PyQt5 - Detecting When Other Window Closes - python

I’m using PyQt5 and I need to have my main window detect when a different window closes. I read here Emit a signal from another class to main class that creating a signal class to serve as an intermediary should work. However I haven’t gotten my example to work.
In my example, clicking the button opens a QWidget window. When the QWidget is closed, the main window is supposed to change from a blue background to a red background. However, the main window remains blue using the script below.
What am I doing wrong?
from PyQt5.QtWidgets import QPushButton
from PyQt5.QtCore import QObject, pyqtSignal
import os, sys
class MySignal(QObject):
signal = pyqtSignal()
class MyMainWindow(QMainWindow):
def __init__(self):
super().__init__()
#Make mainwindow
self.setGeometry(100,100,300,200)
self.setStyleSheet('background-color: blue')
# Make widget and button objects and set connection
self.widget = MyWidget()
self.btn = QPushButton(self)
self.btn.setText('Click')
self.btn.move(175, 150)
self.btn.setStyleSheet('background-color: white')
self.btn.clicked.connect(self.widget.showWidget)
# Make signal object and set connection
self.mySignal = MySignal()
self.mySignal.signal.connect(self.changeToRed)
# Let's start
self.show()
def changeToRed(self):
self.setStyleSheet('background-color: red')
def closeEvent(self, event):
os._exit(0)
class MyWidget(QWidget):
def __init__(self):
super().__init__()
self.setGeometry(500, 100, 200, 200)
self.sig = MySignal()
def showWidget(self):
self.show()
def closeEvent(self, event):
self.sig.signal.emit()
self.close()
if __name__ == '__main__':
app = QApplication(sys.argv)
ex = MyMainWindow()
app.exec()```

The reason your code fails is that the connected signal object in MyMainWindow is not the same as the one you create in MyWidget, so the signal is never emitted. Here is a modified solution using signals in the normal way:
from PyQt5.QtWidgets import QPushButton, QMainWindow, QWidget, QApplication
from PyQt5.QtCore import QObject, pyqtSignal
import os, sys
class MyMainWindow(QMainWindow):
def __init__(self):
super().__init__()
#Make mainwindow
self.setGeometry(100,100,300,200)
self.setStyleSheet('background-color: blue')
# Make widget and button objects and set connection
self.widget = MyWidget()
self.btn = QPushButton(self)
self.btn.setText('Click')
self.btn.move(175, 150)
self.btn.setStyleSheet('background-color: white')
self.btn.clicked.connect(self.widget.showWidget)
self.widget.sig.connect(self.changeToRed)
# Let's start
self.show()
def changeToRed(self):
self.setStyleSheet('background-color: red')
class MyWidget(QWidget):
sig = pyqtSignal()
def __init__(self):
super().__init__()
self.setGeometry(500, 100, 200, 200)
def showWidget(self):
self.show()
def closeEvent(self, event):
self.sig.emit()
if __name__ == '__main__':
app = QApplication(sys.argv)
ex = MyMainWindow()
app.exec()

All you need is to add the connection between the close event and the function that turn your main screen red:
self.widget.closeEvent = self.changeToRed
this line should be in your main Window class
change your changeToRed function so it will except the event too:
def changeToRed(self, e):

Related

Pyqt5 - how to go back to hided Main Window from Secondary Window?

