CODEWITHSUNDEEP
pythonpyqt4
Jason Chen
Jason Chen

Reputation: 19

How to show the window which is called out by another .py file?

I have two .py files, respectively simple_1.py and simple_2.py. How to display the window of simple_2.py when I click the button in simple_1.py ?
The same , how do display the window of simple_1.py when I click the button in simple_2.py ?
And when the window is called out by another .py file , now this window has to turn off at the same time.
Here is my simple_1.py code:

# -*- coding: utf-8 -*- 
#simple_1.py

import sys
from PyQt4 import QtCore, QtGui 
from simple import  Ui_Form
class StartQT4(QtGui.QWidget): 
    def __init__(self, parent=None): 
        QtGui.QWidget.__init__(self, parent)                                            
        self.ui = Ui_Form()            
        self.ui.setupUi(self) 

        self.button() 

    def button(self):
        button1= QtGui.QPushButton('show simple_2', self)
        button1.setGeometry(80, 80,100, 50) 
        self.connect(button1, QtCore.SIGNAL('clicked()'),
            self.buttonClicked)

    def buttonClicked(self):
        #to show window of simple_2.py 


if __name__ == "__main__": 
    app = QtGui.QApplication(sys.argv) 
    myapp = StartQT4() 
    myapp.show() 
    sys.exit(app.exec_())

Here is my simple_2.py code:

# -*- coding: utf-8 -*- 
#simple_2.py

import sys
from PyQt4 import QtCore, QtGui 


class Apple(QtGui.QWidget):
    def __init__(self,parent=None):
        super().__init__()
        self.widget = QtGui.QWidget()
        self.resize(250, 150)
        self.setWindowTitle('simple2')
        self.button() 


    def button(self):
        button1= QtGui.QPushButton('show simple_1', self)
        button1.setGeometry(80, 80, 100, 50) 
        self.connect(button1, QtCore.SIGNAL('clicked()'),
            self.buttonClicked)

    def buttonClicked(self):
         #to show window of simple_1.py 


if __name__ == "__main__":         
    app = QtGui.QApplication(sys.argv)
    mywidget = Apple()
    mywidget.show()
    sys.exit(app.exec_())

Upvotes: 1

Views: 50

Answers (1)

derricw
derricw

Reputation: 7036

There are several errors in your code that prevent the examples you've posted from running. I will ignore them and answer your actual question, which from what I can tell is this:

How can I make my two Qt widgets create and display instances of each other?

First I'd suggest changing the widgets into QDialogs, which have a convenient exec_() method. You do this by inheriting from QtGui.QDialog instead of from QtGui.QWidget like so:

class Apple(QtGui.QDialog):

The next thing you will need to do is import and run your custom QDialogs in the button callbacks.

def buttonClicked(self):
    from simple_1 import StartQT4  # imports your dialog from your other file
    sqt = StartQT4()  # creates an instance of it
    self.close()  # closes the current dialog
    sqt.exec_()  # runs the newly created dialog

Upvotes: 1

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