扩展的JPanel类没有显示JPanel的属性

Question

我正在使用扩展JPanel的类RefreshablePanel

public class RefreshablePanel extends JPanel {
    static String description="";
    protected void paintComponent(Graphics g){
        g.drawString(description, 10, 11);
}
    void updateDescription(String dataToAppend){    
        description = description.concat("\n").concat(dataToAppend);
       }    
}

JPanel descriptionPanel = new JPanel();
scrollPane_2.setViewportView(descriptionPanel);
descriptionPanel.setBackground(Color.WHITE);
descriptionPanel.setLayout(null);

现在当我这样做时

RefreshablePanel descriptionPanel = new RefreshablePanel();
scrollPane_2.setViewportView(descriptionPanel);
descriptionPanel.setBackground(Color.WHITE);
descriptionPanel.setLayout(null); 

Answer 1

protected void paintComponent(Graphics g){
        super.paintComponent(g);
        g.drawString(description, 10, 11);
}

覆盖paintComponent（）方法时，应始终调用super.paintComponent（）。

Answer 2

这种情况发生变化的原因是，当您覆盖paintComponent时，您必须始终将super.paintComponent(g)作为第一行：

protected void paintComponent(Graphics g) {
    super.paintComponent(g);
    g.drawString(description, 10, 11);
}

paintComponent超类中的JPanel方法绘制背景，因此如果插入super.paintComponent(g)，则在绘制任何自定义内容之前将绘制背景。