在java中实现一个定时器

Question

import java.awt.Graphics; 
import javax.swing.JFrame; 
import javax.swing.JPanel; 
import javax.swing.JLabel; 
import java.awt.*; 
import javax.swing.*; 
import java.util.Timer; 
import java.awt.event.*; 
import java.awt.event.*; 
import javax.swing.*; 

class autos extends JLabel 
{ 
    @SuppressWarnings("serial") 
    int z=100,i=50; 
    public static void main(String[] args) 
    { 

     JFrame frame=new JFrame("Rectangle"); 
     frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
     frame.setVisible(true); 
     frame.setSize(1000,1000); 
     frame.add(new autos());  
    } 

    @Override 
    public void paintComponent(Graphics g) 
    { 
     for(i=1;i<=7;i++) 
     { 
      g.drawRect(z,100,100,100); 
      z=z+120; 
      //timer delay 
     } 

    } 
} 

你好，我试图创建一个Java程序，一个接一个地拖延（不是所有的一次）绘制mre矩形。

由于睡眠和TimeUnit将冻结paintComponent，我有点无知。我试图用一个计时器来延迟，但我失败了。我不明白如何使用计时器在这种情况下。

如何在矩形之间进行时间延迟？

Answer 1

你应该考虑看看How to use Swing Timers和Concurrency in Swing

你开始将Swing Timer想象为伪循环，每当它跳动时，就会更新一些值，当您调用repaint和您的paintComponent方法时，您可以适当地更新UI

import java.awt.Dimension; 
import java.awt.EventQueue; 
import java.awt.Graphics; 
import java.awt.Graphics2D; 
import java.awt.event.ActionEvent; 
import java.awt.event.ActionListener; 
import java.awt.event.MouseAdapter; 
import java.awt.event.MouseEvent; 
import javax.swing.JFrame; 
import javax.swing.JPanel; 
import javax.swing.Timer; 
import javax.swing.UIManager; 
import javax.swing.UnsupportedLookAndFeelException; 

public class Test { 

    public static void main(String[] args) { 
     new Test(); 
    } 

    public Test() { 
     EventQueue.invokeLater(new Runnable() { 
      @Override 
      public void run() { 
       try { 
        UIManager.setLookAndFeel(UIManager.getSystemLookAndFeelClassName()); 
       } catch (ClassNotFoundException | InstantiationException | IllegalAccessException | UnsupportedLookAndFeelException ex) { 
        ex.printStackTrace(); 
       } 

       JFrame frame = new JFrame("Testing"); 
       frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
       frame.add(new TestPane()); 
       frame.pack(); 
       frame.setLocationRelativeTo(null); 
       frame.setVisible(true); 
      } 
     }); 
    } 

    public class TestPane extends JPanel { 

     private int count = 0; 

     public TestPane() { 
      Timer timer = new Timer(500, new ActionListener() { 
       @Override 
       public void actionPerformed(ActionEvent e) { 
        count++; 
        if (count == 7) { 
         ((Timer)e.getSource()).stop(); 
        } 
        repaint(); 
       } 
      }); 
      addMouseListener(new MouseAdapter() { 
       @Override 
       public void mouseClicked(MouseEvent e) { 
        timer.stop(); 
        count = 0; 
        timer.start(); 
       } 
      }); 
     } 

     @Override 
     public Dimension getPreferredSize() { 
      return new Dimension((120 * 7) + 100, 300); 
     } 

     @Override 
     protected void paintComponent(Graphics g) { 
      super.paintComponent(g); 
      Graphics2D g2d = (Graphics2D) g.create(); 
      super.paintComponent(g); 
      int x = 100; 
      for (int rect = 1; rect <= count; rect++) { 
       g2d.drawRect(x, 100, 100, 100); 
       x += 120; 
       //timer delay 
      } 
      g2d.dispose(); 
     } 
    } 
} 

PS-单击面板，以得到定时器启动

Answer 2

你并不需要一个循环做到这一点job.You可以使用摆动计时器是这样的：

int delay=500;// delay time in ms 
Timer t=new Timer(delay,new ActionListener() { 

    @Override 
    public void actionPerformed(ActionEvent e){ 
     autos.this.getGraphics().drawRect(z,100,100,100); // draw rectangle in your panel's graphics 
     repaint(); // this line calls paint component to show changes 
     i++; 
     z+=120; 
     if (i>70) t.stop(); // for loop's condition 
    } 
});