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。将每个字符追加到StringBuilder，直到获得预期的字符数。

Answer 2

如果您使用Apache Commons Lang，最简单的方法是

RandomStringUtils.random(20, 0, 0, true, true, null, new SecureRandom());

Answer 3

以下是我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 4

使用UUIDs：

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

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

https://MYDOMAIN/SUBDIRECTORY/Report/ViewReportPage/reportPath=%2FGearUpAlumni%2FAlumniData&page=1&

示例：

    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);

Answer 7

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 8

我对此随机生成的字符串的规范是生成一个35个字符长的加密密钥（a-zA-Z2-7个字符仅用于前25个字符）。每第5个字符后使用-，最后5个字符应为10_000到19_999之间的整数。

这是我所做的代码（在Kotlin中）。

companion object {

    /**
     * Letters lower & upper case excluding:
     *
     *   - l, o, B, O
     *
     * Numbers excluding:
     *
     *   - 0, 1, 8, 9
     * */
    val acceptedChars: CharArray
        get() = charArrayOf(
            'a', 'b', 'c', 'd', 'e',
            'f', 'g', 'h', 'i', 'j',
            'k', 'm', 'n', 'p', 'q',
            'r', 's', 't', 'u', 'v',
            'w', 'x', 'y', 'z', 'A',
            'C', 'D', 'E', 'F', 'G',
            'H', 'I', 'J', 'K', 'L',
            'M', 'N', 'P', 'Q', 'R',
            'S', 'T', 'U', 'V', 'W',
            'X', 'Y', 'Z', '2', '3',
            '4', '5', '6', '7'
        )

    val acceptedInts: CharArray
        get() = charArrayOf(
            '0', '1', '2', '3', '4',
            '5', '6', '7', '8', '9'
        )

    fun generate(): EncryptionKey {
        val random = SecureRandom()
        val stringBuilder = StringBuilder()
        for (i in 0 until 35) {
            when (i) {
                5, 11, 17, 23, 29 -> {
                    stringBuilder.append("-")
                }
                30 -> {
                    stringBuilder.append("1")
                }
                31, 32, 33, 34 -> {
                    val index = random.nextInt(acceptedInts.size)
                    stringBuilder.append(acceptedInts[index])
                }
                else -> {
                    val index = random.nextInt(acceptedChars.size)
                    stringBuilder.append(acceptedChars[index])
                }
            }
        }
        return EncryptionKey(stringBuilder.toString())
    }
}

产生的结果：

value: 'nK6UI-DWYvu-dbmhD-KPe5X-22dPT-10027', length: '35'
value: 'NIFvi-aX4GW-3xCYV-YSAVs-tASIK-15301', length: '35'
value: 'SpNkT-qxHR7-hSMkK-hVxpp-AqLFh-19409', length: '35'
value: 'bNvi2-svqX7-cfEw5-LNYDn-C2FtW-16197', length: '35'
value: 'hhjLX-KmRQU-KbHyU-CkNyD-5ASk6-14537', length: '35'
value: 'Xd2cj-braCm-FaE4E-Jvn2G-2Dv5J-12243', length: '35'
value: '7beFb-aeSe2-iHXZe-mTUHT-aEbry-17349', length: '35'
value: 'NExMa-xCAbU-VkpyS-xeEkj-QUayd-16311', length: '35'
value: '52HWN-EX7wV-csbhj-InhtU-gbV46-18606', length: '35'
value: 'n3RTZ-whpjQ-ZjW5n-tTyfR-eLDSF-14003', length: '35'
value: 'aJEks-ccKdU-KGJdh-Rz4ck-tR7Uq-12199', length: '35'
value: 'nMcUF-ctbcy-FEfq7-VJhRx-pCKej-16369', length: '35'

Answer 9

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);

如何有效地在Java中生成安全的随机字母数字字符串？

9 个答案: