c++ - 鼠标事件选择器openscenegraph

Question

我想使用 OpenSceneGraph Pickhandler 以便在用鼠标单击时打印节点的名称。我制作了一个 PickHandler 头文件，并包含了我认为实现此目的的正确代码。

运行应用程序没有错误后，单击时不会显示节点名称。我错过了什么重要的事情吗？

bool PickHandler::handle( const osgGA::GUIEventAdapter& ea, osgGA::GUIActionAdapter& aa )
{
  `if( ea.getEventType() != osgGA::GUIEventAdapter::RELEASE &&
  ea.getButton()    != osgGA::GUIEventAdapter::LEFT_MOUSE_BUTTON )
  {
return false;
  }

  osgViewer::View* viewer = dynamic_cast<osgViewer::View*>( &aa );

  if( viewer )
  {
osgUtil::LineSegmentIntersector* intersector
    = new osgUtil::LineSegmentIntersector( osgUtil::Intersector::WINDOW, ea.getX(), ea.getY() );`if( ea.getEventType() != osgGA::GUIEventAdapter::RELEASE &&
  ea.getButton()    != osgGA::GUIEventAdapter::LEFT_MOUSE_BUTTON )
 {
return false;
}

osgViewer::View* viewer = dynamic_cast<osgViewer::View*>( &aa );

if( viewer )
{
osgUtil::LineSegmentIntersector* intersector
    = new osgUtil::LineSegmentIntersector( osgUtil::Intersector::WINDOW, ea.getX(), ea.getY() );

osgUtil::IntersectionVisitor iv( intersector );

osg::Camera* camera = viewer->getCamera();
if( !camera )
  return false;

camera->accept( iv );

if( !intersector->containsIntersections() )
  return false;

auto intersections = intersector->getIntersections();

std::cout << "Got " << intersections.size() << " intersections:\n";

for( auto&& intersection : intersections )
  std::cout << "  - Local intersection point = " << intersection.localIntersectionPoint << "\n";
}

return true;
}

Answer 1

您需要提取节点名称才能打印它。如果您不使用任何自定义节点，则使用 intersection.drawable->getName()。确保为该特定的 osg::Geometry 设置名称，否则默认情况下该名称为空。

您的案例的打印代码类似于:

for( auto&& intersection : intersections ) {
   std::cout << "  - Local intersection point = " << intersection.localIntersectionPoint << "\n";
   std::cout << "Intersection name = " << intersection.drawable->getName() << std::endl; 
}