如何处理AWT框架中的关闭按钮

问题描述 投票:0回答:2

我开始介绍一些基本的Java2D示例 - 特别是在AWT下 - 遵循“计算机图形学Java2D / Java3D简介”一书的第一个例子我编写了以下两个类,一个作为“主”类:

package com.example.test;

public class Test {

    public static void main( String[] args ) {

            MyChart terminal = new MyChart();
            terminal.setTitle("The first Java 2D program");
            terminal.setSize(350,80);
            terminal.setVisible(true);

    }

}

另一个扩展AWT框架:

package com.example.test;

import java.awt.Frame;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Insets;
import java.awt.event.WindowAdapter;
import java.awt.event.WindowEvent;

public class MyChart extends Frame {


        public void MyChart() {

            this.addWindowListener(new WindowAdapter(){  
                public void windowClosing(WindowEvent e) {  
                    dispose();  
                }  
            });

        }

        public void paint(Graphics g) {
            Graphics2D g2d =  (Graphics2D) g;
            g2d.drawString("Hello world!", 30, 50);
        };


}

这会在屏幕上生成预期的“Hello World”窗口,但是当我单击关闭按钮(Windows 10)时,没有任何反应。我试图弄清楚在添加监听器时我做错了什么但没有成功。

任何提示或建议甚至进一步调试都受到欢迎。

我在Java下:1.8.0_191和Windows 10如上所述。

java java-8 awt java-2d
2个回答
1
投票

你没有在主类中调用MyChart函数。下面的代码工作正常。

public class Test {

        public static void main( String[] args ) {
            System.out.println("Terminal Group");
            MyChart terminal = new MyChart();
            terminal.MyChart();//I added this.
            terminal.setTitle("The first Java 2D program");
            terminal.setSize(350,80);
            terminal.setVisible(true);
    }

}

public class MyChart extends Frame {
        public void MyChart() {
            this.addWindowListener(new WindowAdapter(){  
                public void windowClosing(WindowEvent e) {  
               dispose();
               }  
            });

        }

        public void paint(Graphics g) {
            Graphics2D g2d =  (Graphics2D) g;
            g2d.drawString("Hello world!", 30, 50);
        };
}

1
投票

构造函数定义错误,void存在且必须删除。

原始代码(WRONG构造函数,使用void):

    public void MyChart() {

        this.addWindowListener(new WindowAdapter(){  
            public void windowClosing(WindowEvent e) {  
                dispose();  
            }  
        });

    }

更正代码(无空格):

public MyChart() {

        this.addWindowListener(new WindowAdapter(){  
            public void windowClosing(WindowEvent e) {  
                dispose();  
            }  
        });

}
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