Okay... This has been bugging me for hours. I have a qtmainwindow with a menubar. I've managed to connect an action in tje menubar to an independent Qwidget. But as soon as the Qwidget appears it disappears. I'm using the latest version of pyqt.
Here's the code:
Import sys
from PyQt4 import QtGui, QtCore
Class Main(QtGui.QtMainWindow) :
def __init__(self) :
QtGui.QtMainWindow.__init__(self)
self.setGeometry(300,300,240,320)
self.show()
menubar = self. menuBar()
filemenu = menubar. addMenu('&File')
new = QtGui.QAction(QtGui.QIcon('new.png'), 'New', self)
new.triggered.connect(self.pop)
filemenu.addAction(new)
def pop(self) :
pop = Pop()
class Pop(QtGui.QWidget) :
def __init__(self) :
QtGui.QWidget.__init__(self)
self.setGeometry(300,300,240,320>
self.setWindowTitle('Pop up')
self.show()
Update the pop(self) method as:
def pop(self):
self.window = Pop()
you need to store object of newly created window in a member variable, other wise as soon as the method finishes with the execution, local variables will be destroyed by the Python Garbage Collector.
if you implement this code, you will see the window gets created and disappears immediately.
import sys
from PyQt5 import QtGui, QtWidgets,QtCore
from PyQt5.QtWidgets import QApplication, QMainWindow
app = QtWidgets.QApplication(sys.argv)
window = QtWidgets.QWidget()
window.setGeometry(50,50,500,500)
window.setWindowTitle("GUI window")
window.show()
To solve that problem write "sys.exit(app.exec_())" after window.show() and the window will stay on the screen.
Related
This question already has answers here:
Variable scopes in Python classes
(4 answers)
Closed 1 year ago.
I was making a program with Python 3.7 & PyQt5 5.15.4, and I have a sudden crash issue.
This is a simplified version which shows this phenomena.
import sys
from PyQt5.QtWidgets import QWidget, QPushButton, QDialog, QVBoxLayout, QApplication
class WindowA(QWidget):
def __init__(self):
super().__init__()
self.make_layout()
def make_layout(self):
vbox = QVBoxLayout()
self.setLayout(vbox)
class LoginDialog(QDialog):
def __init__(self):
super().__init__()
self.btn_login = QPushButton('Login')
self.btn_login.clicked.connect(self.btn_login_clicked)
vbox = QVBoxLayout()
vbox.addWidget(self.btn_login)
self.setLayout(vbox)
def btn_login_clicked(self):
self.accept()
main_window = WindowA()
main_window.show()
if __name__ == '__main__':
app = QApplication(sys.argv)
login_dialog = LoginDialog()
login_dialog.show()
sys.exit(app.exec_())
This is what I intended:
The program starts with LoginDialog, which contains one login button.
If I push the button, WindowA appears.
However, if I run the code, when I push the login button in LoginDialog, WindowA flashes and the program terminates with no error message.
There is no crash, both windows close so the program is finished running.
LoginDialog closes because QDialog.accept() causes the dialog to close.
WindowA gets garbage collected as soon as the function returns. Use an instance variable so it stays in scope.
def btn_login_clicked(self):
self.main_window = WindowA()
self.main_window.show()
I have written python pyqt code to open a new window with a label from another window on a button click. The issue is ,new window exits as soon as it opens.How do i fix this.
The code I wrote is
import sys
from PyQt4 import QtGui,QtCore
class Window(QtGui.QWidget):
def __init__(self):
super(Window,self).__init__()
self.btn=QtGui.QPushButton('button',self)
self.btn.clicked.connect(display)
self.show()
class display(QtGui.QWidget):
def __init__(self):
super(display,self).__init__()
self.lab=QtGui.QLabel()
self.lab.setText("hi")
self.show()
def main():
App=QtGui.QApplication(sys.argv)
Gui=Window()
sys.exit(App.exec_())
main()
You need to keep a reference to the QWidget object for your second window. Currently when you click the button, the clicked signal is fired and it calls disp1. That creates the widget, but then it is immediately garbage collected.