If I click Back from the second window, the program will just exit. How do I go back to mainwindow in this case? I assume I will need some more code in that clickMethodBack function.
import os
import PyQt5
from PyQt5 import QtCore, QtWidgets
from PyQt5.QtWidgets import QMainWindow, QWidget, QLabel, QPushButton
import time
from PyQt5.QtCore import QSize
class GUI_Window():
def __init__(self):
self.main_window()
return
def main_window(self):
app = PyQt5.QtWidgets.QApplication(sys.argv)
self.MainWindow = MainWindow_()
self.MainWindow.show()
app.exec_()
return
class MainWindow_(QMainWindow):
def __init__(self):
QMainWindow.__init__(self)
self.TestAButton = QPushButton("TestA", self)
self.TestAButton.clicked.connect(self.TestA_clickMethod)
self.TestAButton.move(20, 0)
self.CloseButton = QPushButton("Close", self)
self.CloseButton.clicked.connect(self.Close_clickMethod)
self.CloseButton.move(20, 40)
self.TestingA = TestA_MainWindow()
def TestA_clickMethod(self):
self.TestAButton.setEnabled(False)
time.sleep(0.2)
self.TestingA.show()
self.hide()
try:
if self.TestingA.back == True:
self.show()
except:
None
def Close_clickMethod(self):
self.Test_Choice = 'Exit'
self.close()
class TestA_MainWindow(QWidget):
def __init__(self):
super().__init__()
self.setMinimumSize(QSize(980,700))
self.setWindowTitle("TestA")
self.Back_Button = False
self.closeButton = QPushButton("Close", self)
self.closeButton.clicked.connect(self.clickMethodClose)
self.returnButton = QPushButton("Back", self)
self.returnButton.clicked.connect(self.clickMethodBack)
self.returnButton.move(0,30)
def clickMethodClose(self):
self.Back_Button = False
self.close()
def clickMethodBack(self):
self.returnButton.setEnabled(False)
time.sleep(0.5)
self.back = True
self.close()
# Run if Script
if __name__ == "__main__":
main = GUI_Window() # Initialize GUI
Your code has two very important issues.
you're using a blocking function, time.sleep; Qt, as almost any UI toolkit, is event driven, which means that it has to be able to constantly receive and handle events (coming from the system or after user interaction): when something blocks the event queue, it completely freezes the whole program until that block releases control;
you're checking for the variable too soon: even assuming the sleep would work, you cannot know if the window is closed after that sleep timer has ended;
The solution is to use signals and slots. Since you need to know when the second window has been closed using the "back" button, create a custom signal for the second window that will be emitted whenever the function that is called by the button is closed.
from PyQt5 import QtCore, QtWidgets
class MainWindow(QtWidgets.QMainWindow):
def __init__(self):
super().__init__()
central = QtWidgets.QWidget()
layout = QtWidgets.QHBoxLayout(central)
self.testButton = QtWidgets.QPushButton('Test A')
self.closeButton = QtWidgets.QPushButton('Close')
layout.addWidget(self.testButton)
layout.addWidget(self.closeButton)
self.setCentralWidget(central)
self.testButton.clicked.connect(self.launchWindow)
self.closeButton.clicked.connect(self.close)
def launchWindow(self):
self.test = TestA_MainWindow()
self.test.backSignal.connect(self.show)
self.hide()
self.test.show()
class TestA_MainWindow(QtWidgets.QWidget):
backSignal = QtCore.pyqtSignal()
def __init__(self):
super().__init__()
layout = QtWidgets.QHBoxLayout(self)
self.closeButton = QtWidgets.QPushButton('Close')
self.backButton = QtWidgets.QPushButton('Back')
layout.addWidget(self.closeButton)
layout.addWidget(self.backButton)
self.closeButton.clicked.connect(self.close)
self.backButton.clicked.connect(self.goBack)
def goBack(self):
self.close()
self.backSignal.emit()
def GUI_Window():
import sys
app = QtWidgets.QApplication(sys.argv)
mainWindow = MainWindow()
mainWindow.show()
sys.exit(app.exec_())
if __name__ == '__main__':
GUI_Window()
Notes:
I removed the GUI_Window class and made a function, as using a class for that is not really useful;
you should always prefer layout managers instead of setting manual geometries;
widgets should not be added to a QMainWindow as direct children, and a central widget should always be used (see the creation and use of central in the example); read more about it in the documentation;
only classes and constants should be capitalized, while variables, attributes and functions should always have names starting with a lowercase letter;

Opening new window while out of focus using "keyboard"

