Generate random 6 digit number

Question

I've been searching for a couple of hours and I just can't seem to find a answer to this question. I want to generate a random number with 6 digits. Some of you might tell me to use this code:

        Random generator = new Random();
        int r = generator.Next(100000, 1000000);

But that limits my 6 digits to all be above 100 000 in value. I want to be able to generate a int with 000 001, 000 002 etc. I later want to convert this integer to a string.

Solution 1

If you want a string to lead with zeroes, try this. You cannot get an int like 001.

    Random generator = new Random();
    String r = generator.Next(0, 1000000).ToString("D6");

Solution 2

You want to have a string:

Random r = new Random();
var x = r.Next(0, 1000000);
string s = x.ToString("000000");

For example,

x = "2124"
s = "002124"

Solution 3

private static string _numbers = "0123456789";
Random random = new Random();


private void DoWork()
{
   StringBuilder builder = new StringBuilder(6);
   string numberAsString = "";
   int numberAsNumber = 0;

   for (var i = 0; i < 6; i++)
   {
      builder.Append(_numbers[random.Next(0, _numbers.Length)]);
   }

   numberAsString = builder.ToString();
   numberAsNumber = int.Parse(numberAsString);

}

Solution 4

As stated in a comment, a "six digit number" is a string. Here's how you generate a number from 0-999999, then format it like "000482":

Random r = new Random();
int randNum = r.Next(1000000);
string sixDigitNumber = randNum.ToString("D6");

Solution 5

I agree with the comment above that 000 001 can't be an integer, but can be a string with:

Random generator = new Random();
int r = generator.Next(1, 1000000);
string s = r.ToString().PadLeft(6, '0');

View more solutions