How to generate an event for "both the mouse buttons clicked together"?

[seguprasad]

Hi All,

I have a requirement like following,
I need to create a box by clicking both the mouse buttons together, hold and drag should create a box on the plot.

Can you please provide me the solution that I can catch the event for "both the mouse buttons clicked to gether" in Qt? I am not able find this in QEvent or QMouseEvent.

Thanks,
Segu

[babu198649]

//QMouseEvent * mouseEvent
if( (mouseEvent->button() == Qt::LeftButton) &&
(mouseEvent->button() =Qt::RightButton))
{
//your code
}

also see this function
Qt::MouseButtons QMouseEvent::buttons () const


Hope this helps

[wysota]

No, it won't work. button() will return only one of the buttons. Using QMouseEvent::buttons() seems better though.

[babu198649]

the right click followed by leftclick(vice-versa) where the former is still active .does this not work

in other words

if( (mouseEvent->button() == Qt::LeftButton)
{
if(mouseEvent->button() == Qt::RightButton))
{
//your code
}
}

[wysota]

I don't know what you mean But your code is faulty - you are missing one equality sign ('=') before Qt::RightButton.

[babu198649]

ok thanks for pointing out error i will try and will post my result i have edited the bug in the previous post.

[wysota]

To me the proper way of doing what the thread author wants is:

void XXX::mousePressEvent(QMouseEvent *e){
  if(e->buttons() & (Qt::LeftButton|Qt::RightButton))
    doSomething();
  else
    baseClass::mousePressEvent(e);
}

void XXX::mousePressEvent(QMouseEvent *e){
  if(e->buttons() & (Qt::LeftButton|Qt::RightButton))
    doSomething();
  else
    baseClass::mousePressEvent(e);
}

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One might also use == instead of & depending on the effect one wants to obtain.

[babu198649]

i tried u r code it enters if condition even if one button is clicked .but the author of post wants only when both the buttons are active.
this can be done by

if(event->buttons() == (Qt::LeftButton | Qt::RightButton))
code();

i have tested it and it works

but one question why do we use or operator and that too unary.