I am trying to track my keypresses, using the module "keyboard" while a PySide2 Widget is not in focus, which works fine. However when I try to create a new Widget using a "keyboard" shortcut the program crashes. Opening a window on a button press works fine. I can also call non UI functions using "keyboard" eg. the print function without any problem.
Do you know a way to fix this and open a new window using "keyboard" or any other method, while a PySide2 window is not in focus. In this example I want to open a new window on "CTRL+D". The Problem exists both for PySide2 and PyQt5.
This is my shortened code:
import sys
import json
import os
import keyboard
from PySide2.QtWidgets import QApplication, QWidget, QGridLayout, QKeySequenceEdit, QLabel, QPushButton, QShortcut
from PySide2.QtCore import Qt, QObject, Signal, Slot # Qt.Key_W beispielsweise
#from PyQt5.QtWidgets import QApplication, QWidget, QGridLayout, QKeySequenceEdit, QLabel, QPushButton, QShortcut
#from PyQt5.QtCore import Qt, QObject, pyqtSignal as Signal, pyqtSlot as Slot # Qt.Key_W beispielsweise
class ConfigWindow(QWidget):
def __init__(self):
super().__init__()
self.initUi()
self.init_shortcuts()
self.show()
def initUi(self):
self.setGeometry(300,300, 400, 250)
self.setWindowTitle("Settings")
grid = QGridLayout()
self.setLayout(grid)
self.keyseq = QKeySequenceEdit("CTRL+D")
grid.addWidget(self.keyseq, 0, 0)
s_button = QPushButton("Safe")
grid.addWidget(s_button, 1, 0)
cl_button = QPushButton("Close")
grid.addWidget(cl_button, 1, 1)
cl_button.clicked.connect(self.close)
open_button = QPushButton("openw")
grid.addWidget(open_button, 2, 0)
open_button.clicked.connect(self.call_item_parser)
def keyPressEvent(self, event): #event:PySide2.QtGui.QKeyEvent
if event.key() == Qt.Key_Escape:
self.close()
# shortcuts are listened to, while program is running
def init_shortcuts(self):
str_value = self.keyseq.keySequence().toString()
print("Binding _price_keyseq to {}".format(str_value))
keyboard.add_hotkey(str_value, self.call_item_parser)
# keyboard.add_hotkey(str_value, print, args=("this works")) # this would work
def call_item_parser(self):
self.h_w = ParseWindow()
self.h_w.setWindowTitle("New Window")
self.h_w.setGeometry(100, 100, 100, 100)
self.h_w.show()
class ParseWindow(QWidget):
def __init__(self):
super().__init__()
app = QApplication(sys.argv)
w = ConfigWindow()
sys.exit(app.exec_())
The problem is caused because the callback registered in keyboard is executed in a secondary thread as can be verified by modifying the following part of the code and printing threading.current_thread(). In Qt it is forbidden to create any widget in another thread since they are not thread-safe.
def call_item_parser(self):
print(threading.current_thread())
self.h_w = ParseWindow()
self.h_w.setWindowTitle("New Window")
self.h_w.setGeometry(100, 100, 100, 100)
