[SOLVED] Get parameter from QStackedWidget
-
Hi Guys,
I need to get some parameters (public member variables) from a QStackedwidget, which is a member of my first class:
// class 1 constructor
stackedWidget = new QStackedWidget;
stackedWidget ->addWidget(new WidgetPage); // is a class inherited from QWidget// class 1 method, here I want to get a parameter of the current selected widget
QWidget curWidget = pagesWidget->currentWidget();
QString value = curWidget->findChild<WidgetPage>()->comboBox1->currentText();The code compiles, but last line give me an "access violation while reading" error.
What could be the problem with this? Is there a better approach to get members from the widgets placed on my QStackedWidget?
Best regards,
Eddi0406edit:
I think I have to cast the currentWidget() return widget to my specific type. Then I should be able to use the member variables of WidgetPage. But how can I know, which is the right destination? I have different widgets in my stackwidget.
-
Yup, What if the findChild results in a empty pointer?? Then you simply call the currenText anyway. Not great programming. When you do a get function with pointers or any conversion. First check if the conversion worked.
And I do not think what you want is coded here. Shouldn't you need to do a dynamic cast?
@
QWidget * curWidget = pagesWidget->currentWidget() // So we have the pointer
WidgetPage * MyPointer (dynamic_cast<WidgetPage*>(curWidget));
if (WidgetPage != NULL)
{
// So we have a valid pointer to your class
QString Value(WidgetPage->comboBox1->currentText());
}
@
Damm, Something bugged me after a while. Getting a value straight from you GUI elements is not really a great thing. When using designer the GUI elements are always private and that is the best way to go! Use getters/Setters as interface for your WidgetPage
-
That's the line I was looking for, thank you very much!
You are absolutely right, of course I have to check whether the pointer is valid or not.
