Question

这是我第一次使用java.security.SecureRandom，我希望有人批评以下代码，以确保我正确地执行此操作。该代码应该生成任意长度的加密安全随机密码。任何意见都将不胜感激。

import java.util.*;
import java.security.SecureRandom;

public class PassGen{

    private static final String VALID_PW_CHARS = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789!@#$%^&*()-_=+{}[]|:;<>?,./";
    private static final int DEFAULT_PASSWORD_LENGTH = 12;
    private static final Random RANDOM = new SecureRandom();


    // main class
    public static void main(String args[]) throws Exception {


        // Set password length
        int pwLength;
        if (args.length < 1)
            pwLength = DEFAULT_PASSWORD_LENGTH;
        else
            pwLength = Integer.parseInt(args[0]);


        // generate password
        String pw = "";
        for (int i=0; i<pwLength; i++) {
            int index = (int)(RANDOM.nextDouble()*VALID_PW_CHARS.length());
            pw += VALID_PW_CHARS.substring(index, index+1);
        }

        System.out.println("pw = " + pw);
  }
}

Answer 1

使用StringBuilder而不是反复连接字符串。此外，您应该使用string.charAt(index)而不是使用子字符串表示单个字符：

import java.util.*;
import java.security.SecureRandom;

public class PassGen{

        private static final String VALID_PW_CHARS = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789!@#$%^&*()-_=+{}[]|:;<>?,./";
        private static final int DEFAULT_PASSWORD_LENGTH = 12;
        private static final Random RANDOM = new SecureRandom();


        // main class
        public static void main(String args[]) throws Exception {


                // Set password length
                int pwLength;
                if (args.length < 1)
                        pwLength = DEFAULT_PASSWORD_LENGTH;
                else
                        pwLength = Integer.parseInt(args[0]);

                StringBuilder pw = new StringBuilder();


                // generate password
                for (int i=0; i<pwLength; i++) {                        
                        int index = (int)(RANDOM.nextInt(VALID_PW_CHARS.length());
                        pw.append(VALID_PW_CHARS.charAt(index)));
                }

                System.out.println("pw = " + pw.toString());
        }
}

此外，您正在生成double并且不限制索引值。我做了一个有效字符数组长度的mod来解决这个问题。

Answer 2

您可以使用org.apache.commons.lang.RandomStringUtils（http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/RandomStringUtils.html）使用char数组和java.security.SecureRandom生成密码：

public String generatePassword()
{
    return RandomStringUtils.random(DEFAULT_PASSWORD_LENGTH, 0, VALID_PW_CHARS.length(), false,
            false, VALID_PW_CHARS.toCharArray(), new SecureRandom());
}

在pom.xml中

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.4</version>
</dependency>

使用java SecureRandom类创建随机密码

2 个答案: