So, I am new to python programming. I have started to implement a UI in pyqt5 and there I have a button and I want to react when the user clicks it.
According to this Link I should simply write btn.clicked.connect(self.buton_pressed) however I get the message "Cannot find reference connect in function". (The surrounding code is at the end of the question)
So I googled a bit and all I found is that it should just work that way. I just don't get why it does not. I found this Stackoverflow question which also describes the old variant of how to do it. That did not work either, after some googeling I found out that it is no longer supported in pyqt5 or in some other package.
The function where I try to connec to the event:
def __add_button(self, text: str, layout: QLayout):
btn = QPushButton(text, self)
layout.addWidget(btn)
btn.clicked.connect(self.button_pressed)
# TODO: fix this.
return btn
The code where the GUI is generated and the function called, in the __init__ function
lblhm = QLabel("Hauptmessung", self)
layout.addWidget(lblhm)
self.__hm_b = self.__add_button("Messung öffnen", layout)
self.__hm_config_b = self.__add_button("Configruation öffnen", layout)
lblzm = QLabel("Zusatzmessung", self)
layout.addWidget(lblzm)
self.__zm_b = self.__add_button("Messung öffnen", layout)
self.__zm_config_b = self.__add_button("Configuration öffnen", layout)
The button_pressed function is not yet implemented, but it is supposed to open a openFile dialog for file selection.
According to this post i could just connect after returning the function, but then i would have to write it 4 times which is not very nice. Isn't the signal bound to the object not to the variable?
Thankful for any help :)
It's hard to understand your problem since you don't provide us with a working example, i.e. a peace of code one can run "as is". Something like this:
from PyQt4 import QtCore, QtGui
class MyWindow(QtGui.QWidget):
def __init__(self):
super().__init__()
layout = QtGui.QVBoxLayout()
self.setLayout(layout)
lblhm = QtGui.QLabel("Hauptmessung", self)
layout.addWidget(lblhm)
self.__hm_b = self.__add_button("Messung öffnen", layout)
self.__hm_config_b = self.__add_button("Configruation öffnen", layout)
lblzm = QtGui.QLabel("Zusatzmessung", self)
layout.addWidget(lblzm)
self.__zm_b = self.__add_button("Messung öffnen", layout)
self.__zm_config_b = self.__add_button("Configuration öffnen", layout)
def button_pressed(self):
print('Button pressed')
def __add_button(self, text: str, layout: QtGui.QLayout):
btn = QtGui.QPushButton(text, self)
layout.addWidget(btn)
btn.clicked.connect(self.button_pressed)
return btn
if __name__== '__main__':
import sys
app = QtGui.QApplication(sys.argv)
wnd = MyWindow()
wnd.show()
sys.exit(app.exec_())
There's no problem with this code under PyQt4. Does it work for with PyQt5?
Related
I am trying to decouple entirely my GUI from my controller class, and for some reason I can't seem to manage to connect my buttons from outside of my GUI class itself.
Here's a small example of what I mean :
import sys
from PySide6 import QtWidgets
class Gui(QtWidgets.QWidget):
def __init__(self, parent=None):
super(Gui, self).__init__(parent)
layout = QtWidgets.QHBoxLayout(self)
self.button = QtWidgets.QPushButton("Do stuff")
layout.addWidget(self.button)
class Controller(object):
def do_stuff(self):
print("something")
def startup(parent):
ctrl = Controller()
gui = Gui(parent)
gui.button.clicked.connect(ctrl.do_stuff)
return gui
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
dialog = QtWidgets.QDialog()
gui = startup(dialog)
dialog.show()
sys.exit(app.exec_())
I would expect this code, when run, to display a GUI with one push button (which it does), and when pressing the push button, I'd expect the word "something" to get printed. However this doesn't seem to be the case.
I might just be too tired, but I can't find the solution.
What am I missing?
Thanks a lot in advance!
ctrl = None
gui = None
def startup(parent):
global ctrl
global gui
ctrl = Controller()
gui = Gui(parent)
gui.button.clicked.connect(ctrl.do_stuff)
return gui
try this, and it does work. when the variable is in the function, it will be destroyed before the function is finished. the global variable is not a good coding style but is a simple way to figure out your confusion.
