[英]Draw squares between two points
我正在做一些練習,現在我已經堅持了幾個小時(對Java來說很新)。 無論如何,這是我應該做的:當我運行程序時,我將在屏幕中間有一個正方形,然后當我點擊該屏幕中的某個地方時,將在我點擊的地方繪制另一個正方形 - 在這兩點之間應該有10個方格。 所以無論我在哪里點擊,都應該總是畫10個方格。
但是,我無法讓它正常運作。
這是我到目前為止所做的事情:
import se.lth.cs.ptdc.window.SimpleWindow;
import se.lth.cs.ptdc.square.Square;
public class PrintSquares2 {
public static void main(String[] args) {
SimpleWindow w = new SimpleWindow(600, 600, "PrintSquares2");
int posX = 300;
int posY = 300;
int loop = 0;
System.out.println("Skriv rotation");
Square sq1 = new Square(posX,posY,200);
sq1.draw(w);
w.waitForMouseClick();
int destX = w.getMouseX();
int destY = w.getMouseY();
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
SimpleWindow.delay(10);
//sq1.erase(w);
int jumpX = (destX - posX) / 10;
int jumpY = (destY - posY) / 10;
System.out.println(jumpX);
while (posX < destX)
{
posX = posX+10;
SimpleWindow.delay(100);
loop++;
System.out.println("Loop: " + loop);
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
Square sq2 = new Square(posX,posY,200);
sq2.draw(w);
}
while (posX > destX)
{
posX = posX-10;
SimpleWindow.delay(100);
loop++;
System.out.println("Loop: " + loop);
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
sq1.draw(w);
Square sq2 = new Square(posX,posY,200);
sq2.draw(w);
}
while (posY < destY)
{
posY = posY+10;
SimpleWindow.delay(100);
loop++;
System.out.println("Loop: " + loop);
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
sq1.draw(w);
Square sq2 = new Square(posX,posY,200);
sq2.draw(w);
}
while (posY > destY)
{
posY = posY-10;
SimpleWindow.delay(100);
loop++;
System.out.println("Loop: " + loop);
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
sq1.draw(w);
Square sq2 = new Square(posX,posY,200);
sq2.draw(w);
}
SimpleWindow.delay(10);
sq1.draw(w);
//SimpleWindow.clear(w);
}
}
我很確定我過於復雜,因為這應該是非常基本的。
最終結果應如下所示: 最終結果
這是我解決它的方式:
我對se.lth.cs.ptdc.square.Square
上的文檔並不十分了解,但我會假設它根據左上角的坐標和邊長來繪制一個正方形。
所以你有第一個方格左上角的坐標和最后一個方格中心的坐標。 得到最后一個方格的左上角的坐標並不困難:
lastX = centerX - side/2
lastY = centerY - side/2
完成后,您會發現起點和終點之間的差異:
diffX = posX - lastX
diffY = posY - lastY
然后再抽出9個方塊:
for (int i=1; i<10; i++){
squareX = posX + (diffX/10)*i;
squareY = posY + (diffY/10)*i;
Square square = new Square(squareX,squareY,200);
square.draw(w);
}
實際上你做的第一部分是正確的,只是搞砸了那些不必要的檢查。 希望能幫助到你。
-
問候,svz。
同時更新X和Y:
int jumpX = (destX - posX) / 10;
int jumpY = (destY - posY) / 10;
if (posX > destX) {
int temp = destX;
destX = posX;
posX = temp;
}
while (posX <= destX)
{
SimpleWindow.delay(100);
loop++;
System.out.println("Loop: " + loop);
System.out.println("Dest X: " + destX + " Dest Y: " + destY);
System.out.println("Pos X: " + posX + " Pos Y: " + posY);
Square sq2 = new Square(posX,posY,200);
sq2.draw(w);
posX = posX+jumpX;
posY = posY+jumpY;
}
SimpleWindow.delay(10);
sq1.draw(w);
這是你一次向兩個方向移動(在對角線上)。
static final int Steps = 10;
private void test() {
int x1 = 100;
int y1 = 100;
int x2 = 300;
int y2 = 500;
double dx = (double)(x2 - x1) / (double) Steps;
double dy = (double)(y2 - y1) / (double) Steps;
double x = x1;
double y = x2;
for ( int i = 0; i < Steps; i++) {
// Simulate the drawing of the square.
System.out.println("("+x+","+y+")");
x += dx;
y += dy;
}
}
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