我使用JavaFX制作了一个国际象棋棋盘,并将所有图块设置为GridPane的子代,是否可以使用GridPane索引访问这些图块并更改其属性?
我试图通过按图块矩阵访问一个图块来更改其颜色属性,但它并没有更改GridPane中显示的图块。
.getChildren()方法返回节点列表,一旦它成为节点,我就无法访问tile对象的方法。
这是我的图块课程:
package chess.Board;
import static chess.Chess.TILE_W;
import static chess.Chess.TILE_Y;
import javafx.scene.paint.Color;
import javafx.scene.shape.Rectangle;
public class Tile extends Rectangle{
private final int width = TILE_W;
private final int height = TILE_Y;
private final int x;
private final int y;
private Color color;
public void setColor(Color color) {
this.color = color;
}
public Color getColor(){
return this.color;
}
public Tile(Color c, int x, int y){
this.color = c;
this.x = x;
this.y = y;
setFill(c);
setHeight(this.height);
setWidth(this.width);
relocate(x*TILE_W, y*TILE_Y);
}
}
这是我的董事会班级:
package chess.Board;
import static chess.Chess.HEIGHT;
import static chess.Chess.TILE_W;
import static chess.Chess.TILE_Y;
import static chess.Chess.WIDTH;
import javafx.scene.Parent;
import javafx.scene.layout.GridPane;
import javafx.scene.paint.Color;
public class Board {
private Tile[][] tilesMatrix;
private void setMatrix(){
Tile[][] tilesCreator = new Tile[WIDTH][HEIGHT];
for(int y = 0; y < HEIGHT; y++){
for(int x = 0; x < WIDTH; x++){
if(y%2 == 0){
if(x%2 == 0)
tilesCreator[x][y] = new Tile(Color.ANTIQUEWHITE, x, y);
else
tilesCreator[x][y] = new Tile(Color.DARKGREEN, x, y);
}
else{
if(x%2 == 0)
tilesCreator[x][y] = new Tile(Color.DARKGREEN, x, y);
else
tilesCreator[x][y] = new Tile(Color.ANTIQUEWHITE, x, y);
}
}
}
this.tilesMatrix = tilesCreator;
}
public Tile[][] getMatrix(){
return this.tilesMatrix;
}
public Parent getBoard(){
GridPane board = new GridPane();
board.setPrefSize(WIDTH*TILE_W, HEIGHT*TILE_Y);
for(int y = 0; y < HEIGHT; y++)
for(int x = 0; x < WIDTH; x++)
board.add(getMatrix()[x][y], x, y);
return board;
}
public Board(){
setMatrix();
}
}
答案 0 :(得分:0)
通过矩阵访问切片是可以的,但是使用getChildren()也是可以的:您只需键入将Node投射到Tile即可。除非更改了Tile的 fill 属性(从Shape继承),否则瓦片的颜色不会改变。 Tile的构造函数会调用setFill(),但不会调用setColor()。