How to generate a random string, and specify the l

Question

There is a library to generate Random numbers, so why isn't there a library for generation of random strings?

In other words how to generate a random string, and specify desired length, or better, generate unique string on specification you want i.e. specify the length, a unique string within my application is enough for me.

I know I can create a Guid (Globally Unique IDentifier) but those are quite long, longer they need to be.

int length = 8;
string s = RandomString.NextRandomString(length)
uniquestringCollection = new UniquestringsCollection(length)
string s2 = uniquestringCollection.GetNext();

Answer 1

Random string using Concatenated GUIDs

This approach uses the minimum number of concatenated GUIDs to return a random string of the desired number of characters.

    /// <summary>
    /// Uses concatenated then SubStringed GUIDs to get a random string of the
    /// desired length. Relies on the randomness of the GUID generation algorithm.
    /// </summary>
    /// <param name="stringLength">Length of string to return</param>
    /// <returns>Random string of <paramref name="stringLength"/> characters</returns>
    internal static string GetRandomString(int stringLength)
    {
        StringBuilder sb = new StringBuilder();
        int numGuidsToConcat = (((stringLength - 1) / 32) + 1);
        for(int i = 1; i <= numGuidsToConcat; i++)
        {
            sb.Append(Guid.NewGuid().ToString("N"));
        }

        return sb.ToString(0, stringLength);
    }

Example output with a length of 8:

39877037
2f1461d8
152ece65
79778fc6
76f426d8
73a27a0d
8efd1210
4bc5b0d2
7b1aa10e
3a7a5b3a
77676839
abffa3c9
37fdbeb1
45736489

Example output with a length of 40 (note the repeated '4' - in a v4 GUID, thee is one hex digit that will always be a 4 (effectively removing 4 bits) -- the algorithm could be improved by removing the '4' to improve randomness, given that the returned length of the string can be less than the length of a GUID (32 chars)...):

e5af105b73924c3590e99d2820e3ae7a3086d0e3
e03542e1b0a44138a49965b1ee434e3efe8d063d
c182cecb5f5b4b85a255a397de1c8615a6d6eef5
676548dc532a4c96acbe01292f260a52abdc4703
43d6735ef36841cd9085e56f496ece7c87c8beb9
f537d7702b22418d8ee6476dcd5f4ff3b3547f11
93749400bd494bfab187ac0a662baaa2771ce39d
335ce3c0f742434a904bd4bcad53fc3c8783a9f9
f2dd06d176634c5b9d7083962e68d3277cb2a060
4c89143715d34742b5f1b7047e8107fd28781b39
2f060d86f7244ae8b3b419a6e659a84135ec2bf8
54d5477a78194600af55c376c2b0c8f55ded2ab6
746acb308acf46ca88303dfbf38c831da39dc66e
bdc98417074047a79636e567e4de60aa19e89710
a114d8883d58451da03dfff75796f73711821b02

C# Fiddler Demo: https://dotnetfiddle.net/Y1j6Dw

Answer 2

A library for generating random strings wouldn't be very useful. It would either be too simple, so that you often need to replace it anyway, or too complex in order to be able to cover any possible situation, that you replace it because it's to complicated to use.

Creating random strings is quite easy by using the random geneator for numbers, given the exact details of your needs. It's just more efficient to write the code specifically for each situation.

If you want a unique string, there is two possibilities. You can either keep every random string that is created, so that you can check for uniqueness, or you can make it really long so that it's incredibly unlikely that there would be duplicates. A GUID does the latter (which explains why it's so long), so there is already an implementation for that.

Answer 3

I can't recall where I got this, so if you know who originally authored this, please help me give attribution.

private static void Main()
{
    const string AllowedChars =
        "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#@$^*()";
    Random rng = new Random();

    foreach (var randomString in RandomStrings(AllowedChars, 1, 16, 25, rng))
    {
        Console.WriteLine(randomString);
    }

    Console.ReadLine();
}

private static IEnumerable<string> RandomStrings(
    string allowedChars,
    int minLength,
    int maxLength,
    int count,
    Random rng)
{
    char[] chars = new char[maxLength];
    int setLength = allowedChars.Length;

    while (count-- > 0)
    {
        int length = rng.Next(minLength, maxLength + 1);

        for (int i = 0; i < length; ++i)
        {
            chars[i] = allowedChars[rng.Next(setLength)];
        }

        yield return new string(chars, 0, length);
    }
}

Answer 4

How about-

    static Random rd = new Random();
    internal static string CreateString(int stringLength)
    {
        const string allowedChars = "ABCDEFGHJKLMNOPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz0123456789!@$?_-";
        char[] chars = new char[stringLength];

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

        return new string(chars);
    }

Answer 5

I commonly use code like the following to generate a random string:

internal static class Utilities {
  static Random randomGenerator = new Random();
  internal static string GenerateRandomString(int length) {
    byte[] randomBytes = new byte[randomGenerator.Next(length)];
    randomGenerator.NextBytes(randomBytes);
    return Convert.ToBase64String(randomBytes);
  }
}

This will create a Base64 encoded string of the random bytes generated by the random object. It isn't thread safe so multiple threads will have to lock around it. Additionally I use a static Random object so two calls to the method at the same time won't get the same initial seed.

Answer 6

How about something like this...

static string GetRandomString(int lenOfTheNewStr)
{
    string output = string.Empty;
    while (true)
    {
        output = output + Path.GetRandomFileName().Replace(".", string.Empty);    
        if (output.Length > lenOfTheNewStr)
        {
            output = output.Substring(0, lenOfTheNewStr);
            break;
        }
    }
    return output;
}

Output

static void Main(string[] args)
{            
    for (int x = 0; x < 10; x++)
    {
        string output = GetRandomString(20);
        Console.WriteLine(output);
    }
    Console.ReadKey();
}

 

r323afsluywgozfpvne4
qpfumdh3pmskleiavi3x
nq40h0uki2o0ptljxtpr
n4o0rzwcz5pdvfhmiwey
sihfvt1pvofgxfs3etxg
z3iagj5nqm4j1f5iwemg
m2kbffbyqrjs1ad15kcn
cckd1wvebdzcce0fpnru
n3tvq0qphfkunek0220d
ufh2noeccbwyfrtkwi02

Online Demo: https://dotnetfiddle.net/PVgf0k

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^{References
• https://www.dotnetperls.com/random-string}