Access the UI, from multiple windows.

mrjj

@Loc888
Hi
Why not just
Settings_Window* MainWindow::Get_Window_Data()
{
return Settings_Widget; // Settings_Widget is * (pointer) ?
}

Loc888

@mrjj

It run at least, but is not working. If i check something, then press the button with this code, nothing happened. Probably is another copy.
I am tired, tomorrow i try something else.

Is this the way how i should use it?

MainWindow B;


if(B.Get_Window_Data()->ui->Checker_Box_01->isChecked())
{

    //Do something

}

mrjj

Hi
Yes but im old school and always check pointers
so i would do

MainWindow B; // this should not be a second instance but the one you (might) create in main.cpp

Settings_Window * win=B.Get_Window_Data();

if (!win) {
qDebug() << "NULL ptr from Get_Window_Data";
return;
}
if(win->ui->Checker_Box_01->isChecked())
{
    //Do something
}
if(win->ui->Checker_Box_XX->isChecked())
{
    //Do something
}

However, the UI variable is private so unless the using class is friend, its not allowed.
So you make make UI public ( bad design )
or provide access functions for the widgets.

jsulm

@Loc888 said in Access the UI, from multiple windows.:

B.Get_Window_Data()->ui->Checker_Box_01->isChecked()

From software design point of view this is so bad!
You should not expose internal details of your MainWindow like that.
Add public methods to MainWindow which return needed invormation without exposing internal implementation details.
Example:

class MainWindow...
{
public:
    bool isSomethingActivated()
    {
        return ui->Checker_Box_01->isChecked();
    }
}

MainWindow B;
if(B.isSomethingActivated())
{
    //Do something
}

Now, the user of MainWindow does not have to know anything about how the MainWindow is designed (what UI elements it has for example) - it just calls a public interface method to get the information. Usn't this much nicer? One more advantage of this approach: if you later change your MainWindow UI the caller of the MainWindow will not be affected.
Example:

class MainWindow...
{
public:
    bool isSomethingActivated()
    {
        // You decided that the condition should be different
        return ui->Checker_Box_01->isChecked() && ui->Checker_Box_02->isChecked();
    }
}

// No need to change the caller of MainWindow
MainWindow B;
if(B.isSomethingActivated())
{
    //Do something
}

mrjj

@jsulm
And we come full circle \o/
I told OP that 4 days ago, but he seems unwilling to
create access functions :)
Maybe seeing your good example, OP will feel the joy of good design
and do it the right way.

Loc888

@mrjj

It makes me a little bit mad, why everything need's to be complicated.
So, in main window can i use something like that:

  B.Get_Window_Data()->ui->Checker_Box_01->isChecked()

But in other window, i need to do it in other way?

jsulm

@Loc888 You should no access internal implementation details of one window (class) from another. It is not related to Qt, this are simply software design basics...
And what is complicated about it? Is my example really complicated? What you're trying to do is much more complicated and error prone. Why do you want to know in window A how window B is constructed? Why not simply define simple APIs to communicate between windows? To be honest code like what you're trying to write would not pass code reviews in the company where I'm working.

Loc888

@mrjj Listen, i show you my code, cuz i am getting a little bit crazy, i can't create it in main.cpp, becasuse then how can i show it?

In MainWindow.h i have

    Settings_Window* Get_Window_Data();

     Settings_Window* Settings_Widget = new Settings_Window;

MainWindow.cpp

Settings_Window* MainWindow::Get_Window_Data()
{

    return Settings_Widget;

}

Window1

MainWindow B;

Settings_Window S;


if(B.Get_Window_Data()->ui->Checker_01->isChecked())
{

  
}

Loc888

@jsulm I never gonna work for anybody from software design. I am doing it more like hobby.

Loc888

@mrjj This method is not working

Settings_Window* MainWindow::Get_Window_Data()
{

return Settings_Widget;

}

If i press the checkerbox, it should be setted to true

void Settings_Window::on_CheckerBox_clicked()
{

    B->Get_Window_Data()->ui->CheckerBox->setChecked(true);

}

Then when i go to window one, and press the button

void Window1::on_Button001_clicked()
{

MainWindow* B = new MainWindow;


if(B->Get_Window_Data()->ui->Auto_Reset_Data_Timer->isChecked())
{

    //do something

  }

}

I don't understand all your methods, so if anyone can, please correct this stuff, cuz i have enough.

mrjj

In what way not working ?
it just returns the pointer to your
Settings_Widget so only way it can fail is to return null.

I wonder here
void Window1::on_Button001_clicked()
{

MainWindow* B = new MainWindow; << you create new one. ?

so you dont have one already ?
Or is that on purpose ?
So each time you click you make a new window ?

Loc888

@mrjj I have another button to activate and show that window.

MainWindow* B = new MainWindow;

No,i have already one window.
I just use this to get MainWindow method, and try to set checkerBox as true but from ther.

If i copy the same code from that button, and paste it to Window1, it's working, but is not the way how it should work, i dont want to set it ther....

