Generating Alphanumeric random string in Java [duplicate]

Question

I am using String Builder from another answer, but I can't use anything but alpha/numeric, no whitespace, punctuation, etc. Can you explain how to limit the character set in this code? Also, how do I insure it is ALWAYS 30 characters long?

 Random generator = new Random(); StringBuilder stringBuilder = new StringBuilder(); int Length = 30; char tempChar ; for (int i = 0; i < Length; i++){ tempChar = (char) (generator.nextInt(96) + 32); stringBuilder.append(tempChar);

I have looked at most of the other answers, and can't figure out a solution to this. Thanks. Don't yell at me if this is a duplicate. Most of the answers don't explain which part of the code controls how long the generated number is or where to adjust the character set.

I also tried stringBuilder.Replace(' ', '1'), which might have worked, but eclipse says there is no method for Replace for StringBuilder.

"Most of the answers don't explain which part of the code controls how long the generated number is or where to adjust the character set." ... pointing out the obvious is usually considered rude. — Brian Roach
– Brian Roach, Commented Aug 5, 2013 at 23:31
I knew there would be people who would take umbrage with this question. Sorry if I offended, but the answers below have definitely added to the wealth and depth of knowledge available on the site. I do not think this is a duplicate, since the answers have more detailed explanations of the process. — PrivusGuru
– PrivusGuru, Commented Aug 7, 2013 at 15:42

zapl · Accepted Answer · 2013-08-06 01:19:01Z

If you want to control the characterset and length take for example

public static String randomString(char[] characterSet, int length) { Random random = new SecureRandom(); char[] result = new char[length]; for (int i = 0; i < result.length; i++) { // picks a random index out of character set > random character int randomCharIndex = random.nextInt(characterSet.length); result[i] = characterSet[randomCharIndex]; } return new String(result); }

and combine with

char[] CHARSET_AZ_09 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".toCharArray();

to specify the characterset.

It's not based on StringBuilder since you know the length and don't need all the overhead.

It allocates a char[] array of the correct size, then fills each cell in that array with a randomly chosen character from the input array.

more example use here: http://ideone.com/xvIZcd

@PrivusGuru btw the code you posted generates numbers between (0..95) + 32 => 32..127. The characters that correspond to those numbers are here asciitable.com for example. Doing it the way your code does would mean limiting the random numbers to 48..57 OR 65..90 OR 97..122 which would be rather painful. (or you discard characters that are not in that range like e.g. @Sello proposed)
@Virtu No, it's random. That includes generating the same string 10 times in a row, although with very small probability.

aroth · Accepted Answer · 2013-08-05 23:33:58Z

Here's what I use:

public static String randomStringOfLength(int length) { StringBuffer buffer = new StringBuffer(); while (buffer.length() < length) { buffer.append(uuidString()); } //this part controls the length of the returned string return buffer.substring(0, length); } private static String uuidString() { return UUID.randomUUID().toString().replaceAll("-", ""); }

Sello · Accepted Answer · 2013-08-06 00:04:57Z

You may try this:

 //piece int i = 0; while(i < length){ char temp =(char) (generator.nextInt(92)+32); if(Character.isLetterOrDigit(temp)) { stringBuilder.append(temp); ++i; } } System.out.println(stringBuilder);

Should achieve your goal

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Generating Alphanumeric random string in Java [duplicate]

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