self.h_w.show()
print(threading.current_thread())
app = QApplication(sys.argv)
w = ConfigWindow()
sys.exit(app.exec_())
Output:
<_MainThread(MainThread, started 140144979916608)>
Binding _price_keyseq to ctrl+a
<Thread(Thread-10, started daemon 140144220817152)>
One possible solution is to use a signal to send the information to the main thread, and invoke the callback in the main thread.
import sys
from functools import partial
import platform
import threading
import keyboard
from PySide2.QtCore import Qt, QObject, Signal, Slot
from PySide2.QtGui import QKeySequence
from PySide2.QtWidgets import (
QApplication,
QWidget,
QGridLayout,
QKeySequenceEdit,
QPushButton,
)
class KeyBoardManager(QObject):
activated = Signal(str)
def __init__(self, parent=None):
super().__init__(parent)
self._callbacks = dict()
self.activated.connect(self._handle_activated)
#property
def callbacks(self):
return self._callbacks
def register(self, shortcut, callback, *, args=(), kwargs=None):
self.callbacks[shortcut] = (callback, args, kwargs or {})
keyboard.add_hotkey(shortcut, partial(self.activated.emit, shortcut))
#Slot(str)
def _handle_activated(self, shortcut):
values = self.callbacks.get(shortcut)
if values is not None:
callback, args, kwargs = self._callbacks[shortcut]
callback(*args, **kwargs)
class ConfigWindow(QWidget):
def __init__(self):
super().__init__()
self.initUi()
self.init_shortcuts()
self.show()
def initUi(self):
self.setGeometry(300, 300, 400, 250)
self.setWindowTitle("Settings")
grid = QGridLayout(self)
self.keyseq = QKeySequenceEdit("CTRL+A")
grid.addWidget(self.keyseq, 0, 0)
s_button = QPushButton("Safe")
grid.addWidget(s_button, 1, 0)
cl_button = QPushButton("Close")
grid.addWidget(cl_button, 1, 1)
cl_button.clicked.connect(self.close)
open_button = QPushButton("openw")
grid.addWidget(open_button, 2, 0)
open_button.clicked.connect(self.call_item_parser)
def keyPressEvent(self, event): # event:PySide2.QtGui.QKeyEvent
if event.key() == Qt.Key_Escape:
self.close()
# shortcuts are listened to, while program is running
def init_shortcuts(self):
self.keyboard_manager = KeyBoardManager()
str_value = self.keyseq.keySequence().toString()
if platform.system() == "Linux":
str_value = str_value.lower()
print("Binding _price_keyseq to {}".format(str_value))
self.keyboard_manager.register(str_value, self.call_item_parser)
def call_item_parser(self):
print(threading.current_thread())
self.h_w = ParseWindow()
self.h_w.setWindowTitle("New Window")
self.h_w.setGeometry(100, 100, 100, 100)
self.h_w.show()
class ParseWindow(QWidget):
pass
def main():
print(threading.current_thread())
app = QApplication(sys.argv)
w = ConfigWindow()
sys.exit(app.exec_())
if __name__ == "__main__":
main()
Output:
<_MainThread(MainThread, started 140037641176896)>
Binding _price_keyseq to ctrl+a
<_MainThread(MainThread, started 140037641176896)>

