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 ? 
Filed under: Tech on June 19th, 2007
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”));
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}
Nice, thanks.
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
LINQ technique:
Random rand = new Random();
return new string(Enumerable.Repeat(0, 9).Select(i => (char)rand.Next(65, 90)).ToArray());
how to rand insert words in string?
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.