这个actionPerformed如何与我的按钮交互?

时间:2013-01-22 23:55:23

标签: java swing button

我试图让用户选择他们想要在我的GUI上绘制的形状。我有一系列按钮:圆形,方形和矩形。我的actionListener工作,因为它将一个字符串打印到我的控制台,但它不会在我的GUI上显示形状。如何使用actionCommand在我的面板上绘制该形状。

public void paintComponent(Graphics g) {
    g2D = (Graphics2D) g;
    //Rectangle2D rect = new Rectangle2D.Double(x, y, x2-x, y2-y);
    //g2D.draw(rect);
      repaint();
}

public void actionPerformed(ActionEvent arg0) { 
    if(arg0.getActionCommand().equals("Rect")){     
        System.out.println("hello");
        Rectangle2D rect = new Rectangle2D.Double(x, y, x2-x, y2-y);
        g2D.draw(rect); //can only be accessed within paintComponent method
        repaint();
    }

2 个答案:

答案 0 :(得分:3)

如果您首先绘制矩形,然后要求重新绘制,矩形将消失。

您应该将新形状存储在临时变量中并将其渲染到paintComponent中。

private Rectangle2D temp;

// inside the actionPerformed
    temp = new Rectangle2D.Double(x, y, x2-x, y2-y);
    repaint();

// inside the paintComponent
    if(temp != null) {
        g2D.draw(temp);
    }

答案 1 :(得分:2)

使rect成为字段nto局部变量。在actionPerformed中创建正确的rect并调用repaint()。然后将调用paintComponent()。它应该是这样的

public void paintComponent(Graphics g) {
    g2D = (Graphics2D) g;
    g2D.draw(rect);
}