How to print test output to a label in pyqt gui

what I am trying to do with a python script: Use a pytest test method to print a text line to a label in the pyqt GUI.
When running the main python file, the GUI starts and a click on the "test" button runs the test without blocking the GUI (see full code example below). But I have no clue how to proceed now.
Code:
import sys
import pytest
from PyQt5 import QtWidgets, QtCore
from PyQt5.QtWidgets import QApplication, QVBoxLayout, QWidget
class Window(QtWidgets.QMainWindow):
signal_start_background_job = QtCore.pyqtSignal()
def __init__(self):
super(Window, self).__init__()
layout = QVBoxLayout()
self.button = QtWidgets.QPushButton("test", self)
self.label = QtWidgets.QLabel("console output")
layout.addWidget(self.button)
layout.addWidget(self.label)
widget = QWidget()
widget.setLayout(layout)
self.setCentralWidget(widget)
self.worker = WorkerObject()
self.thread = QtCore.QThread()
self.worker.moveToThread(self.thread)
self.signal_start_background_job.connect(self.worker.background_job)
self.button.clicked.connect(self.start_background_job)
def start_background_job(self):
self.thread.start()
self.signal_start_background_job.emit()
class WorkerObject(QtCore.QObject):
#QtCore.pyqtSlot()
def background_job(self):
pytest.main(["-s", "-k test_something"])
if __name__ == "__main__":
app = QApplication(sys.argv)
window = Window()
window.show()
sys.exit(app.exec_())
def test_something():
print("unit test some stuff")
assert 0 == 0
Instead of using pytest directly you could use QProcess to launch it and then capture the output:
import os
import sys
from PyQt5 import QtWidgets, QtCore
from PyQt5.QtWidgets import QApplication, QVBoxLayout, QWidget
CURRENT_DIR = os.path.dirname(os.path.realpath(__file__))
class Window(QtWidgets.QMainWindow):
def __init__(self):
super(Window, self).__init__()
self.button = QtWidgets.QPushButton("test", self)
self.label = QtWidgets.QLabel("console output")
self.textedit = QtWidgets.QTextEdit(readOnly=True)
widget = QWidget()
layout = QVBoxLayout(widget)
layout.addWidget(self.button)
layout.addWidget(self.label)
layout.addWidget(self.textedit)
self.setCentralWidget(widget)
self.process = QtCore.QProcess()
self.process.setProgram(sys.executable)
self.process.readyReadStandardError.connect(self.on_readyReadStandardError)
self.process.readyReadStandardOutput.connect(self.on_readyReadStandardOutput)
self.button.clicked.connect(self.on_clicked)
#QtCore.pyqtSlot()
def on_clicked(self):
self.process.setWorkingDirectory(CURRENT_DIR)
self.process.setArguments(["-m", "pytest", "-s", "-k", "test_something"])
self.process.start()
#QtCore.pyqtSlot()
def on_readyReadStandardError(self):
err = self.process.readAllStandardError().data().decode()
self.textedit.append(err)
#QtCore.pyqtSlot()
def on_readyReadStandardOutput(self):
out = self.process.readAllStandardOutput().data().decode()
self.textedit.append(out)
if __name__ == "__main__":
app = QApplication(sys.argv)
window = Window()
window.show()
sys.exit(app.exec_())
I'm thinking you need to check out sys.stdout, and route that to an io object that you can route to the label in your widget. Then I would set a timer and every 0.1 seconds or so set the text of your label to that object. Alternatively you can implement a widget that grabs the stdout text in qt, example here: https://stackoverflow.com/a/1220002/6615517 I haven't tried it but the other answers to that question should help. You'll probably want to clear the label on each test to prevent it from getting too long.

How to get the name of a QMenu Item when clicked?

