CODEWITHSUNDEEP
pythonpyqt5
findusl
findusl

Reputation: 2644

How to react to a button click in pyqt5

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 :)

Upvotes: 2

Views: 25766

Answers (1)

Alexander Lutsenko
Alexander Lutsenko

Reputation: 2160

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?

Upvotes: 4

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