[Eric] PyQt-Slot for QtWidget (QComboBox) created during runtime

[Christos Sevastiadis]

Dear George,

I think that you should create a slot for the specific QComboBox object and
not for the cell of the QTableWidget object.
For example, for a QComboBox created in the Ui_MainWindow.setupUI(), the
corresponding slot created in the MainWindow.setupUI() should have the
following form, for integer or string type index.

    @pyqtSlot(int)
    def on_spk_ver_rep_QComboBox_currentIndexChanged(self, index):
        """
        Slot documentation goes here.

        @param index DESCRIPTION
        @type int
        """
        # TODO: not implemented yet
        raise NotImplementedError

    @pyqtSlot(str)
    def on_spk_ver_rep_QComboBox_currentIndexChanged(self, p0):
        """
        Slot documentation goes here.

        @param p0 DESCRIPTION
        @type str
        """
        # TODO: not implemented yet
        raise NotImplementedError

Christos



On Sun, 23 Sep 2018 at 15:40, GM <mossl at gmx.net> wrote:

> Hi @all,
>
> I've created a dialog-window with an QTableWidget. Users can add a row
> by using a pushbutton. In a cell of the new row appears a QComboBox. How
> can changes of this QComboBox trigger a function?
>
> It's no problem for to text-only-cells (I can use:
>
>      @pyqtSlot(int, int)
>      def on_tableWidgetX_cellChanged(self, row, column):
>          print("cell changed: ", row, column)
>
> which works as intended.)
>
> Functions like
>
>      @pyqtSlot(QComboBox)
>      def on_tableWidgetX_itemChanged(self, item):
>          print("OK1")
>
>      @pyqtSlot(QComboBox)
>      def on_tableWidgetX_itemClicked(self, item):
>          print("Ok2")
>
> never recieve a signal.
>
> I think the QComboBox can not send a aignal. Originally the "Generate
> Dialog Code..."-dialog suggests to pass a "QTableWidgetItem*" to the
> pyqtSlot, but this Item is created during runtime, so it can not be
> found in the Ui_Dialog...
>
> Any suggestions?
>
>
> george
>
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