0

所以我创建了一个程序，它的目的是发送随机击键。现在我已经实现了一个带有开始/停止按钮的非常基本的 GUI。该程序是在整数“running”等于1时运行的。所以我决定让按钮将running的值从0更改为1，开始循环。

这是代码：

public class AutoKeyboard extends JFrame {

public static int running = 0; // program will not run until this is 1
Random r = new Random();

public static int randInt(int min, int max) { // returns random number
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}

private JLabel label;
private JButton button;

public AutoKeyboard() {
setLayout(new FlowLayout());
label = new JLabel("Not Running");
add(label);


button = new JButton("Start");
add(button);

event f = new event();
button.addActionListener(f);
}


public class event implements ActionListener {

    public void actionPerformed(ActionEvent f) {
        label.setText("Running");
        System.out.println("Running");
        running = 1; // changes running to 1? but doesn't start the program?
}
}


public static void main(String[] args) throws InterruptedException {

AutoKeyboard gui = new AutoKeyboard();
gui.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
gui.setSize(180, 80);
gui.setVisible(true);
gui.setResizable(false);
gui.setTitle("Anti AFK");

    while (running == 1) { // if running is 1, do this
try { 

int delay = randInt(4864,7834); // 336415, 783410 15 97
Robot robot = new Robot(); 
int keypress = randInt(65, 86);

Thread.sleep(delay);
robot.keyPress(keypress);

} catch (AWTException e) { 
e.printStackTrace(); 
} 
} 
}
}

我的问题是，每当我按下我的 JButton“开始”时，它似乎并没有将 int running 更改为 1，并且程序没有启动。每当我在代码中手动更改 int 时，程序都会运行。所以问题是 JButton 没有更新变量。为什么？我真的很困惑。

感谢大家阅读。

4

3 回答 3

1

试试这个代码

import java.awt.AWTException;
import java.awt.FlowLayout;
import java.awt.Robot;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Random;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;

public class NewClass1 extends JFrame {

public int running; // program will not run until this is 1
Random r = new Random();
public void performtast(){
     while (running == 1) { // if running is 1, do this
try { 

int delay = randInt(4864,7834); // 336415, 783410 15 97
Robot robot = new Robot(); 
int keypress = randInt(65, 86);

Thread.sleep(delay);
robot.keyPress(keypress);

}       catch (InterruptedException ex) { 
            Logger.getLogger(NewClass1.class.getName()).log(Level.SEVERE, null, ex);
        } catch (AWTException e) { 
e.printStackTrace(); 
} 
}
}
public static int randInt(int min, int max) {
    // returns random number
    Random rand = new Random();
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}

private JLabel label;
private JButton button;

public NewClass1() {
setLayout(new FlowLayout());
label = new JLabel("Not Running");
add(label);


button = new JButton("Start");
add(button);

event f = new event();
button.addActionListener(f);
}


public class event implements ActionListener {
 @Override
    public void actionPerformed(ActionEvent f) {
        label.setText("Running");
        System.out.println("Running");
        running = 1; // changes running to 1? but doesn't start the program?
        performtast();
}
}




public static void main(String[] args) throws InterruptedException, AWTException {

NewClass1 gui = new NewClass1();
gui.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
gui.setSize(180, 80);
gui.setVisible(true);
gui.setResizable(false);
gui.setTitle("Anti AFK");

}
}
于 2013-10-12T12:27:30.483 回答
0

您必须通知您的代码运行为 1

 running = 1; // changes running to 1? but doesn't start the program?

之后您正在制作1 但不运行代码。

你要做的是

public void actionPerformed(ActionEvent f) {
        label.setText("Running");
        System.out.println("Running");
        running = 1; // changes running to 1? but doesn't start the program?
        //ok everything set, now do my task.
       //  now call the method here to do your task, which uses *running * 
}
于 2013-10-12T12:17:43.247 回答
0

没有方法告诉程序它应该运行，例如 checkForPaused()。

public static void checkForPaused() {
    synchronized (GUI_INITIALIZATION_MONITOR) {
        while (isPaused()) {
            try {
                GUI_INITIALIZATION_MONITOR.wait();
            } catch (Exception e) {

            }
        }   
    }        
}
于 2013-10-12T12:21:37.130 回答