Question

Newbie here, I put a dock widget in a main window, and there is a button in this dock widget panel, now I want to connect, this button with a function defined in the main window, it threw out an error, what should I do? Thanks

connect
(
    perfectPanel_->btn_AAA,
    SIGNAL(clicked()),
    this,
    SLOT(on_actionAAA_triggered()),
    Qt::UniqueConnection
);

Error message is

$PWD/ui_perfectPanel.h: In constructor ‘xixi::xixi()’:
$PWD/ui_perfectPanel.h:71:18: error: ‘QPushButton* Ui_perfectPanel::btn_AAA’ is inaccessible
$PWD/xixi/xixi.cpp:51:25: error: within this context

Note that I have already managed to connect this with a toolbar button in the main window (xixi.cpp), it works great.

Answer 1

This happens because your dock class, perfectPanel, inherits privately from the generated ui class Ui::perfectPanel:

class perfectPanel : public QWidget, private Ui::perfectPanel

You could make that inheritance public, but shouldn’t. Instead you should make the signal part of the perfectPanel class, and route the internal signal from the button to that external signal:

class perfectPanel ... {
...
signals:
    void AAA_clicked();
};

perfectPanel::perfectPanel() {
     setupUi(this);
     connect(btn_AAA, SIGNAL(clicked()), this, SIGNAL(AAA_clicked()));
}

(And in case you would ask, yes, you can connect 2 signals together).

Then you simply connect the new signal inside your main window class:

connect(perfectPanel_, 
    SIGNAL(AAA_clicked()), 
    this,
    SLOT(on_actionAAA_triggered()),
    Qt::UniqueConnection
);