Pick random char

Question

i have some chars:

chars = \"$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&\".ToCharArray();

now i\'m lookin

Answer 1

Instead of 26 please use size of your CHARS buffer.

int num = random.Next(0, chars.Length)

Then instead of

let = (char)('a' + num)

use

let = chars[num]

Answer 2

You can try this :

 public static string GetPassword()
 {
string Characters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
 Random rnd = new Random();
int index = rnd.Next(0,51);
string char1 = Characters[index].ToString();
return char1;
  }

Now you can play with this code block as per your wish. Cheers!

Answer 3

This might work for you:

public static char GetLetter()
{
    string chars = "$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&";
    Random rand = new Random();
    int num = rand.Next(0, chars.Length -1);
    return chars[num];
}

Answer 4

Getting Character from ASCII number:

private string GenerateRandomString()
{
Random rnd = new Random();
string txtRand = string.Empty;
for (int i = 0; i <8; i++) txtRand += ((char)rnd.Next(97, 122)).ToString();
return txtRand;
}

Answer 5

I wish This code helps you :

 string s = "$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&";
            Random random = new Random();
            int num = random.Next(0, s.Length -1);
            MessageBox.Show(s[num].ToString());