关于使用StringTokenizer为Player对象设置姓名的技术咨询

Let's break down the issues in your code first, then walk through how to fix them:

Key Problems in the Current Code #

• NullPointerException waiting to happen: You're initializing StringTokenizer lineTokens = new StringTokenizer(fullname); when fullname is null. StringTokenizer requires a non-null input string, so this will crash immediately.
• Chaotic input flow: You prompt for "Name:" but then read three separate inputs (name, surname, fullname) in sequence—this doesn't match what a user would expect. Plus, you set the player's name to name + surname (no space between them) before even handling the fullname input, making the StringTokenizer loop completely useless.
• Unfinished/incorrect token logic: The while (lineTokens.hasMoreTokens()) block references a tokenizer initialized with a null value, and the code cuts off mid-implementation, leaving invalid syntax.

Corrected Implementation Options #

Option 1: Read Full Name and Split with StringTokenizer #

If you want users to enter their full name in one line (e.g., "John Doe"), this approach is clean and uses StringTokenizer as intended:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PlayerRegistration {
    private static BufferedReader input = new BufferedReader(new InputStreamReader(System.in));

    public static Player newPlayer(Integer playerNum) throws IOException {
        Player player = new Player();
        System.out.println("Player " + (playerNum + 1) + " registration");
        
        System.out.print("Enter full name (first last): ");
        String fullname = input.readLine();
        
        // Initialize StringTokenizer ONLY after we have a valid, non-null fullname
        StringTokenizer lineTokens = new StringTokenizer(fullname);
        
        // Handle edge cases where user might enter only one name
        String firstName = lineTokens.hasMoreTokens() ? lineTokens.nextToken() : "";
        String lastName = lineTokens.hasMoreTokens() ? lineTokens.nextToken() : "";
        
        // Combine with a space for proper formatting
        player.setName(firstName + " " + lastName);
        
        return player;
    }

    // Dummy Player class for context (replace with your actual Player class)
    static class Player {
        private String name;
        public void setName(String name) { this.name = name; }
        public String getName() { return name; }
    }
}

Option 2: Read First and Last Name Separately #

If you want to prompt for first and last name individually (no need for StringTokenizer here), this is a simpler approach:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class PlayerRegistration {
    private static BufferedReader input = new BufferedReader(new InputStreamReader(System.in));

    public static Player newPlayer(Integer playerNum) throws IOException {
        Player player = new Player();
        System.out.println("Player " + (playerNum + 1) + " registration");
        
        System.out.print("Enter first name: ");
        String firstName = input.readLine();
        
        System.out.print("Enter last name: ");
        String lastName = input.readLine();
        
        player.setName(firstName + " " + lastName);
        
        return player;
    }

    // Dummy Player class for context
    static class Player {
        private String name;
        public void setName(String name) { this.name = name; }
        public String getName() { return name; }
    }
}

Additional Best Practices #

• Always initialize StringTokenizer with a valid, non-null string—never pass null to its constructor.
• Handle edge cases (like users entering only one name) to avoid NoSuchElementException from calling nextToken() when no tokens are left.
• Use BufferedReader safely: either declare throws IOException in your method or wrap input calls in a try-catch block if you need to handle errors gracefully (e.g., invalid input, unexpected stream closure).

内容的提问来源于stack exchange，提问作者Caroline Mourão