如何将我的QTableWidget子类链接到我的主程序

Question

Thanks to the following post ( Python Qt: How to catch "return" in qtablewidget ) I found how to subclass QTableWidget to connect an ENTER key event. My problem is that from this subclass, I can't find the way to reach all the other widgets.

Here is what I did.

Using QtDesigner I built this simple interface with a QTableWidget and a TextField . I promoted the QTableWidget to my own MyTableWidget (from the above post).

The main code is as follow:

import sys
from PyQt4 import QtGui, QtCore, Qt
from my_interface import Ui_MainWindow

class AppTemplateMain(QtGui.QMainWindow):

   variable1 = 'my variable 1'

   #initialize app
   def __init__(self, parent=None):
       QtGui.QMainWindow.__init__(self, parent)
       self.ui = Ui_MainWindow()
       self.ui.setupUi(self)

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

   exit_code=app.exec_()
   sys.exit(exit_code)

and the subclass of QTableWidget is defined in its own file mytablewidget.py

from PyQt4 import QtGui
from PyQt4.QtCore import Qt

class MyTableWidget(QtGui.QTableWidget):

def __init__(self, parent=None):
    self.parent = parent
    super(MyTableWidget, self).__init__(parent)

def keyPressEvent(self, event):
    key = event.key()

    if key == Qt.Key_Return or key == Qt.Key_Enter:
        print('clicked enter')
    else:
        super(MyTableWidget, self).keyPressEvent(event)

When I click in the table, I get as expected the message 'clicked enter' printed as expected. But I want to have access to the other widgets of the GUI from this subclass.... and to the variable1 defined in AppTemplateMain .

I feel like I'm missing something obvious but can't figure out what. Can someone help me here.

Thanks a lot.

Answer 1

You can create a custom signal in your table, that you emit whenever you want. Using the new style signal and slots :

class MyTableWidget(QtGui.QTableWidget):
    myCustomSignal=pyqtSignal()

    def keyPressEvent(self, event):
        key = event.key()
        if key == Qt.Key_Return or key == Qt.Key_Enter:
            print('clicked enter')
            self.myCustomSignal.emit()  
        ...

In your main code, you connect to this signal:

self.ui.tableWidget.myCustomSignal(self.myFunction)

You can't connect directly to keyPressEvent because, as stated in its name, it's an event, not a signal.

Answer 2

Thanks tmoreau,

I just found the way to do it and here it is (in case some other people are as dump as me with pyqt)

mytablewidget.py rom PyQt4 import QtGui from PyQt4.QtCore import Qt

class MyTableWidget(QtGui.QTableWidget):

parent = None

def __init__(self, parent=None):
    self.parent = parent
    super(MyTableWidget, self).__init__(parent)

def keyPressEvent(self, event):
    key = event.key()

    if key == Qt.Key_Return or key == Qt.Key_Enter:
        print('clicked enter')
    else:
        super(MyTableWidget, self).keyPressEvent(event)

def setUI(self, ui_parent):
    self.ui = ui_parent

and inside main.py

import sys from PyQt4 import QtGui, QtCore, Qt from my_interface import Ui_MainWindow

class AppTemplateMain(QtGui.QMainWindow):

variable1 = 'my variable 1'

#initialize app
def __init__(self, parent=None):
    QtGui.QMainWindow.__init__(self, parent)
    self.ui = Ui_MainWindow()
    self.ui.setupUi(self)

    self.ui.tableWidget.setUI(self)

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

   exit_code=app.exec_()
   sys.exit(exit_code)