如何有效地在Java中生成安全的随机字母数字字符串？

Question

如何有效地在Java中生成安全的随机（或伪随机）字母数字字符串？

Answer 1

初始化包含所有已接受的字符（CHARS_ARRAY）的数组，然后实例化SecureRandom实例，并重复调用nextInt(CHARS_ARRAY.length)以获取char数组中的随机索引。将每个字符附加到StringBuilder，直到获得预期的字符数。

Answer 2

这是我the duplicate question.代码的略微修改版本

public final class RandomString
{

  /* Assign a string that contains the set of characters you allow. */
  private static final String symbols = "ABCDEFGJKLMNPRSTUVWXYZ0123456789"; 

  private final Random random = new SecureRandom();

  private final char[] buf;

  public RandomString(int length)
  {
    if (length < 1)
      throw new IllegalArgumentException("length < 1: " + length);
    buf = new char[length];
  }

  public String nextString()
  {
    for (int idx = 0; idx < buf.length; ++idx) 
      buf[idx] = symbols.charAt(random.nextInt(symbols.length()));
    return new String(buf);
  }

}

Answer 3

使用UUIDs：

UUID random = UUID.randomUUID();
System.out.println( random );

Answer 4

    String chrs = "0123456789abcdefghijklmnopqrstuvwxyz-_ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    SecureRandom secureRandom = SecureRandom.getInstanceStrong();
    // 9 is the length of the string you want
    String customTag = secureRandom.ints(9, 0, chrs.length()).mapToObj(i -> chrs.charAt(i))
      .collect(StringBuilder::new, StringBuilder::append, StringBuilder::append).toString();
    System.out.println(customTag);

例子：

// q3HX6EctP
// WjRrMjQT4
// sX-Piq4DB

Answer 5

为开放密钥加密算法生成公钥，并通过Base64算法将字节序列转换为字符串。

Answer 6

import java.security.SecureRandom;
import java.util.Random;

public class PasswordHelper {        

    public static String generatePassword (int length) {

    //minimum length of 6
    if (length < 4) {
        length = 6;
    }

    final char[] allAllowed = "abcdefghijklmnopqrstuvwxyzABCDEFGJKLMNPRSTUVWXYZ0123456789".toCharArray();

    //Use cryptographically secure random number generator
    Random random = new SecureRandom();

    StringBuilder password = new StringBuilder(); 

    for (int i = 0; i < length; i++) {
        password.append(allAllowed[random.nextInt(allAllowed.length)]);
    }

    return password.toString();

    }

}

Answer 7

http://download.oracle.com/javase/6/docs/api/java/security/SecureRandom.html

来自Javadoc：

SecureRandom random = new SecureRandom();
byte bytes[] = new byte[20];
random.nextBytes(bytes);