How to generate a random string in c# 2.0

19Jun/078

How to generate a random string in c# 2.0

Although this is common knowledge to most dev's,
I justed wanted to share some possible approaches for generating a random string in C#.

Approach 1: Create a static method in a utility class that will return a defined length of random characters.

One possible implementation could be:

static class Generator
{

private static Random _random = new Random();

public static string RandomString(int size)
{

StringBuilder builder = new StringBuilder();
for(int i=0; i < size; i++)
{

//26 letters in the alfabet, ascii + 65 for the capital letters
builder.Append(Convert.ToChar(Convert.ToInt32(Math.Floor(26 * _random.NextDouble() + 65))));

}
return builder.ToString();

}

}

Advantage:

can produce multiple length strings
can easily be updated to provide string with lowercase/uppercase/numbers in it

Approach 2: GUID

Just use a GUID if may contain numbers, letters and the character '-'
string randomString = Guid.NewGuid().ToString();

Advantage:

Simple, always unique

Disadvantage: always 36 chars in length

Approach 3 (humor): Be Lame

Be lame and try the following

string randomString = System.IO.Path.ChangeExtension(System.IO.Path.GetRandomFileName(), null);

DisAdvantage: only produces strings of 8 characters and the idea is really way off...

Any one else fancy to post a (humorlike approach to a) random string generating solution ?

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Comments (8) Trackbacks (0) ( subscribe to comments on this post )

den Ben
July 12th, 2007 - 16:56

Regarding your second approach. A Guid can be formatted when calling .ToString(). Example: Guid x = Guid.NewGuid();
System.Diagnostics.Debug.WriteLine(“N: ” + x.ToString(“N”));
System.Diagnostics.Debug.WriteLine(“D: ” + x.ToString(“D”));
System.Diagnostics.Debug.WriteLine(“P: ” + x.ToString(“P”));
System.Diagnostics.Debug.WriteLine(“B: ” + x.ToString(“B”)); produces Guid x = Guid.NewGuid();
System.Diagnostics.Debug.WriteLine(“N: ” + x.ToString(“N”));
System.Diagnostics.Debug.WriteLine(“D: ” + x.ToString(“D”));
System.Diagnostics.Debug.WriteLine(“P: ” + x.ToString(“P”));
System.Diagnostics.Debug.WriteLine(“B: ” + x.ToString(“B”));

( REPLY )
den Ben
July 12th, 2007 - 16:56

oops… copy pasted wrong… this is the output:
N: a1e9332e94f14b2a87ddaeaeef4d205e
D: a1e9332e-94f1-4b2a-87dd-aeaeef4d205e
P: (a1e9332e-94f1-4b2a-87dd-aeaeef4d205e)
B: {a1e9332e-94f1-4b2a-87dd-aeaeef4d205e}

( REPLY )
Marquin
August 4th, 2008 - 16:22

Nice, thanks.

( REPLY )
Walter van hoof
January 17th, 2009 - 15:59

Hey

Your statement

builder.Append(Convert.ToChar(Convert.ToInt32(Math.Floor(26 * _random.NextDouble() + 65))));

seems a little bit bad practice. What do you think about

builder.Append((char)_random.Next(‘A’, ‘Z’ + 1));

I think this is much more readable and much shorter. And by the way, for strings with lowercase letters, you don’t need a different methode. Just call:
Generator.RandomString(NUMBER_OF_CHAR).ToLower()

Have much fun
Walter

( REPLY )
Josh Comley
April 1st, 2009 - 11:15

LINQ technique:

Random rand = new Random();
return new string(Enumerable.Repeat(0, 9).Select(i => (char)rand.Next(65, 90)).ToArray());

( REPLY )
best sale
May 16th, 2010 - 05:11

how to rand insert words in string?

( REPLY )
Mayiko
June 24th, 2010 - 09:43

Thx!
I was looking for a way to generate a unique string in this format: ###-###-###-###

With your first approach I now know how I can achieve this.

( REPLY )
Nick Olsen
August 28th, 2010 - 05:40

Interesting post. I have seen approach 3 before as well and it never really made sense to me either!

Here is a blog post that provides a class for generating random words, sentences and paragraphs:

http://nickstips.wordpress.com/2010/08/26/c-random-text-generator/

The full code can be found here:

http://randomtext.codeplex.com

( REPLY )