Instead do this to keep a reference:
import sys
from PyQt4 import QtGui,QtCore
class Window(QtGui.QWidget):
def __init__(self):
super(Window,self).__init__()
self.btn=QtGui.QPushButton('button',self)
self.btn.clicked.connect(self.open_new_window)
self.show()
def open_new_window(self):
# creates the window and saves a reference to it in self.second_window
self.second_window = disp1()
class displ(QtGui.QWidget):
def __init__(self):
super(displ,self).__init__()
self.lab=QtGui.QLabel()
self.lab.setText("hello")
self.show()
def main():
App=QtGui.QApplication(sys.argv)
Gui=Window()
sys.exit(App.exec_())
main()
When passing a function as parameter, maybe it's better not to include the parentheses? Try
sys.exit(App.exec_)
Instead of
sys.exit(App.exec_())
Disclaimer: New to both python and qt designer
QT Designer 4.8.7
Python 3.4
PyCharm 5.0.3
Question - How do I add controls to the main form or a scroll area widget on the main form (created in QT Designer) programmatically?
I have created a MainWindow in qt designer and added my widgets. The following is the entire test program in PyCharm:
import sys
from PyQt4 import QtGui, QtCore, uic
from PyQt4.QtGui import *
from PyQt4.QtCore import *
qtCreatorFile = "programLauncher.ui"
Ui_MainWindow, QtBaseClass = uic.loadUiType(qtCreatorFile)
class MyApp(QtGui.QMainWindow, Ui_MainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
Ui_MainWindow.__init__(self)
self.setupUi(self)
# Cannot resize or maximize
self.setFixedSize(1045, 770)
# Add button test
self.dateLabel = QtGui.QLabel("Test")
self.pushButton = QtGui.QPushButton('Test button')
# self.scrollArea_programs.addWidget()
grid = QtGui.QGridLayout()
# self.scrollArea_programs.addWidget(self.pushButton)
grid.addWidget(self.dateLabel,0,0)
grid.addWidget(self.pushButton,0,1)
self.setLayout(grid)
self.pushButton_exit.clicked.connect(self.closeEvent)
def closeEvent(self):
QtGui.QApplication.quit()
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
window = MyApp()
window.show()
sys.exit(app.exec_())
As you can see I tried to add controls to a grid but nothing shows up when the program runs - I have also tried to add a control to the scroll area. Can someone help me to just add 1 control to the scroll area at run time - so then I can know the proper way to do it or "a" proper way to do this.
Thanks in advance
Without having access to your programLauncher.ui and making minimal changes to your posted code, you can add your UI elements to the window like so:
from PyQt4 import QtGui
import sys
class MyApp(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
# Cannot resize or maximize
self.setFixedSize(1045, 770)
widget = QtGui.QWidget(self)
self.setCentralWidget(widget)
# Add button test
self.dateLabel = QtGui.QLabel("Test")
self.pushButton = QtGui.QPushButton('Test button')
grid = QtGui.QGridLayout()
grid.addWidget(self.dateLabel, 0, 0)
grid.addWidget(self.pushButton, 0, 1)
widget.setLayout(grid)
self.pushButton.clicked.connect(self.closeEvent)
def closeEvent(self, event):
QtGui.QApplication.quit()
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
window = MyApp()
window.show()
sys.exit(app.exec_())
This will get the controls on the screen, although the layout leaves a lot to be desired. You may have to make modifications to this based on what's in your .ui file. One thing that you'll want to note in this example is that the QMainWindow needs a central widget (widget in the example above). You then set the layout on that widget.
You can use the designer to create your .ui file
The you can load it in your .py using something like:
from PyQt4 import QtCore, QtGui, uic
class my_win(QtGui.QMainWindow):
def __init__(self):
self.ui = uic.loadUi('my_ui.ui',self)
then you can access all your widgets with something like
self.ui.actionQuit.triggered.connect(QtGui.qApp.quit)
or
self.ui.my_button.triggered.connect(self.do_someting)
Thanks to JCVanHamme (the programLauncher.ui hint) and also outside help I now learned most of what I need to know to access MainWindow at run time. So for anyone interested in this beginner tip:
Take a blank form in QT Designer
Add a control
Run pyuic4 batch file
Take a look at the generated .py file to learn EVERYTHING about how to add controls.
Don't let the power go to your head - cheers
In the below mentioned example when I click on 'Help' submenu under 'View' menu multiple times its creating multiple windows. Can anyone tell me how to resolve this issue?
import sys
from PySide import Qt Gui
from PySide.QtCore import Qt
class Window(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.menu_bar()
def menu_bar(self):
helpAction = QtGui.QAction('&Help', self)
helpAction.setShortcut('Ctrl+H')
helpAction.triggered.connect(self.add_helpWindow)
menu = self.menuBar().addMenu('View')
menu.addAction(helpAction)
def add_helpWindow(self):
window = QtGui.QMainWindow(self)
window.setWindowTitle('New Window')
window.show()
if __name__ == '__main__':
import sys
app=QtGui.QApplication.instance()
if not app:
app = QtGui.QApplication(sys.argv)
window = Window()
window.resize(300, 300)
window.show()
sys.exit(app.exec_())
You help window is just a QMainWindow, which is not modal and there are no restrictions on the number that can exist. Hence why if you select the help option multiple times, you get multiple windows.
