Java random number with given length

Question

I need to genarate a random number with exactly 6 digits in Java  I know i could loop 6 times over a randomicer but is there a nother way to do this in the standard Java SE    EDIT - Follow up question   Now that I can generate my 6 digits i got a new problem  the whole ID I m trying to create is of the syntax 123456-A1B45  So how do i randomice the last 5 chars that can be either A-Z or 0-9  I m thinking of using the char value and randomice a number between 48 - 90 and simply drop any value that gets the numbers that represent 58-64  Is this the way to go or is there a better solution   EDIT 2   This is my final solution  Thanks for all the help guys   protected String createRandomRegistryId String handleId           syntax we would like to generate is DIA123456-A1B34           String val    DI                 char  1   random A-Z     int ranChar   65    new Random    nextInt 90-65       char ch    char ranChar              val    ch                numbers  6   random 0-9     Random r   new Random        int numbers   100000    int  r nextFloat     899900       val    String valueOf numbers        val     -          char or numbers  5   random 0-9 A-Z     for int i   0  i lt 6            int ranAny   48    new Random    nextInt 90-65            if   57  lt  ranAny  amp  amp  ranAny lt   65            char c    char ranAny                val    c          i                          return val

User · Answer

If you need to specify the exact charactor length, we have to avoid values with 0 in-front.

Final String representation must have that exact character length.

String GenerateRandomNumber(int charLength) {
        return String.valueOf(charLength < 1 ? 0 : new Random()
                .nextInt((9 * (int) Math.pow(10, charLength - 1)) - 1)
                + (int) Math.pow(10, charLength - 1));
    }

User · Answer

try this   public int getRandomNumber int min  int max        return  int  Math floor Math random      max - min   1     min

User · Answer

To generate a 6-digit number   Use Random and nextInt as follows   Random rnd   new Random    int n   100000   rnd nextInt 900000     Note that n will never be 7 digits  1000000  since nextInt 900000  can at most return 899999      So how do I randomize the last 5 chars that can be either A-Z or 0-9    Here s a simple solution      Generate random id  for example 283952-V8M32 char   chars    ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789  toCharArray    Random rnd   new Random    StringBuilder sb   new StringBuilder  100000   rnd nextInt 900000      -    for  int i   0  i  lt  5  i        sb append chars rnd nextInt chars length      return sb toString

User · Answer

Would that work for you   public class Main    public static void main String   args        Random r   new Random System currentTimeMillis         System out println r nextInt 100000    0 000001          result e g  0 019007

User · Answer

int rand    new Random    getNextInt 900000    100000    EDIT  Fixed off-by-1 error and removed invalid solution

User · Answer

For the follow-up question  you can get a number between  36 5 and 36 6 and convert it in base 36  UPDATED   using this code   http   javaconfessions com 2008 09 convert-between-base-10-and-base-62-in 28 html  It s written BaseConverterUtil toBase36 60466176 r nextInt 2116316160    but in your use case  it can be optimized by using a StringBuilder and having the number in the reverse order ie 71 should be converted in Z1 instead of 1Z  EDITED

User · Answer

Generate a random number  which is always between 0-1  and multiply by 1000000  Math round Math random   1000000

User · Answer

Generate a number in the range from 100000 to 999999      pseudo code int n   100000   random float     900000    For more details see the documentation for Random

[java] Java random number with given length

Examples related to java

Examples related to random