How to generate a secure random alphanumeric strin

问题:

How do you generate a secure random (or pseudo-random) alphanumeric string in Java efficiently?

回答1:

Initialize an array containing all the accepted chars (CHARS_ARRAY), then instantiate a SecureRandom instance, and call nextInt(CHARS_ARRAY.length) repeatedly to get a random index in your char array. Append each char to a StringBuilder until you get the expected number of chars.

回答2:

Here's a slightly modified version of my code from the duplicate question.

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);
  }

}

回答3:

Using UUIDs:

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

回答4:

For what purpose ?

Generate a public key for an open-key encryption algorithm and convert the byte sequence to string via Base64 algorithm.

回答5:

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

From the Javadoc:

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