CODEWITHSUNDEEP
c#randomchar
user2139293
user2139293

Reputation:

Pick random char

i have some chars:

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

now i'm looking for a method to return a random char from these.

I found a code which maybe can be usefull:

static Random random = new Random();
        public static char GetLetter()
        {
            // This method returns a random lowercase letter
            // ... Between 'a' and 'z' inclusize.
            int num = random.Next(0, 26); // Zero to 25
            char let = (char)('a' + num);
            return let;
        }

this code returns me a random char form the alphabet but only returns me lower case letters

Upvotes: 28

Views: 89256

Answers (11)

jordanhill123
jordanhill123

Reputation: 4192

This might work for you:

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

Upvotes: 15

Nitika Chopra
Nitika Chopra

Reputation: 1405

Your Code is good, you just need to change 'a' to 'A' 

static Random random = new Random();
        public static char GetLetter()
        {
            // This method returns a random lowercase letter
            // ... Between 'a' and 'z' inclusize.
            int num = random.Next(0, 26); // Zero to 25
            char let = (char)('A' + num);
            return let;
        }

This code same as mentioned in the question , just change char let = (char)('A' + num); , it returns upper case letters.

Thanks!!!

Upvotes: 4

Decon21
Decon21

Reputation: 55

I'm not sure how efficient it is as I'm very new to coding, however, why not just utilize the random number your already creating? Wouldn't this "randomize" an uppercase char as well?

    int num = random.Next(0,26);           
    char let = (num > 13) ? Char.ToUpper((char)('a' + num)) : (char)('a' + num);

Also, if you're looking to take a single letter from your char[], would it be easier to just use a string?

    string charRepo = "$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&"; 
    Random rando = new Random();
    int ranNum = rando.Next(0, charRepo.Length);

    char ranChar = charRepo[ranNum];

Upvotes: 3

user5710892
user5710892

Reputation:

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!

Upvotes: 0

Yaugen Vlasau
Yaugen Vlasau

Reputation: 2228

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]

Upvotes: 3

Shirish Joshi
Shirish Joshi

Reputation: 9

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;
}

Upvotes: 1

Volodymyr Machula
Volodymyr Machula

Reputation: 1604

I had approximate issue and I did it by this way:

public static String GetRandomString()
{
    var allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789";
    var length = 15;

    var chars = new char[length];
    var rd = new Random();

    for (var i = 0; i < length; i++)
    {
        chars[i] = allowedChars[rd.Next(0, allowedChars.Length)];
    }

    return new String(chars);
}

Upvotes: 3

Saman Gholami
Saman Gholami

Reputation: 3512

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());

Upvotes: 1

Soner G&#246;n&#252;l
Soner G&#246;n&#252;l

Reputation: 98868

You can use it like;

char[] chars = "$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&".ToCharArray();
Random r = new Random();
int i = r.Next(chars.Length);
Console.WriteLine(chars[i]);

Here is a DEMO.

Upvotes: 5

Matt
Matt

Reputation: 6963

private static void Main(string[] args)
        {
            Console.WriteLine(GetLetters(6));
            Console.ReadLine();
        }

        public static string GetLetters(int numberOfCharsToGenerate)
        {
            var random = new Random();
            char[] chars = "$%#@!*abcdefghijklmnopqrstuvwxyz1234567890?;:ABCDEFGHIJKLMNOPQRSTUVWXYZ^&".ToCharArray();

            var sb = new StringBuilder();
            for (int i = 0; i < numberOfCharsToGenerate; i++)
            {
                int num = random.Next(0, chars.Length);
                sb.Append(chars[num]);
            }
            return sb.ToString();
        }

Upvotes: 2

Jon Skeet
Jon Skeet

Reputation: 1504172

Well you're nearly there - you want to return a random element from a string, so you just generate a random number in the range of the length of the string:

public static char GetRandomCharacter(string text, Random rng)
{
    int index = rng.Next(text.Length);
    return text[index];
}

I'd advise against using a static variable of type Random without any locking, by the way - Random isn't thread-safe. See my article on random numbers for more details (and workarounds).

Upvotes: 45

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