I mean:

MainWindow* B = new MainWindow;

B->Get_Window_Data()->ui->CheckerBox->setChecked(true);


if(B->Get_Window_Data()->ui->Auto_Reset_Data_Timer->isChecked())
{

    //do something

  }

When i say "Is not working", i mean when i click the button, checkBox should return true, and when i compare it in another window, checkBox returns false, i think because i set it to another copy... How,i don't know. It should return Settings_Widget in window1.

mrjj

Hi
Im sorry if we frustrated you. We just know from experience
what works bad over time so that is why we seem so keen to do it the right way.
Lets start over.
ok, so you have a normal main.cpp

int main(int argc, char* argv[]) {
  QApplication a(argc, argv);
  MainWindow w;
  w.show();

  return a.exec();
}

so w is mainwindow
and it has the true Settings_Window embedded.

So we need Window1 to have access to MainWins Settings_Window and not a new copy.
So Window1 should NOT new one it self but get the one from Mainwin.

I assume Mainwindow also creates and show Window1
You can change its constructor to take a Settings_Window * and it can then store it for accessing.

Like

#include "Settings_Window.h"  // so it knows the TYPE ( make to use right name if not that)

class Window1 : public QMainWindow
{
    Q_OBJECT
Settings_Window *GivenToMe;
public:
    explicit Window1 (QWidget *parent = 0, Settings_Window *TheTrueOne) {
GivenToMe = TheTrueOne; // store the pointer for later
}
    ~Window1 ();
....
};

// then in MainWin where you show the Window1
void MainWindow::ShowSettings() {
Window1 * Win1= new Window1(this,   & S_Window ); // the & to make it pointer
Win1->show();
}

now inside
void Window1::SomeButtonClick() {
// you now have GivenToMe that is the Settings_Widget from MainWindow
now you can call public functions in it
if ( GivenToMe -> SomeFunc() ) ...

}

If you do not want to create public access functions but go directly via UI you have to make it public
so in Settings_Window.h

private:
Ui::Settings_Windowui;
must be
public:
Ui::Settings_Windowui;

to allow
if ( GivenToMe -> ui->CheckerBox->isChecked() ) ...

--

What we tried to make you do was to add public access function in mainwindow
so you would say
if ( GivenToMe ->GetSettingForX() ) ( and it just returns the ui->Checker_01->isChecked() )

but if you really not want that, then you can make ui public and the syntax you want is then possible
even if its not best practice/bad idea/design.

Hope this helps. :)

Loc888

@mrjj Thank you, i like exemples rly :)

Because this is the only thing can help at least when you rly dont know what to do. Tomorrow i will try in this way.

Loc888

@mrjj

I have few errors:

default argument missing for parameter 2 of 'Window1::Window1(QWidget*, Settings_Window*)'
explicit Window1(QWidget *parent = 0, Settings_Window *TheTrueOne)
^

redefinition of 'Window1::Window1(QWidget*, Settings_Window*)'
Window1::Window1(QWidget *parent, Settings_Window *TheTrueOne = 0) :
^

'Window1::Window1(QWidget*, Settings_Window*)' previously defined here
explicit Window1(QWidget *parent = 0, Settings_Window *TheTrueOne)
^

Can you correct it? Because i tried what i could, but doesn't help.

Window1.h

#ifndef WINDOW1_H
#define WINDOW1_H

#include <QWidget>
#include "Settings_Window.h"

namespace Ui {
class Window1;
}

class Window1 : public QWidget
{
    Q_OBJECT

    Settings_Window *GivenToMe;

public:
    explicit Window1(QWidget *parent = 0, Settings_Window *TheTrueOne)
{

    GivenToMe = TheTrueOne; // store the pointer for later

}


~Window1();

private:
Ui::Window1 *ui;
};

#endif // WINDOW1_H

And this is Window1.cpp

#include "window1.h"
#include "ui_window1.h"
#include "settings_window.h"

Window1::Window1(QWidget *parent, Settings_Window *TheTrueOne = 0) :
    QWidget(parent),
    ui(new Ui::Window1)
{
    ui->setupUi(this);
}

Window1::~Window1()
{
    delete ui;
}

SGaist

Hi,

Default arguments can only be set in the function declaration. So move all = 0 to the header.

As a matter of good practice, the parent parameter should always be last. Question of coherency with the Qt library and what other developers expects.

Loc888

@SGaist Ok, i try that.

Loc888

@SGaist I still have some errors the:

Error: redefinition of 'Window1::Window1(Settings_Window*, QWidget*)'
Window1::Window1(Settings_Window *TheTrueOne,QWidget *parent) : QWidget(parent),
^

SGaist

Do you have matching declaration and implementation ?

Loc888

@SGaist

Header :

    explicit Window1(Settings_Window *TheTrueOne = 0, QWidget *parent = 0)
    {

        GivenToMe = TheTrueOne;

    }

Source:

Window1::Window1(Settings_Window *TheTrueOne,QWidget *parent) :         
QWidget(parent),ui(new Ui::Window1)
{
ui->setupUi(this);
}

error: redefinition of 'Window1::Window1(Settings_Window*, QWidget*)'
Window1::Window1(Settings_Window *TheTrueOne,QWidget *parent) : QWidget(parent),ui(new Ui::Window1)
^

I i remove ther Settings_Window, then i have no type specified error.