兩點之間畫正方形

Question

我正在做一些練習，並且我一直堅持這幾個小時（對Java來說很新）。 無論如何，這是我應該做的： 當我運行程序時，屏幕中間會有一個正方形，然後當我點擊該屏幕內的某個位置時，會在我點擊的地方繪製另一個正方形在這兩點之間應該有10個方格。所以無論我點擊哪裏，總是應該在10個方格之間繪製。

但是，我不能讓它正常工作。

這是我已經設法到目前爲止做：

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); 


    } 

}

我敢肯定，我一切都過於複雜，因爲這應該是非常基本的。

最終的結果應該是這樣的： End result

Answer 1

這是我已經解決了它的方式：

我聽不太懂的se.lth.cs.ptdc.square.Square的文檔，但我會假設它根據其左上角的座標和側面大小繪製一個正方形。

因此，您有第一個方形左上角的座標和最後一個方形中心的座標。有，它不是很難得到最後一個方塊的左上角的座標：
lastX = centerX - side/2
lastY = centerY - side/2

之後，你有你找到起點和終點之間的區別：
diffX = posX - lastX
diffY = posY - lastY

，之後只需畫出9個正方形：

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); 
} 

其實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。

Answer 2

更新X和Y在同一時間：

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); 

Answer 3

這裏是你如何在兩個方向同時移動（在對角線）。

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; 
    } 
}