toolbox

[mickey]

hi, I inserted a toolBox from designer in MainForm; now from mainForm I'd like to control its width and do something like this:


//paintGL()
int widthWidget = this->width();
mainForm->toolBox->setMinimumWidth(widthWidget/4);
mainForm->toolBox->setMaximumWidth(widthWidget/4);

(but it doesn't work, when I enlarge window app maximumToolBoxWidth block the window app width too...)
I need set width of toolBox 1/4 of Widget (and when I resize my window app toolBox should be maintain 1/4 width)
How can I do this? Thanks

[zlatko]

Have 2 choice :
- reinplement resizeEvent for your main window, and set toolbox size there
- use layouts(better way)

[wysota]

You can use stretchFactors to manipulate widget size relative to its siblings in a layout.

[mickey]

Ok, but I'd like avoid this; is there a way to retrieve the screen resolution (from QT)? Thanks

[wysota]

QDesktopWidget

[mickey]

Thanks, this seem works; but return height 704 instead of 768; the task Bar or window isn't included in height...How can I do? Thanks


QDesktopWidget* qw;
QRect rect = qw->availableGeometry(0);
std::cout << "Screen Width: " << rect.width() << endl;
std::cout << "Screen Height: " << rect.height() << endl;

[zlatko]

Use QScreen instead QDesktopWidget

[wysota]

Thanks, this seem works; but return height 704 instead of 768; the task Bar or window isn't included in height...How can I do?



QDesktopWidget* qw;
QRect rect = qw->screenGeometry(0);
std::cout << "Screen Width: " << rect.width() << endl;
std::cout << "Screen Height: " << rect.height() << endl;

The docs say:


const QRect & QDesktopWidget::availableGeometry ( int screen = -1 ) const
Returns the available geometry of the screen with index screen. What is available will be subrect of screenGeometry() based on what the platform decides is available (for example excludes the Dock and Menubar on Mac OS X, or the taskbar on Windows).
See also screenNumber() and screenGeometry().