[FIXED] Signal Not coming though - PySide

Issue

I am sending a signal from another class to update a PySide QTableWidget but nothing is coming though. I have made this very simple for this demonstration:

This is in the controller module called Records.py

class Records(QDialog, randomDialog.Ui_watchingDialog):

    signal = 1
    atSig = Signal(int)

    def add_button_clicked(self):

        # Do some stuff
        self.signal = 1
        self.atSig.emit(self.signal)
        # Do some other Stuff

This sits out side the controller called main.py

from controller import Records

class main(QMainWindow, pyMainWindow.Ui_mainWindow):

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

        signal_records = Records.Records()

        signal_records.atSig.connect(self.showNewData)


    def showNewData(self, signal):
        if signal == 1:
            print "It worked!"
        else:
            print "Problem"

How come this signal is not coming through? No error messages are being thrown and neither of the print statements aren't being called. How can I fix this?

Solution

If It's possibly, please define this before caller to receive;

.
.
@Slot (int)
def showNewData(self, signal):
.
.

Or it not, Please check your caller def add_button_clicked(self). I work in pyqt4 (same pySide) and (I cut some part out and put some path for test in your) code, It's work.

import sys  
from PyQt4 import QtCore  
from PyQt4 import QtGui  

class QRecordsDialog (QtGui.QDialog):
    addButtonClickedSignal = QtCore.pyqtSignal(int)

    def __init__ (self, parent = None):
        super(QRecordsDialog, self).__init__(parent)
        self.myQPushButton = QtGui.QPushButton('Test Signal', self)
        self.myQHBoxLayout = QtGui.QHBoxLayout()
        self.myQHBoxLayout.addWidget(self.myQPushButton)
        self.setLayout(self.myQHBoxLayout)
        self.myQPushButton.clicked.connect(self.addButtonClicked)

    def addButtonClicked (self):
        self.addButtonClickedSignal.emit(1)

class QMainWindow (QtGui.QMainWindow):
    def __init__ (self, parent = None):
        super(QMainWindow, self).__init__(parent)
        myQRecordsDialog = QRecordsDialog(self)
        myQRecordsDialog.addButtonClickedSignal.connect(self.showNewData)
        myQRecordsDialog.show()

    @QtCore.pyqtSlot(int)
    def showNewData (self, signal):
        if signal == 1:
            print "It worked !"
        else:
            print "Problem ?"

myQApplication = QtGui.QApplication(sys.argv)
myQMainWindow = QMainWindow()
myQMainWindow.show()
sys.exit(myQApplication.exec_())

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If you want to modify your PyQt code to use the PySide naming scheme, that can be done using a simple definition:

QtCore.Signal = QtCore.pyqtSignal
QtCore.Slot = QtCore.pyqtSlot

Reference this

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Regards,

Answered By - Kitsune Meyoko