Unable to add a widget to a QScrollArea object dynamically

[jsulm]

@sunshk1227 Try to call show() on new buttons:

def _add(self):
    b = QPushButton('test')
    self._list._layout.addWidget(b)
    b.show()

[sunshk1227]

@jsulm Hi, thank you for your help! But it seems that it does not work.

(By the way, if I do NOT comment it out, a new button will be displayed on the screen when I click the "Add" button...)

[jsulm]

@sunshk1227 Strange.
Not a solution for the problem, but an improvement idea: instead of exposing internals of the MyList class (_layout) to the outside world you should make it private and add a method to add new widgets (like buttons).

class MyList(QScrollArea):
    ...
    def addWidget(self, widget):
        self.__layoutaddWidget(widget)

This way MyButton does not have to know where to add new buttons to the list:

def _add(self):
        self._list.addWidget(QPushButton('test'))

You could improve your design even more if you remove self._list from MyButton completely and instead emit a signal in MyButton and connect it to a slot in MyList. This way MyButton and MyList do not have to know anything about each other.

[sunshk1227]

@jsulm Thank you very much for your suggestions! Got it. But I still have a question... should I further improve my design like this? (In this case, could you please tell me how to connect a function with an additional argument, where I marked by '?' in my code?)
Thank you again for your advice!

class MyList(QScrollArea):
    ...
    def _addWidget(self, widget):
        self._layout.addWidget(widget)


class MyButton(QPushButton):
    def __init__(self, text):
        super().__init__(text=text)


class MainWindow(QWidget):
    def __init__(self):
        ...
	self._my_list = MyList()
        self._my_button = MyButton(text='Add')
        self._my_button.clicked.connect(self._my_list._addWidget(?))
	...
    ...
...

[jsulm]

@sunshk1227 said in Unable to add a widget to a QScrollArea object dynamically:

self._my_button.clicked.connect(self._my_list._addWidget(?))

This is wrong, should be:

self._my_button.clicked.connect(self._my_list._addWidget)

In this line you only connect the signal to the slot, you do not call the slot. Slot will be called as soon as signal is emitted.

class MyButton(QPushButton):
    newWidget = pyqtSignal()
    def __init__(self, text):
        super().__init__(text=text)
        self.clicked.connect(self._add)
   
    def _add(self):
            newWidget.emit(QPushButton('test'))

self._my_button.newWidget.connect(self._my_list._addWidget)

Not sure this 100% correct as I don't use PyQt/PiSide :-)

[sunshk1227]

I figure out the reason with the help of @T.sagiv.
The solution is that I have to adjust the scrollarea. Just replace self.setMinimumSize(1024, 540) with self.setWidgetResizable(True).

[JonB]

@jsulm said in Unable to add a widget to a QScrollArea object dynamically:

Not sure this 100% correct as I don't use PyQt/PiSide :-)

newWidget.emit(QPushButton('test'))

Whazzis? :)

[jsulm]

@JonB You're the Python guy here :-) It was meant to emit a signal with a parameter.

[JonB]

@jsulm
Yeah but I was suprised at your parameter! Yours is emit newWidget(new QPushButton("test")), is that what you intended? I didn't really read what OP wanted :-; So you want to emit a signal which accepts an unplaced new widget, presumably the slot does something with it? Interesting!

[jsulm]

@JonB Yes, the emitter sends new widget to add via signal. The receiver adds it to its UI.