在一行中绘制多个矩形

Question

我正在尝试制作一个游戏，如果球碰到一个矩形，则矩形的值会下降，一旦达到0，它将消失。 我尚未实现计数变量，但遇到了一个问题，我试图在顶部绘制一列数组，但是当我尝试绘制多个矩形并将它们留在此处时，我的循环仅渲染一个对象球会对此做出反应。 我该怎么办？

package graphics;

import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;

public class Ballz extends LASSPanel
{
//Variables for the circle (Global Variables)

int circleX, circleY, circleSize;
double ballDirection,circleDX, circleDY;


//Colors
Color circleColor;
Color backgroundColor;
Dimension size;

int turn = 1;
int click =0;
int go =0;
int move = 0;
boolean roundEnd;



//Variables for the map
int brickWH;
int[] rect = new int[12];



public Ballz()
{
    //Ball Variables
    size = new Dimension(0,0);


    circleX = 190;
    circleY = 545;
    circleDX = 0;
    circleDY=0;

    circleSize = 15;

    circleColor = new Color (245,245,245);
    backgroundColor = new Color (28,28,28);

    //Game Variables
    roundEnd = true;

    brickWH = 18;
}

public void update()
{
    //Get size of screen
    getSize(size);




    circleX += circleDX;
    circleY += circleDY;



    //Screen borders
    if (circleX >= (size.width) - circleSize)
    {
        circleDX = -circleDX;
        circleX = size.width-circleSize;

    }
    if (circleX <= 0)
    {
        circleDX = -circleDX;
        circleX = 0;
    }
    if (circleY >= (size.height -circleSize))
    {
        circleDX =0;
        circleDY = 0;
        circleX = 190;
        circleY = 545;
        roundEnd = true;
    }
    if (circleY <= 0)
    {
        circleDY = -circleDY;
        circleY = 0;
    }


    circleX += circleDX;
    circleY += circleDY;


    click = getMouseButton(0);

    if (click ==1)
    {
        circleDY = -5;
        circleDX = -3;
    }


    //Rectangle Loop


    for (int i =0; i<rect.length; i++)
    {
        rect[i] = rect[i] +50;
    }





    repaint();

}

public void paint(Graphics g)
{


    //Game Colors
    g.setColor(circleColor);
    g.fillOval(circleX, circleY, circleSize, circleSize);


    for (int i=0; i<rect.length; i++)
    {
        g.fillRect(rect[i], 2, brickWH, brickWH);

    }

    setBackground(backgroundColor); 







}

}

Answer 1

好吧，在看了更多代码之后，让我们开始...

int[] rect = new int[12];

这将创建一个int数组，该数组最初初始化为全0 s

接下来，您要做...

for (int i =0; i<rect.length; i++)
{
    rect[i] = rect[i] +50;
}

基本上所有这些工作就是将50加0并将其分配回数组元素，因此当您执行此操作时...

for (int i=0; i<rect.length; i++)
{
    g.fillRect(rect[i], 2, brickWH, brickWH);

}

它只是将每个“砖”画在最后一个砖的顶部，因为它们都处于相同的水平位置

对我而言，真正，真正，非常突出的是...

for (int i =0; i<rect.length; i++)
{
    rect[i] = rect[i] +50;
}

您是否打算让每个积木相隔50个像素？ 在这种情况下，您可能应该做更多类似的事情...

int xPos = 0;
for (int i =0; i<rect.length; i++)
{
    rect[i] = xPos;
    xPos += 50;
}

相反，如果假设它们放在一起，那么您想要更多类似的东西...

for (int i =0; i<rect.length; i++)
{
    rect[i] = brickWH * i;
}