第二次实例化 class 时出错

Question

I'm new to python and PyQt and was developing my first app using it, and I've been stuck in a problem when trying to instantiate a class I made again. I've got the following error:

Traceback (most recent call last):
  File "ConfiguradorAnx.py", line 16, in <lambda>
     self.ProductInfo.clicked.connect(lambda: self.newWindow(InfoProduct))
TypeError: 'InfoProduct' object is not callable
Aborted

The code goes like this:

from PyQt5 import QtCore, QtGui, QtWidgets, uic
import sys

class StartWindow(QtWidgets.QMainWindow):   #This function should inherit the class
                                            #used to make the ui file  
    def __init__(self):
        super(StartWindow,self).__init__()   #Calling the QMainWindow constructor
        uic.loadUi('Janela_inicial.ui',self)

        #defining quit button from generated ui
        self.QuitButton = self.findChild(QtWidgets.QPushButton, 'QuitButton')
        self.QuitButton.clicked.connect(QtCore.QCoreApplication.instance().quit)

        #defining product info button
        self.ProductInfo = self.findChild(QtWidgets.QPushButton, 'ProductInformation')
        self.ProductInfo.clicked.connect(lambda: self.newWindow(InfoProduct))
        self.show() #Show the start window

    def newWindow(self, _class):
        self.newWindow = _class()
        del self.newWindow

class InfoProduct(QtWidgets.QMainWindow):
    def __init__(self):
        super(InfoProduct,self).__init__()
        uic.loadUi('informacao_prod.ui',self)
        self.QuitButton = self.findChild(QtWidgets.QPushButton, 'pushButton')
        self.QuitButton.clicked.connect(lambda: self.destroy())
        self.show()

def main():
    app = QtWidgets.QApplication(sys.argv)  #Creates a instance of Qt application
    InitialWindow = StartWindow()
    app.exec_() #Start application

if __name__ == '__main__':
    main()

The first time I click on self.ProductInfo button it works and the InfoProduct window opens, but when I close the window and click on the same button again, I've got the error. I can't figure out what is it that I'm missing, I hope you guys could help!

Cheers!

Answer 1

You're overwriting the newWindow function in its execution:

def newWindow(self, _class):
    self.newWindow = _class()

By doing this, the result is that the next time you click the button, the lambda will try to call self.newWindow(InfoProduct) , but at that point self.newWindow is an instance of InfoProduct , which obviously is not callable.

The solution is simple (and very important) use different names for the function and the variable that points to the instance:

        self.ProductInfo.clicked.connect(lambda: self.createNewWindow(InfoProduct))

    def createNewWindow(self, _class):
        self.newWindow = _class()

Two small side notes:

• There's no need to use findChild , as loadUi already creates python instance attributes for widgets: you already can access self.QuitButton , etc.
• Avoid using capitalized names for variables and attributes. Read more about this and other code styling suggestions on the Style Guide for Python Code (aka, PEP-8).