I hope this question isn't too far-fetched. I'm good with Selenium and I've been working with PyQt4 recently. I want to use them both together with a program I'm currently working on and it'd work out a lot more smoothly if I could embed the controllable browser into a Qt4 frame or widget. Can this be done? And if so, how?
It doesn't have to be done with Selenium, I just want to be able to control the browser or at least show a webpage in a Qt widget or frame.
So after some research into methods other people have used, I figured it out.
The code I used came from a "very simple browser" module I obtained from here
I modified the code to be more customizable for my future self.
Here's my modified version of the code:
from PyQt4 import QtCore, QtGui, QtWebKit
class Browser(QtGui.QMainWindow):
def __init__(self, size=[800,600], frame=None, centralWidget=None, default_url='https://www.google.com', backButton=True, forwardButton=True, topBar=True):
"""
Initialize the browser GUI and connect the events
"""
self.showBackButton = backButton
self.showForwardButton = forwardButton
self.showTopBar = topBar
QtGui.QMainWindow.__init__(self)
self.resize(size[0],size[1])
if (centralWidget == None):
self.centralwidget = QtGui.QWidget(self)
else:
self.centralwidget = centralWidget
self.mainLayout = QtGui.QHBoxLayout(self.centralwidget)
self.mainLayout.setSpacing(0)
self.mainLayout.setMargin(1)
if (frame == None):
self.frame = QtGui.QFrame(self.centralwidget)
else:
self.frame = frame
self.gridLayout = QtGui.QVBoxLayout(self.frame)
self.gridLayout.setMargin(0)
self.gridLayout.setSpacing(0)
self.horizontalLayout = QtGui.QHBoxLayout()
if (self.showTopBar):
self.tb_url = QtGui.QLineEdit(self.frame)
if (self.showBackButton):
self.bt_back = QtGui.QPushButton(self.frame)
if (self.showForwardButton):
self.bt_ahead = QtGui.QPushButton(self.frame)
if (self.showBackButton):
self.bt_back.setIcon(QtGui.QIcon().fromTheme("go-previous"))
if (self.showForwardButton):
self.bt_ahead.setIcon(QtGui.QIcon().fromTheme("go-next"))
if (self.showBackButton):
self.horizontalLayout.addWidget(self.bt_back)
if (self.showForwardButton):
self.horizontalLayout.addWidget(self.bt_ahead)
if (self.showTopBar):
self.horizontalLayout.addWidget(self.tb_url)
self.gridLayout.addLayout(self.horizontalLayout)
self.html = QtWebKit.QWebView()
self.gridLayout.addWidget(self.html)
self.mainLayout.addWidget(self.frame)
#self.setCentralWidget(self.centralwidget) --- Not needed when embedding into a frame
if (self.showTopBar):
self.connect(self.tb_url, QtCore.SIGNAL("returnPressed()"), self.browse)
if (self.showBackButton):
self.connect(self.bt_back, QtCore.SIGNAL("clicked()"), self.html.back)
if (self.showForwardButton):
self.connect(self.bt_ahead, QtCore.SIGNAL("clicked()"), self.html.forward)
self.connect(self.html, QtCore.SIGNAL("urlChanged(const QUrl)"), self.url_changed)
self.default_url = default_url
if (self.showTopBar):
self.tb_url.setText(self.default_url)
self.open(self.default_url)
def browse(self):
"""
Make a web browse on a specific url and show the page on the
Webview widget.
"""
if (self.showTopBar):
url = self.tb_url.text() if self.tb_url.text() else self.default_url
self.html.load(QtCore.QUrl(url))
self.html.show()
else:
pass
def url_changed(self, url):
"""
Triggered when the url is changed
"""
if (self.showTopBar):
self.tb_url.setText(url.toString())
else:
pass
def open(self, url):
self.html.load(QtCore.QUrl(url))
self.html.show()
It could use some work at the moment, but I've tested it out and it's doing exactly what I need it to do. I tested it out with the following chunk of code that runs when the script is executed
if (__name__ == "__main__"):
app = QtGui.QApplication(sys.argv)
window = QtGui.QMainWindow()
window.resize(800,600)
myFrame = QtGui.QFrame(window)
myFrame.resize(200,200)
myFrame.move(10,10)
main = Browser(centralWidget=myFrame, default_url='https://www.google.com/', forwardButton=False, backButton=False, topBar=False)
window.show()
sys.exit(app.exec_())
Like I said, it could use work, but it does exactly what I needed it to do. Now I can embed it into a frame (with the size of my choosing) to use within another application.
