等距屏幕到地图问题 [英] Isometric screen to map issue

问题描述

我知道关于等轴测图的很多建议，但是我已经阅读了大多数建议，但并没有解决我的问题. 我为C#重写了代码，以简化操作(此代码将在Android平台上使用) 我需要将屏幕线接到等轴测坐标上.

I know that about isometric map a lot of advice but I've read most of them and didn't solve my problem. I rewrite code for C# for more simplicity (this code will be used on Android platform) I need to get screen cords to isometric coords.

在这里，我为我使用64:32的1:2瓦片，并使用此代码构建菱形贴图

Here we go I used 1:2 tiles for me 64x32, I build diamond map using this code

private void drawIsoGrid(PaintEventArgs e)
{
    for(int y=0;y<20;y++)
        for(int x=0;x<20;x++)
        {
            float rx = (x - y) * (surface.Width) / 2 - globX;
            float ry = (x + y) * (surface.Height) / 2 - globY;
            e.Graphics.DrawImage(surface,rx,ry);
        }

我还使用全局锚来滚动地图 代码在这里

I also use global anchor for scroll my map code here

protected override void OnMouseMove(MouseEventArgs e)
{
    mouseCoordsX = e.X;
    mouseCoordsY = e.Y;
    if(e.Button==MouseButtons.Left)
    {
        globX += prevX - e.X;
        globY += prevY - e.Y;
        this.Invalidate();
    }
    prevX = e.X;
    prevY = e.Y;            
}

主要问题是如何在鼠标下方平铺哪个公式对我有用.

the main question is how to get tile under the mouse which formula will be useful for me.

推荐答案

由于尚未得到回答，并且这是等距屏幕"的最佳结果之一，因此我想我会回答这个问题(而不是提到我刚刚完成此操作.)

Since this hasn't been answered, and it is one of the top results for "isometric screen" here, I figured I'd answer this (not to mention that I just finished this).

由于您具有一个从iso网格映射到屏幕坐标的函数，并且它是带有反函数的线性变换，因此我们可以向后进行操作以获取另一个函数.因此，让我们做到这一点.

Since you have a function that maps from an iso grid to screen coordinates, and it's a linear transform with an inverse, we can just go backwards to get the other function. So let's do just that.

我们要去:

rx = (x - y) * (surface.Width) / 2 - globX
ry = (x + y) * (surface.Height) / 2 - globY

收件人:

x = <something>
y = <something>

同时解决这些问题最简单.将全局变量添加到两侧:

It's easiest to solve these at the same time. Add the globals to both sides:

rx + globX = (x - y) * (surface.Width) / 2
ry + globY = (x + y) * (surface.Height) / 2

除以(surface.Width) / 2和(surface.Height) / 2:

(rx + globX) / (surface.Width / 2)  = x - y
(ry + globY) / (surface.Height / 2) = x + y

几乎完成了，让我们将两个方程式加在一起以消除y的情况:

Almost done, let's add both equations together to get rid of the y's:

(rx + globX) / (surface.Width / 2) + (ry + globY) / (surface.Height / 2) = 2 * x

现在要摆脱x，从第二个方程中减去第一个方程:

Now to get rid of the x's subtract the first equation from the second:

(ry + globY) / (surface.Height / 2) - (rx + globX) / (surface.Width / 2) = 2 * y

将两个方程式都除以2，我们已初步完成:

Divide both equations through by 2, and we're tentatively done:

x = ((rx + globX) / (surface.Width / 2) + (ry + globY) / (surface.Height / 2)) / 2
y = ((ry + globY) / (surface.Height / 2) - (rx + globX) / (surface.Width / 2)) / 2

很酷，现在您有了屏幕上的网格坐标.让我们清理一下，因为我们基本上具有与a / b相同的(a / (b / c)) / c，所以可以摆脱c的情况，在这种情况下是2的情况:

Cool, now you have the grid coordinates in terms of the screen. Let's clean some of this up, since we have basically (a / (b / c)) / c which is the same as a / b, we can get rid of the c's, in this case the 2's:

x = (rx + globX) / surface.Width + (ry + globY) / surface.Height
y = (ry + globY) / surface.Height - (rx + globX) / surface.Width

因此，您应该能够编写一个函数，该函数获取屏幕的x和y位置并返回x和y网格位置.我对c#并不十分熟悉，所以我不知道它如何处理应该为int的float值，但是由于您是在android上运行的，因此我认为这并不重要.

So you should be able to write up a function that takes the screen x and y positions and returns the x and y grid positions. I am not incredibly familiar with c# so I don't know how it handles float values that should be int's, but since you're running this on android, I suppose it doesn't really matter.

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