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<6;){
int ranAny = 48 + (new Random()).nextInt(90-65);
if(!(57 < ranAny && ranAny<= 65)){
char c = (char)ranAny;
val += c;
i++;
}
}
return val;
}
int rand = (new Random()).getNextInt(900000) + 100000;
EDIT: Fixed off-by-1 error and removed invalid solution.
try this:
public int getRandomNumber(int min, int max) {
return (int) Math.floor(Math.random() * (max - min + 1)) + min;
}
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
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));
}
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 < 5; i++)
sb.append(chars[rnd.nextInt(chars.length)]);
return sb.toString();
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:
Generate a random number (which is always between 0-1) and multiply by 1000000
Math.round(Math.random()*1000000);
Source: Stackoverflow.com