CODEWITHSUNDEEP
c#arrays
Bryan Beltran
Bryan Beltran

Reputation: 7

C# - Random output from array to textbox

I'm kinda new to array and I made a char array (with 174 things in it) but I don't know how to output it in a randomize way. I'm trying to make a secured code for my system. I wanted to output 13 char from that 174 char array into a textbox, but I don't think I get the logic. Thank you in advance! Here is the code that only outputs 1 char per button click:

Random rnd = new Random();
int randomnum = rnd.Next(0, 174);
for (int x = 0; x <= 13; x++)
{
    textBox11.Text = chararray[randomnum];
}

Upvotes: 0

Views: 446

Answers (2)

Zakk Diaz
Zakk Diaz

Reputation: 1093

I've attached a screenshot of this code working. I had a small typo

This will change the seed for random

int seed = 1;

Create an instance of Random, we don't need to recreate it every time we need to use it

Random r = new Random(seed);

This only initializes the characters

char[] _myChars = new char[170];
for(var i = 0; i < _myChars.Length; i++)
{
    _myChars[i] = (char)(i%26 + 65);
}

This is the query you're looking for, it will query the characters and order them by a random order with r.Next()

var output = _myChars.OrderBy(o => r.Next()).Take(13).ToList();

This is only for displaying the output, you would want to use the output in your textbox

for(var i = 0; i < output.Count; i++)
{
    Console.WriteLine(output[i]);
}
Console.ReadLine();

Working code

Upvotes: 1

p.s.w.g
p.s.w.g

Reputation: 148990

Your code is almost there, but there are a few issues:

  1. You need to append the new character to the end of the string as apposed to just setting the Text value directly. You can do this easily with += instead of =.

  2. You need to pick a different random character for each iteration of the loop, so move your call to .Next inside the for.

Putting this together you'd have something like this:

Random rnd = new Random();

for (int x = 0; x <= 13; x++)
{
    int randomnum = rnd.Next(0, 174);
    textBox11.Text += chararray[randomnum];
}

Note however, that if this is for the purpose of security, using Random isn't great. You should probably use something like the RNGCryptoServiceProvider. For example:

using (var rng = new RNGCryptoServiceProvider())
{
    byte[] password = new byte[10];
    rng.GetBytes(password);
    textBox11.Text = Convert.ToBase64String(password).Remove(13);
}

Upvotes: 2

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