java - JPanel 的扩展类未显示 JPanel 的属性

Question

我正在使用一个扩展 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);
}

当您覆盖 paintComponent() 方法时，您应该始终调用 super.paintComponent()。

Answer 2

之所以更改，是因为当您覆盖时paintComponent，您必须始终将super.paintComponent(g)其作为第一行调用：

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

超类中的paintComponent方法JPanel绘制背景，所以如果你 insert super.paintComponent(g)，背景将在你绘制任何自定义之前绘制。