Regarding my modifications: I made it possible to keep/remove the back button, forward button and top bar (for the URL). But the webbrowser is still controllable using the "open" function.
And if you wanted to open another webpage, it's as simple as the following
main.open('https://your.webpage.here.com')
I'm trying to do something quite simple: add a menu bar with an Exit action that will close a QMainWindow when it is selected. However, when I actually click Exit, it doesn't close the application. A SSCCE:
from PyQt4 import QtGui, QtCore
import sys
class Window(QtGui.QMainWindow):
def __init__(self, parent=None):
super(Window, self).__init__(parent)
widget = QtGui.QWidget()
self.setCentralWidget(widget)
self.menu_bar = QtGui.QMenuBar(self)
menu = self.menu_bar.addMenu('File')
exit_action = QtGui.QAction('Exit', self)
exit_action.triggered.connect(lambda:
self.closeEvent(QtGui.QCloseEvent()))
menu.addAction(exit_action)
self.setMenuBar(self.menu_bar)
def closeEvent(self, event):
print('Calling')
print('event: {0}'.format(event))
event.accept()
app = QtGui.QApplication(sys.argv)
form = Window()
form.show()
sys.exit(app.exec_())
What is really confusing me is that when I click Exit from the File menu, I get the following output:
Calling
event: <PyQt4.QtGui.QCloseEvent object at 0x024B7348>
and the application does not exit.
If I click the top-right corner X, I get the same thing (down to the same memory address for the event object):
Calling
event: <PyQt4.QtGui.QCloseEvent object at 0x024B7348>
and the application does exit.
This is on Windows 7 64-bit, Python 2.7.2, PyQt 4.8.6.
Document says,
The QCloseEvent class contains parameters that describe a close event.
Close events are sent to widgets that the user wants to close, usually
by choosing "Close" from the window menu, or by clicking the X title
bar button. They are also sent when you call QWidget.close() to close
a widget programmatically.
Your can call directly with signal close not by QCloseEvent, please call self.close().
from PyQt4 import QtGui, QtCore
import sys
class Window(QtGui.QMainWindow):
def __init__(self, parent=None):
super(Window, self).__init__(parent)
widget = QtGui.QWidget()
self.setCentralWidget(widget)
self.menu_bar = QtGui.QMenuBar(self)
menu = self.menu_bar.addMenu('File')
exit_action = QtGui.QAction('Exit', self)
exit_action.triggered.connect(self.close)
menu.addAction(exit_action)
self.setMenuBar(self.menu_bar)
def closeEvent(self, event):
print('Calling')
print('event: {0}'.format(event))
event.accept()
app = QtGui.QApplication(sys.argv)
form = Window()
form.show()
sys.exit(app.exec_())
The close event doesn't actually make the window close, it's just triggered when the window is already closing. To actually make the window close, you need to call self.close(), which will have the side effect of triggering a QCloseEvent. So simply use this:
exit_action.triggered.connect(self.close)
The documentation of close describes the interaction between close and closeEvent:
bool QWidget.close (self)
This method is also a Qt slot with the C++ signature bool close().
Closes this widget. Returns true if the widget was closed; otherwise
returns false.
First it sends the widget a QCloseEvent. The widget is hidden if it
accepts the close event. If it ignores the event, nothing happens. The
default implementation of QWidget.closeEvent() accepts the close
event.
I have a login screen dialog written using pyqt and python and it shows a dialog pup up when it runs and you can type in a certin username and password to unlock it basicly. It's just something simple I made in learning pyqt. I'm trying to take and use it somewhere else but need to know if there is a way to prevent someone from using the x button and closing it i would like to also have it stay on top of all windows so it cant be moved out of the way? Is this possible? I did some research and couldn't find anything that could help me.
