Javafx如何通过tilePane移动对象

时间:2015-01-19 17:31:16

标签: javafx

我有一个tilePanes网格,其中对象(动物)随机放置在它上面,作为图像。在它们实际移动之前,我需要找到一种方法来检查该特定单元格(北,南,东,西)旁边的四个槽/单元格,以查看其中是否有食物来源,如果为真,请移至细胞。如果为false则尝试下一个方向,或者如果全部为假,则只需随机移动。

此刻他们只是随意移动,如果幸运的是,细胞上有食物来源,他们就会吃。这就是我现在拥有的,它确实有用

private void makeAnimalsMove() {
    Random random = new Random();

    // Mark all animals that they haven't moved yet
    for (Slot[] row : slots) {
        for (Slot slot : row) { 
            for (Animal animal : slot.getAnimals()) { 
                animal.setMoved(false); 
            }
        }
    }

    // Now we move only those who needs to be moved
    for (int row = 0; row < slots.length; row++) {
        for (int column = 0; column < slots[row].length; column++) {
            final Slot slot = slots[row][column];

            for (final Animal animal : slot.getAnimals()) {
                if (animal.hasMoved()) {
                    continue;
                }

                int[][] directions = {
                    {row - 1, column}, // north
                    {row, column + 1}, // east
                    {row + 1, column}, // south
                    {row, column - 1}, // west
                };

                int[] selectedDirection = directions[random.nextInt(directions.length)];

                // Move the animal to the chosen direction if possible
                final int rowDirection = selectedDirection[0];
                final int columnDirection = selectedDirection[1];

                if (rowDirection >= 0 && rowDirection < slots.length && columnDirection >= 0 && columnDirection < slots[rowDirection].length) {
                    Platform.runLater(new Runnable() { 
                    @Override
                    public void run() { 
                        slot.removeObject(animal);
                        slots[rowDirection][columnDirection].addObject(animal);
                        }
                    });
                }

                // Decrease the animal's life
                animal.setMoved(true);
                animal.setLifeSpan(animal.getLifeSpan() - 1);
            }
        }
    }
}

对于'吃'部分有一个单独的方法,如果细胞含有食物来源,将被称为。我只是不确定如何在移动之前检查四个单元格?

1 个答案:

答案 0 :(得分:0)

为了我的解决方案,我建议您从TilePane切换到GridPane

您可以轻松绘制元素的网格:

GridPane grid = new GridPane();
grid.add(child, columnIndex, rowIndex);

稍后会检测特定单元格中是否存在某些内容,例如使用此帮助程序:

private static boolean isCellOccupied(GridPane gridPane, int column, int row)
{
    return gridPane
            .getChildren()
            .stream()
            .filter(Node::isManaged)
            .anyMatch(
                    n -> Objects.equals(GridPane.getRowIndex(n), row)
                            && Objects.equals(GridPane.getColumnIndex(n),
                                    column));
}