I've been searching the internet for RNG for a while now. Everything I saw was either TOO complex or was just not what I was looking for. After reading a few articles I was able to come up with this simple code.
{
Random rnd = new Random(DateTime.Now.Millisecond);
int[] b = new int[10] { 5, 8, 1, 7, 3, 2, 9, 0, 4, 6 };
textBox1.Text = Convert.ToString(b[rnd.Next(10)])
}
Simple explanation,
This works well.
To obtain a random number less than 100 use
{
Random rnd = new Random(DateTime.Now.Millisecond);
int[] b = new int[10] { 5, 8, 1, 7, 3, 2, 9, 0, 4, 6 };
int[] d = new int[10] { 9, 4, 7, 2, 8, 0, 5, 1, 3, 4 };
textBox1.Text = Convert.ToString(b[rnd.Next(10)]) + Convert.ToString(d[rnd.Next(10)]);
}
and so on for 3, 4, 5, and 6 ... digit random numbers.
Hope this assists someone positively.
You can use Random.Next(int maxValue)
:
Return: A 32-bit signed integer greater than or equal to zero, and less than maxValue; that is, the range of return values ordinarily includes zero but not maxValue. However, if maxValue equals zero, maxValue is returned.
var r = new Random();
// print random integer >= 0 and < 100
Console.WriteLine(r.Next(100));
For this case however you could use Random.Next(int minValue, int maxValue)
, like this:
// print random integer >= 1 and < 101
Console.WriteLine(r.Next(1, 101);)
// or perhaps (if you have this specific case)
Console.WriteLine(r.Next(100) + 1);
Source: Stackoverflow.com