Javafx 如何在 GridPane 中定位特定按钮

Question

我正在制作带有按钮的西洋跳棋游戏。要杀死另一块，你必须将你的块沿对角线移动到另一块上，但我不确定如何确保你的块移动到另一块上。我解决这个问题的想法是获取第二个按钮的行和列，这是你的作品移动到的按钮，然后从每个行和列中减去 1，然后从该按钮获取文本以测试它是否是“黑色”或“红色”。第一和第二 = 按钮System.out.println((GridPane.getColumnIndex(second) + " vs " + (GridPane.getColumnIndex(second) - 1)));    if (GridPane.getColumnIndex(second) > 0) {        System.out.println("checking if a button has been jumped");        GridPane.setRowIndex(second, (GridPane.getRowIndex(second) - 1));        GridPane.setColumnIndex(second, (GridPane.getColumnIndex(second) - 1));        System.out.println("this is a printing of the second button name for location " + (GridPane.getColumnIndex(second)) + " " + (GridPane.getRowIndex(second)) + " " + second.getText());        if (second.getText().contains("black")) {            System.out.println("it's a kill");        }         else {            System.out.println("no kill");            GridPane.setRowIndex(second, (GridPane.getRowIndex(second) + 1));            GridPane.setColumnIndex(second, (GridPane.getColumnIndex(second) + 1));        }    }我可以将行和列更改为与另一部分的位置匹配的内容，但是当我从该按钮（第二个）获取文本时，它不会以名称“黑色”或“红色”返回，而只是空白按钮的名称。我的猜测是 GridPane 可能不会像这样工作，我只需要想出另一个解决方案，希望我不必将整个代码重做为二维数组或其他东西。

Answer 1

GridPane您可以通过遍历子节点来获取对节点的引用GridPane。
添加一些简单的计算来找到被点击的两个按钮之间的对角线，如果有的话：

import java.awt.Toolkit;

import javafx.animation.PauseTransition;

import javafx.application.Application;

import javafx.geometry.Insets;

import javafx.scene.Node;

import javafx.scene.Scene;

import javafx.scene.control.Button;

import javafx.scene.layout.GridPane;

import javafx.scene.layout.Pane;

import javafx.scene.layout.VBox;

import javafx.scene.text.Text;

import javafx.stage.Stage;

import javafx.util.Duration;

public class FxMain extends Application {

    private static final int COLS = 5, ROWS = 5;

    private int clickCounter = 0;

    private GridPane grid;

    private Button first, second;

    @Override

    public void start(Stage primaryStage){

        VBox root = new VBox(10);

        root.setPadding(new Insets(10));

        root.getChildren().addAll(makeGrid(), 

                new Text("Click 2 buttons to find the \n diagonally between them"));

        primaryStage.setScene(new Scene(root));

        primaryStage.sizeToScene();

        primaryStage.show();

    }

    private Pane makeGrid() {

        grid = new GridPane();

        for(int rowIndex = 0; rowIndex < ROWS ; rowIndex++) {

            //an array to hold buttons of one row

            Node[] nodes = new Node[COLS];

            for(int colIndex = 0; colIndex < COLS ; colIndex++) {

                Button node= new Button(rowIndex+""+colIndex);

                node.setOnAction(e->buttonCliked(node)); //add action listener

                nodes[colIndex]= node;

            }

            grid.addRow(rowIndex, nodes);

        }

        return grid;

    }

    private void buttonCliked(Button button) {

        if(clickCounter == 0){

            first = button;

        }else{

            second = button;

            markNode(findMidDiagonalButton());

        }

        System.out.println(clickCounter + " " + button.getText()    );

        clickCounter=  ++clickCounter %2 ;  // changes values between 0 1

    }

    //change node background for a short while, and then reset it

    private void markNode(Node node) {

        if(node == null) return;

        String style = node.getStyle();

        node.setStyle("-fx-background-color: cornflowerblue;");

        PauseTransition pause = new PauseTransition(Duration.seconds(1));

        pause.play();

        pause.setOnFinished(e-> node.setStyle(style));

    }

    private Node findMidDiagonalButton() {

        int rowDelta = GridPane.getRowIndex(first) - GridPane.getRowIndex(second);

        int colDelta = GridPane.getColumnIndex(first) - GridPane.getColumnIndex(second);

        if( Math.abs(rowDelta) != 2 ||  Math.abs(colDelta) != 2 ){

            Toolkit.getDefaultToolkit().beep();

            return null;

        }

        int rowsSum = GridPane.getRowIndex(first) + GridPane.getRowIndex(second);

        int colsSum = GridPane.getColumnIndex(first) + GridPane.getColumnIndex(second);

        return  getNodeByRowCol(Math.abs(rowsSum / 2), Math.abs(colsSum / 2) );

    }

    public Node getNodeByRowCol (int row, int col) {

        for (Node node : grid.getChildren()) {

            if(GridPane.getRowIndex(node) == row && GridPane.getColumnIndex(node) == col)

                return node;

        }

        return null;

    }

    public static void main(final String[] args) {

        launch(args);

    }

}

Answer 2

所以这是可能的，我从这篇文章中找到了答案。他只要自己想办法，就能找到具体的位置。 javafx GridPane 检索特定单元格内容

private Node getNodeFromGridPane(GridPane gridPane, int col, int row) {

for (Node node : gridPane.getChildren()) {

    if (GridPane.getColumnIndex(node) == col && GridPane.getRowIndex(node) == row) {

        return node;

    }

}

return null;

}

虽然，为了我自己的目的，我仍然需要弄清楚如何能够测试节点是否包含“红色”或“黑色”，所以我只是添加了这个，现在一切正常！

private Boolean getNodeFromGridPane(GridPane gridPane, int col, int row) {

    for (Node node : gridPane.getChildren()) {

        if (GridPane.getColumnIndex(node) == col && GridPane.getRowIndex(node) == row) {

            if (node.toString().contains("black")) {

                System.out.println("The second button is black = " + node.toString().contains("black"));

                return true;

            }

            if (node.toString().contains("red")) {

                System.out.println("The second button is red = " + node.toString().contains("red"));

                return true;

            }

        }

    }

    return false;

}