单击Java直接射击对象libgdx

Question

I am trying to get my player to shoot a spell and it travels to where ever the player clicked. I can easily accomplish this by doing the following.

    if(position.x >= destination.x - 1 && position.x <= destination.x + 1)
        reachedX = true;
    if(position.y >= destination.y - 1 && position.y <= destination.y + 1)
        reachedY = true;

However if the players origin is at, for example, 0,0 and I click at 10,300 then it travels right and up but when the spell reaches an x of 10 it travels directly upwards. I want the spell to travel at an angle that it will reach the x coordinate at the same time as the y coordinate. Here is an image showing what happens and what I want to happen.

Answer 1

In the first picture it looks like the spell goes 45° until it reaches the right x-coordinate. This sounds like the x and y speed are equals, no matter where the destination point is.
You shuld instead have a direction and depending on that ax and y speed.
For that you first need to get the point, the player is clicking.
Therefore you can implement InputProcessor and it's touchDown(int screenX, int screenY, int pointer, int button) .
The screenX and screenY arguments are given in screen coordinates (pixels) and therefore need to be converted to your world-coordinates. This can be done using the camera or the viewport and it's unproject(Vector2 screenCoords) . This method returns a Vector2 giving the world-coordinates.
Now you need to find out the direction Vector2 between you and the clicked point. The direction Vector is calculated like this:

new Vector2(otherPos.x - myPos.x, otherPos.y - myPos.y).nor();

This returns the normalized direction Vector between the two points.
Now you only need to move the spell by dir.x*spellSpeed*delta in x-direction and dir.y*spellSpeed*delta in y-direction and it should look like in your second picture.