I have a few actions in a QMenu that I'm trying to connect to a single method; there are additional actions that are unrelated.
import sys
from PyQt5.QtWidgets import *
class Window(QWidget):
def __init__(self):
super(Window, self).__init__()
layout = QGridLayout(self)
self.menu_bar = QMenuBar()
self.menu = QMenu('MENU', self)
self.menu_action = QAction('Option #1', self)
self.menu_action.setData('option1')
self.menu_action.triggered.connect(self.actionClicked)
self.menu.addAction(self.menu_action)
self.menu_bar.addMenu(self.menu)
layout.addWidget(self.menu_bar)
def actionClicked(self, action):
print(action)
try:
print(action.text())
except AttributeError as e:
print(e)
try:
print(action.data())
except AttributeError as e:
print(e)
if __name__ == '__main__':
app = QApplication(sys.argv)
window = Window()
window.setGeometry(600, 100, 300, 100)
window.show()
sys.exit(app.exec_())
I was wondering how I can tell which of the actions was clicked when the method is called. Currently I'm trying to use self.menu_action.setData() to give the action a cleaner name for the receiving method, but that does not seem to be working properly.
A possible solution is to use the sender() method that returns the QObject that emits the signal, in this case the QAction:
import sys
from PyQt5 import QtCore, QtWidgets
class Window(QtWidgets.QWidget):
def __init__(self):
super(Window, self).__init__()
layout = QtWidgets.QGridLayout(self)
menubar = QtWidgets.QMenuBar()
filemenu = menubar.addMenu('MENU')
filemenu.addAction('Option #1', self.actionClicked)
filemenu.addAction('Option #2', self.actionClicked)
layout.addWidget(menubar)
#QtCore.pyqtSlot()
def actionClicked(self):
action = self.sender()
print('Action: ', action.text())
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = Window()
window.setGeometry(600, 100, 300, 100)
window.show()
sys.exit(app.exec_())
If you are going to connect all the QActions of the QMenu then you can use the triggered signal of the QMenu, this sends the QAction that was pressed.
import sys
from PyQt5 import QtCore, QtWidgets
class Window(QtWidgets.QWidget):
def __init__(self):
super(Window, self).__init__()
layout = QtWidgets.QGridLayout(self)
menubar = QtWidgets.QMenuBar()
filemenu = menubar.addMenu('MENU')
filemenu.triggered.connect(self.actionClicked)
filemenu.addAction('Option #1')
filemenu.addAction('Option #2')
layout.addWidget(menubar)
#QtCore.pyqtSlot(QtWidgets.QAction)
def actionClicked(self, action):
print('Action: ', action.text())
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = Window()
window.setGeometry(600, 100, 300, 100)
window.show()
sys.exit(app.exec_())
Update:
The QAction triggered signal is different from the QMenu triggered signal, in the case of the first one it sends a Boolean value that indicates if the QAction is checked (by default a QAction is not checkable, if you want to activate it you must use setCheckable(True)) and in the second case it sends the QAction that was pressed that belongs to the QMenu. That's why you always get False, if you do not want to have that problem you have to use my first method using sender(). On the other hand in Qt very rarely you will have to use try-except, in reality you should never use it in the world of Qt since that has an additional cost that Qt does not want, and on the other hand it hides the cause of the errors , as a recommendation use try-except when there is no other solution, and in the case of Qt there will almost always be another option.
import sys
from PyQt5.QtWidgets import *
class Window(QWidget):
def __init__(self):
super(Window, self).__init__()
layout = QGridLayout(self)
self.menu_bar = QMenuBar()
self.menu = QMenu('MENU', self)
self.menu_action = QAction('Option #1', self)
self.menu_action.setData('option1')
self.menu_action.triggered.connect(self.actionClicked)
self.menu.addAction(self.menu_action)
self.menu_bar.addMenu(self.menu)
layout.addWidget(self.menu_bar)
def actionClicked(self, checked):
action = self.sender()
print(action.text())
print(action.data())
if __name__ == '__main__':
app = QApplication(sys.argv)
window = Window()
window.setGeometry(600, 100, 300, 100)
window.show()
sys.exit(app.exec_())

Draw QFrame around QPushButton PyQt5

I am unable to get a QFrame to completely surround a QPushButton Like a Border. It only frames the top and left side of the button. I was wondering what I'm doing wrong with the frame.
import sys
from PyQt5.QtGui import *
from PyQt5.QtWidgets import *
from PyQt5.QtCore import *
class main(QWidget):
def __init__(self):
super().__init__()
layout1 = QVBoxLayout()
btn1 = QPushButton("Test")
frame = QFrame(btn1)
frame.setGeometry(btn1.geometry())
frame.setFrameShape(QFrame.Box)
frame.setFrameShadow(QFrame.Plain)
frame.setLineWidth(4)
layout1.addWidget(btn1)
self.setLayout(layout1)
if __name__ == '__main__':
app = QApplication(sys.argv)
window = main()
window.show()
sys.exit(app.exec_())
The problem is caused because the QFrame does not change the size, instead QPushButton does. In my solution I have resized every time the size is changed in the QPushButton
class FrameButton(QPushButton):
def __init__(self, *args, **kwargs):
QPushButton.__init__(self, *args, **kwargs)
self.frame = QFrame(self)
self.frame.setFrameShape(QFrame.Box)
self.frame.setFrameShadow(QFrame.Plain)
self.frame.setLineWidth(4)
def resizeEvent(self, event):
self.frame.resize(self.size())
QWidget.resizeEvent(self, event)
class main(QWidget):
def __init__(self):
super().__init__()
layout = QVBoxLayout(self)
layout.addWidget(FrameButton("Test"))

Categories

Resources