Edit:
as requested here is the code:
from PyQt4 import QtGui
class Test(QtGui.QDialog):
def __init__(self):
QtGui.QDialog.__init__(self)
self.textUsername = QtGui.QLineEdit(self)
self.textPassword = QtGui.QLineEdit(self)
self.loginbuton = QtGui.QPushButton('Test Login', self)
self.loginbuton.clicked.connect(self.Login)
layout = QtGui.QVBoxLayout(self)
layout.addWidget(self.textUsername)
layout.addWidget(self.textPassword)
layout.addWidget(self.loginbuton)
def Login(self):
if (self.textUsername.text() == 'Test' and
self.textPassword.text() == 'Password'):
self.accept()
else:
QtGui.QMessageBox.warning(
self, 'Wrong', 'Incorrect user or password')
class Window(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
if __name__ == '__main__':
import sys
app = QtGui.QApplication(sys.argv)
if Test().exec_() == QtGui.QDialog.Accepted:
window = Window()
window.show()
sys.exit(app.exec_())
Bad news first, it is not possible to remove the close button from the window, based on the Riverbank mailing system
You can't remove/disable close button because its handled by the
window manager, Qt can't do anything there.
Good news, you can override and ignore, so that when the user sends the event, you can ignore or put a message or something.
Read this article for ignoring the QCloseEvent
Also, take a look at this question, How do I catch a pyqt closeEvent and minimize the dialog instead of exiting?
Which uses this:
class MyDialog(QtGui.QDialog):
# ...
def __init__(self, parent=None):
super(MyDialog, self).__init__(parent)
# when you want to destroy the dialog set this to True
self._want_to_close = False
def closeEvent(self, evnt):
if self._want_to_close:
super(MyDialog, self).closeEvent(evnt)
else:
evnt.ignore()
self.setWindowState(QtCore.Qt.WindowMinimized)
You can disable the window buttons in PyQt5.
The key is to combine it with "CustomizeWindowHint",
and exclude the ones you want to be disabled.
Example:
#exclude "QtCore.Qt.WindowCloseButtonHint" or any other window button
self.setWindowFlags(
QtCore.Qt.Window |
QtCore.Qt.CustomizeWindowHint |
QtCore.Qt.WindowTitleHint |
QtCore.Qt.WindowMinimizeButtonHint
)
Result with QDialog:
Reference: https://doc.qt.io/qt-5/qt.html#WindowType-enum
Tip: if you want to change flags of the current window, use window.show()
after window.setWindowFlags,
because it needs to refresh it, so it calls window.hide().
Tested with QtWidgets.QDialog on:
Windows 10 x32,
Python 3.7.9,
PyQt5 5.15.1
.
I don't know if you want to do this but you can also make your window frameless. To make window frameless you can set the window flag equal to QtCore.Qt.FramelessWindowHint
It's difficult to describe what I want to do here, so here's what I want:
-Click a button that "pops up" a dialog.
-This dialog, however, should be non-modal.
-I'd also like for it to be sort of 'locked' with the parent app so if I moved it, it too would come with.
It seems like i'm more or less trying to describe an in-line popup, but I'm not sure if that's what it's called, or even how I should go about approaching this problem. So, PyQt gurus, how would you make something like this? Thanks in advance.
To get a modeless dialog, open it with show() rather than exec_().
To ensure that the dialog is "locked with the parent app", set a parent window in the dialog's constructor.
from PyQt4 import QtGui
class Window(QtGui.QWidget):
def __init__(self):
QtGui.QWidget.__init__(self)
button = QtGui.QPushButton('Open Dialog', self)
button.clicked.connect(self.handleOpenDialog)
self.resize(300, 200)
self._dialog = None
def handleOpenDialog(self):
if self._dialog is None:
self._dialog = QtGui.QDialog(self)
self._dialog.resize(200, 100)
self._dialog.show()
if __name__ == '__main__':
import sys
app = QtGui.QApplication(sys.argv)
win = Window()
win.show()
sys.exit(app.exec_())
for Qt5:
I ignore why dialog.setWindowModality(QtCore.Qt.NonModal) won't work, but
dialog.run()
dialog.exec_()
will do. run will make the dialog non modal and exec_ will block it until the user input.