如何验证 Java 教科书字段中的用户名？

Question

所以我想验证用户在游戏继续之前输入的用户名（长度应在 6 到 30 个字符之间，并且不应允许数字），我在互联网上搜索并发现了正则表达式。我不习惯，所以我发现了一些问题。代码应该执行以下操作：如果用户没有输入用户名，或者当他单击“开始游戏”按钮时包含不允许的字符，我希望弹出一个对话框并告诉他再试一次。如果一切正常，游戏应该会正常进行，但是当我运行此代码时，不会检查用户名，尽管在文本框中输入了内容，但它会进入游戏，以下是我目前所拥有的。

import java.awt.EventQueue;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.border.EmptyBorder;
import javax.swing.JLabel;
import javax.swing.JOptionPane;

import java.awt.Font;
import javax.swing.JTextField;
import javax.swing.KeyStroke;
import javax.swing.JButton;
import java.awt.event.ActionListener;
import java.awt.event.KeyEvent;
import java.awt.event.ActionEvent;
import java.util.regex.*;

public class StartScreen extends JFrame {

private JPanel contentPane;
private JTextField textbox_name;

/**
 * Launch the application.
 */
public static void main(String[] args) {
    EventQueue.invokeLater(new Runnable() {
        public void run() {
            try {
                StartScreen frame = new StartScreen();
                frame.setVisible(true);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    });
}

/**
 * Create the frame.
 */

public static boolean isValidUsername(String name) 
{ 
    String regex = "^[a-zA-Z]{5,29}$";  //Removed ^[aA-zZ]\\\\w{5,29}$
    Pattern p = Pattern.compile(regex);     
    if (name == null) { 
        return false; 
    }   
    Matcher m = p.matcher(name);    
    return m.matches(); 
 } 

public StartScreen() {
    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    setBounds(100, 100, 455, 191);
    contentPane = new JPanel();
    contentPane.setBorder(new EmptyBorder(5, 5, 5, 5));
    setContentPane(contentPane);
    contentPane.setLayout(null);
    
    JLabel lblEnterName = new JLabel("Enter Name:");
    lblEnterName.setFont(new Font("Lucida Grande", Font.PLAIN, 15));
    lblEnterName.setBounds(91, 48, 102, 29);
    contentPane.add(lblEnterName);
    
    textbox_name = new JTextField();
    textbox_name.setBounds(205, 49, 130, 28);
    contentPane.add(textbox_name);
    textbox_name.setColumns(10);
    
    String field;
    field = textbox_name.getText();
    
    JButton btnStartGame = new JButton("Start Game");
    btnStartGame.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            
            if(isValidUsername(field)) {
                MainScreen window = new MainScreen();
                window.setVisible(true);
                dispose();
            }
            else {
                JOptionPane.showMessageDialog(null, "Please enter your username correctly");
            }
        }
    });
    
    
    
    btnStartGame.setBounds(150, 102, 117, 29);
    contentPane.add(btnStartGame);
}

}

Answer 1

您在问题中发布的代码中的这两行：

String field;
field = textbox_name.getText();

需要inside方法actionPerformed()，因为您想在用户点击按钮btnStartGame之后获取textbox_name的内容。

这是我对您发布的代码的重写。

import java.awt.EventQueue;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.border.EmptyBorder;
import javax.swing.JLabel;
import javax.swing.JOptionPane;

import java.awt.Font;
import javax.swing.JTextField;
import javax.swing.KeyStroke;
import javax.swing.JButton;
import java.awt.event.ActionListener;
import java.awt.event.KeyEvent;
import java.awt.event.ActionEvent;
import java.util.regex.*;

public class StartScreen extends JFrame {

private JPanel contentPane;
private JTextField textbox_name;

/**
 * Launch the application.
 */
public static void main(String[] args) {
    EventQueue.invokeLater(new Runnable() {
        public void run() {
            try {
                StartScreen frame = new StartScreen();
                frame.setVisible(true);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    });
}

/**
 * Create the frame.
 */

public static boolean isValidUsername(String name) 
{ 
    String regex = "^[a-zA-Z]{5,29}$";  //Removed ^[aA-zZ]\\\\w{5,29}$
    Pattern p = Pattern.compile(regex);     
    if (name == null) { 
        return false; 
    }   
    Matcher m = p.matcher(name);    
    return m.matches(); 
 } 

public StartScreen() {
    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    setBounds(100, 100, 455, 191);
    contentPane = new JPanel();
    contentPane.setBorder(new EmptyBorder(5, 5, 5, 5));
    setContentPane(contentPane);
    contentPane.setLayout(null);
    
    JLabel lblEnterName = new JLabel("Enter Name:");
    lblEnterName.setFont(new Font("Lucida Grande", Font.PLAIN, 15));
    lblEnterName.setBounds(91, 48, 102, 29);
    contentPane.add(lblEnterName);
    
    textbox_name = new JTextField();
    textbox_name.setBounds(205, 49, 130, 28);
    contentPane.add(textbox_name);
    textbox_name.setColumns(10);
    
    JButton btnStartGame = new JButton("Start Game");
    btnStartGame.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            String field;  // added this line
            field = textbox_name.getText();  // added this line
            
            if(isValidUsername(field)) {
                MainScreen window = new MainScreen();
                window.setVisible(true);
                dispose();
            }
            else {
                JOptionPane.showMessageDialog(null, "Please enter your username correctly");
            }
        }
    });
    
    btnStartGame.setBounds(150, 102, 117, 29);
    contentPane.add(btnStartGame);
}

顺便说一句，你应该尝试使用layout managers。

Answer 2

我尝试将您的 isValidUsername 作为独立方法，但没有给出有效结果。即使对于有效名称，它也是错误的。我将其更改为以下内容，并且效果很好：

public static boolean isValidUsername(String name) 
{ 
    String regex = "^[a-zA-Z]{5,29}$";  //Removed ^[aA-zZ]\\\\w{5,29}$
    Pattern p = Pattern.compile(regex);     
    if (name == null) { 
        return false; 
    }   
    Matcher m = p.matcher(name);    
    return m.matches(); 
 } 

此外，由于代码是在按钮 btnStartGame 被单击后执行的，因此按钮不可能被禁用（禁用的按钮不会触发事件）。出于这个原因，我认为检查 if(btnStartGame.isEnabled()) 是多余的（我可能错了）。因此，您可以检查：

if(sValidUsername(field)){
  //start game here
} else {
  //display prompt here
}