下面是我的代码,它在窗口中显示两个图像。
现在通过移动鼠标,两个图像一起移动,但我应该如何分别访问它们?
int main()
{
// Let's setup a window
sf::RenderWindow window(sf::VideoMode(640, 480), "SFML ViewTransformation");
// Let's create the background image here, where everything initializes.
sf::Texture BackgroundTexture;
sf::Sprite bd;
sf::Vector2u TextureSize; //Added to store texture size.
sf::Vector2u WindowSize; //Added to store window size.
if(!BackgroundTexture.loadFromFile("bg1.jpg"))
{
return -1;
}
else
{
TextureSize = BackgroundTexture.getSize(); //Get size of texture.
WindowSize = window.getSize(); //Get size of window.
float ScaleX = (float) WindowSize.x / TextureSize.x;
float ScaleY = (float) WindowSize.y / TextureSize.y; //Calculate scale.
bd.setTexture(BackgroundTexture);
bd.setScale(ScaleX, ScaleY); //Set scale.
}
// Create something simple to draw
sf::Texture texture;
texture.loadFromFile("background.jpg");
sf::Sprite background(texture);
sf::Vector2f oldPos;
bool moving = false;
sf::View view = window.getDefaultView();
while (window.isOpen()) {
sf::Event event;
while (window.pollEvent(event)) {
switch (event.type) {
case sf::Event::Closed:
window.close();
break;
case sf::Event::MouseButtonPressed:
if (event.mouseButton.button == 0) {
moving = true;
oldPos = window.mapPixelToCoords(sf::Vector2i(event.mouseButton.x, event.mouseButton.y));
}
break;
case sf::Event::MouseButtonReleased:
if (event.mouseButton.button == 0) {
moving = false;
}
break;
case sf::Event::MouseMoved:
{
if (!moving)
break;
const sf::Vector2f newPos = window.mapPixelToCoords(sf::Vector2i(event.mouseMove.x, event.mouseMove.y));
const sf::Vector2f deltaPos = oldPos - newPos;
view.setCenter(view.getCenter() + deltaPos);
window.setView(view);
oldPos = window.mapPixelToCoords(sf::Vector2i(event.mouseMove.x, event.mouseMove.y));
break;
}
}
}
window.clear(sf::Color::White);
window.draw(bd);
window.draw(background);
window.display();
}
}
我也很乐意听到建议,在那里我可以找到具有适当见解和良好解释的示例。提前谢谢。
问题是你移动的是视图而不是精灵本身:
view.setCenter(view.getCenter() + deltaPos);
window.setView(view);
相反,您应该移动精灵。所以用这条线交换上面的两行,它应该可以工作
bd.setPosition(bd.getPosition() - deltaPos);