CODEWITHSUNDEEP
qt4signalspyqt4qtablewidget
neydroydrec
neydroydrec

Reputation: 7313

QTableWidget.cellActivated() signal not working?

I have tried to re-implement the QTableWidget.cellActivated(row, number) using PyQt4, to catch the coordinates of the currently active cell. The sample below does not work as expected, the method is not activated when a cell is clicked. Have I done something wrong?

class DictionaryTable(QtGui.QTableWidget):

    def __init__(self, parent=None):
        QtGui.QTableWidget.__init__(self, parent)

    def cellActivated(self, row, column):
        print row, column

Upvotes: 1

Views: 4222

Answers (1)

warvariuc
warvariuc

Reputation: 59574

Looking at the docs, i see:

void QTableWidget::cellActivated ( int row, int column ) [signal]

This signal is emitted when the cell specified by row and column has been activated This function was introduced in Qt 4.1.

This is a signal, not an event. So you don't need to [re]implement cellActivated method (it's not a method, it's a class attribute of signal type). You need to connect to the signal:

class DictionaryTable(QtGui.QTableWidget):

    def __init__(self, language_code, parent=None):
        QtGui.QTableWidget.__init__(self, parent)
        self.cellActivated.connect(self.handleCellActivated)

    def handleCellActivated(self, row, column):
        print row, column

Upvotes: 4

Related Questions