Cross-thread commnunication is usually done with signals, because it is an easy way to ensure that everything is done asynchronously and in a thread-safe manner. Qt does this by wrapping the signals as events, which means that an event-loop must be running for everything to work properly.
To fix your example, use the thread's started signal to tell the worker to start working, and then periodically emit a custom signal from the worker to tell the GUI to update the progress bar:
class Worker(QtCore.QObject):
valueChanged = QtCore.pyqtSignal(int)
def loop(self):
for step in range(0, 10):
print step
time.sleep(1)
self.valueChanged.emit((step + 1) * 10)
...
thread = QtCore.QThread()
work.moveToThread(thread)
thread.started.connect(work.loop)
work.valueChanged.connect(view.pbar.setValue)
thread.start()
sys.exit(app.exec_())

Open a second window in PyQt

I'm trying to use pyqt to show a custom QDialog window when a button on a QMainWindow is clicked. I keep getting the following error:
$ python main.py
DEBUG: Launch edit window
Traceback (most recent call last):
File "/home/james/Dropbox/Database/qt/ui_med.py", line 23, in launchEditWindow
dialog = Ui_Dialog(c)
File "/home/james/Dropbox/Database/qt/ui_edit.py", line 15, in __init__
QtGui.QDialog.__init__(self)
TypeError: descriptor '__init__' requires a 'sip.simplewrapper' object but received a 'Ui_Dialog'
I've gone over several online tutorials, but most of them stop just short of showing how to use a non built-in dialog window. I generated the code for both the main window and the dialog using pyuic4. What I think should be the relevant code is below. What am I missing here?
class Ui_Dialog(object):
def __init__(self, dbConnection):
QtGui.QDialog.__init__(self)
global c
c = dbConnection
class Ui_MainWindow(object):
def __init__(self, dbConnection):
global c
c = dbConnection
def launchEditWindow(self):
print "DEBUG: Launch edit window"
dialog = QtGui.QDialog()
dialogui = Ui_Dialog(c)
dialogui = setupUi(dialog)
dialogui.show()
class StartQT4(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
conn = sqlite3.connect('meds.sqlite')
c = conn.cursor()
self.ui = Ui_MainWindow(c)
self.ui.setupUi(self)
def main():
app = QtGui.QApplication(sys.argv)
program = StartQT4()
program.show()
sys.exit(app.exec_())
if __name__ == '__main__':
main()
Bonus question: since it looks like you can't pass arguments in pyqt function callbacks, is setting something which would otherwise be passed as an argument (the poorly named "c") to be global the best way to get information into those functions?
I've done like this in the past, and i can tell it works.
assuming your button is called "Button"
class Main(QtGui.QMainWindow):
''' some stuff '''
def on_Button_clicked(self, checked=None):
if checked==None: return
dialog = QDialog()
dialog.ui = Ui_MyDialog()
dialog.ui.setupUi(dialog)
dialog.setAttribute(QtCore.Qt.WA_DeleteOnClose)
dialog.exec_()
This works for my application, and I believe it should work with yours as well. hope it'll help, it should be pretty straight forward to do the few changes needed to apply it to your case.
have a good day everybody.
Ui_Dialog should inherent from QtGui.QDialog, not object.
class Ui_Dialog(QtGui.QDialog):
def __init__(self, dbConnection):
QtGui.QDialog.__init__(self)
global c
c = dbConnection
class StartQT4(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
Why QtGui.QWidget.__init___ ???
Use insted:
class StartQT4(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QMainWindow.__init__(self, parent)
You must call __init__ methon from base class (name in parenthesis '()')
QDialog have two useful routins:
exec_()
show()
First wait for closing dialog and then you can access any field form dialog. Second show dialog but don't wait, so to work properly you must set some slot/signals connections to respond for dialog actions.
eg. for exec_():
class Dialog(QDialog):
def __init__(self, parent):
QDialog.__init__(parent)
line_edit = QLineEdit()
...
dialog = Dialog()
if dialog.exec_(): # here dialog will be shown and main script will wait for its closing (with no errors)
data = dialog.line_edit.text()
Small tip: can you change your ui classes into widgets (with layouts). And perhaps problem is that your __init__ should be __init__(self, parent=None, dbConnection)
Because when you create new widget in existing one PyQt may try to set it as children of existing one. (So change all init to have additional parent param (must be on second position)).

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