如何在Java的教科书字段中验证用户名

Question

所以我想验证用户在游戏进行前输入的用户名(长度应该在6到30个字符之间，不允许使用数字(，我在互联网上搜索了一下regex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，或者在单击"时包含不允许的字符；开始游戏"；按钮，我想弹出一个对话框，告诉他再试一次。如果一切都是正确的，游戏应该照常进行，但当我运行此代码时，用户名没有被选中，它会进入游戏，尽管在文本框中输入了什么，下面是我到目前为止所拥有的。

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

这两行来自您在问题中发布的代码：

String field;
field = textbox_name.getText();

需要在方法actionPerformed()内，因为您想在用户点击按钮btnStartGame后获得textbox_name的内容。

这是我对你发布的代码的改写。

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);
}

顺便说一下，您应该尝试使用布局管理器。

Answer 2

我尝试将isValidUsername作为一个独立的方法，但它没有给出有效的结果。即使是有效的名字，它也会给出错误的名字。我把它改成了下面的，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，由于代码是在点击按钮btnStartGame后执行的，因此该按钮不可能被禁用(禁用的按钮不会触发事件(。因此，我认为检查if(btnStartGame.isEnabled(((是多余的(我可能错了(。因此，您只需检查：

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