You likely want to use a QMessageBox which has its modal property set. While there is nothing forcing only one dialog to exist at a time, being modal means that the use can only interact with that window so long as it is open. Example:
from Pyside.QtGui import QMessageBox
def add_helpWindow(self):
help_dialog = QMessageBox.information(self, 'Help', 'Some Help Text Here')
help_dialog.setModal(True)
return help_dialog.exec_()
You can also get a more generic dialog box using QDialog, which is the parent class of QMessageBox.
If that's not the behavior you want, you'll need to manually track whether the user has opened that window before, and then connect a signal that is emitted when the user closes the help window to a slot that reset the existence tracker. Here is an example using a non-modal QDialog:
from Pyside.QtGui import QDialog
class Window(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.menu_bar()
self.help_open = False # Tracks if the help dialog is already open
def help_closed(self):
self.help_open = False
...
def add_helpWindow(self):
if not self.help_open:
self.help_open = True
help_dialog = QDialog(self)
# Any other setup code here
help_dialog.setModal(False)
help_dialog.accepted.connect(self.help_closed)
help_dialog.rejected.connect(self.help_closed)
help_dialog.show()
I'm having problems with a "New Window" function in PyQt4/PySide with Python 2.7. I connected a initNewWindow() function, to create a new window, to an action and put it in a menu bar. Once a common function in desktop software. Instead of giving me a new persistent window alongside the other one the new window pops up and closes. The code I'm working on is proprietary so I created an example that does the same thing with the same error below. Is there any way to get this to work? Runs in PySide with Python 2.7. It was written in and tested in Windows.
from PySide.QtCore import QSize
from PySide.QtGui import QAction
from PySide.QtGui import QApplication
from PySide.QtGui import QLabel
from PySide.QtGui import QMainWindow
from PySide.QtGui import QMenuBar
from PySide.QtGui import QMenu
from sys import argv
def main():
application = QApplication(argv)
window = QMainWindow()
window.setWindowTitle('New Window Test')
menu = QMenuBar(window)
view = QMenu('View')
new_window = QAction('New Window', view)
new_window.triggered.connect(initNewWindow)
view.addAction(new_window)
menu.addMenu(view)
label = QLabel()
label.setMinimumSize(QSize(300,300))
window.setMenuBar(menu)
window.setCentralWidget(label)
window.show()
application.exec_()
def initNewWindow():
window = QMainWindow()
window.setWindowTitle('New Window')
window.show()
if __name__ == '__main__':
main()
If a function creates a PyQt object that the application needs to continue using, you will have to ensure that a reference to it is kept somehow. Otherwise, it could be deleted by the Python garbage collector immediately after the function returns.
So either give the object a parent, or keep it as an attribute of some other object. (In principle, the object could also be made a global variable, but that is usually considered bad practice).
Here's a revised version of your example script that demonstrates how to fix your problem:
from PySide import QtGui, QtCore
class Window(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
menu = self.menuBar().addMenu(self.tr('View'))
action = menu.addAction(self.tr('New Window'))
action.triggered.connect(self.handleNewWindow)
def handleNewWindow(self):
window = QtGui.QMainWindow(self)
window.setAttribute(QtCore.Qt.WA_DeleteOnClose)
window.setWindowTitle(self.tr('New Window'))
window.show()
# or, alternatively
# self.window = QtGui.QMainWindow()
# self.window.setWindowTitle(self.tr('New Window'))
# self.window.show()
if __name__ == '__main__':
import sys
app = QtGui.QApplication(sys.argv)
window = Window()
window.resize(300, 300)
window.show()
sys.exit(app.exec_())
When initNewWindow() returns, the window variable is deleted and the window's reference count drops to zero, causing the newly created C++ object to be deleted. This is why your window closes immediately.
If you want to keep it open, make sure to keep a reference around. The easiest way to do this is to make your new window a child of the calling window, and set its WA_DeleteOnClose widget attribute (see Qt::WidgetAttribute).