Generate unique id with 6 of length - logic

Question

I need to generate a unique id with a limit of 6 blocks. That id should contains letters in upper case and number.
How can I do it? I thougth in use the date, but I'm failed.
More details...
That Id just need not repeat, but should be generate alone, whithout base in a last sequence.
I can do this in any language.

Answer 1

I made a blog post to generate a strong SQL Server password. You can adapt this to what you need. For a better format blog post here.
We have the requirement that our passwords have to change every 90 days. I wanted to automate this and it sounded pretty easy to do, but it wasn’t that easy. Why? Because there are a lot of rules for passwords.
First, I have to store it in the web.config. So no XML special characters.
quot "
amp &
apos '
lt <
gt >
Next, If used in an OLE DB or ODBC connection string, a password must not contain the following characters: [] {}() , ; ? * ! #.
Finally, strong passwords must contains characters from at least three of the following categories:
English uppercase characters (A through Z)
English lowercase characters (a through z)
Base 10 digits (0 through 9)
Nonalphabetic characters (for example: !, $, #, %)
So keeping all of these rules in mind I created a simple class. that I can just call
public class PasswordGenerator
{
private static string CHARS_LCASE = "abcdefgijkmnopqrstwxyz";
private static string CHARS_UCASE = "ABCDEFGHJKLMNPQRSTWXYZ";
private static string CHARS_NUMERIC = "23456789";
private static string CHARS_SPECIAL = "*-+_%/";
private static string CHARS_ALL = CHARS_LCASE + CHARS_UCASE + CHARS_NUMERIC + CHARS_SPECIAL;
public static string GeneratePassword(int length)
{
char[] chars = new char[length];
Random rand = new Random();
for (int i = 0; i < length; i++)
{
switch (i)
{
case 0:
chars[i] = CHARS_LCASE[rand.Next(0, CHARS_LCASE.Length)];
break;
case 1:
chars[i] = CHARS_UCASE[rand.Next(0, CHARS_UCASE.Length)];
break;
case 2:
chars[i] = CHARS_NUMERIC[rand.Next(0, CHARS_NUMERIC.Length)];
break;
case 3:
chars[i] = CHARS_SPECIAL[rand.Next(0, CHARS_SPECIAL.Length)];
break;
default:
chars[i] = CHARS_ALL[rand.Next(0, CHARS_ALL.Length)];
break;
}
}
return new string(chars);
}
}
So now I just simply call this for a new password:
PasswordGenerator.GeneratePassword(13)