How generate unique random numbers combination of letters and digits

As the following console app shows, it's easy to generate any number of random alphanumerics of a given length:

using System;

using System.Text;

class Program

{

static Random rand = new Random();

static void Main()

{

// generate and print 10 random alphanumerics of 10 characters each

for(int i = 0; i < 10; i++) Console.WriteLine(RandomAlphaNumeric(10));

Console.ReadKey();

}

static string RandomAlphaNumeric(int numChars)

{

char[] ca = new char[numChars];

// represent digits by random numbers 0 to 9

// represent letters (use lower case say) by numbers 10 to 35

for(int i = 0; i < numChars; i++)

{

int number = rand.Next(36);

if (number < 10)

ca[i] = (char)(48 + number); // 0 to 9 is ASCII 48 to 57

else

ca[i] = (char)(87 + number); // a to z is ASCII 97 to 122

}

return new string(ca);

}

However, unless you're going to keep track of every alphanumeric generated and throw any duplicates away, it's impossible to guarantee that they'll be unique.

Of course, the longer the length, the greater the chance that the alphanumerics will be unique.

This principle is used to generates GUIDs - using the 10 digits and the letters A to F - where a length of 32 means that the probability of duplication is very low and the GUIDs can therefore, be regarded as unique for practical purposes.

For more details on GUIDs see:

http://en.wikipedia.org/wiki/Globally_unique_identifier

http://msdn.microsoft.com/en-us/library/system.guid.aspx

2 Replies

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Vulpes14y ago
As the following console app shows, it's easy to generate any number of random alphanumerics of a given length:

using System;
using System.Text;

class Program
{
static Random rand = new Random();

static void Main()
{
// generate and print 10 random alphanumerics of 10 characters each
for(int i = 0; i < 10; i++) Console.WriteLine(RandomAlphaNumeric(10));
Console.ReadKey();
}

static string RandomAlphaNumeric(int numChars)
{
char[] ca = new char[numChars];

// represent digits by random numbers 0 to 9
// represent letters (use lower case say) by numbers 10 to 35

for(int i = 0; i < numChars; i++)
{
int number = rand.Next(36);
if (number < 10)
ca[i] = (char)(48 + number); // 0 to 9 is ASCII 48 to 57
else
ca[i] = (char)(87 + number); // a to z is ASCII 97 to 122

}

return new string(ca);
}
}

However, unless you're going to keep track of every alphanumeric generated and throw any duplicates away, it's impossible to guarantee that they'll be unique.

Of course, the longer the length, the greater the chance that the alphanumerics will be unique.

This principle is used to generates GUIDs - using the 10 digits and the letters A to F - where a length of 32 means that the probability of duplication is very low and the GUIDs can therefore, be regarded as unique for practical purposes.

For more details on GUIDs see:

http://en.wikipedia.org/wiki/Globally_unique_identifier

http://msdn.microsoft.com/en-us/library/system.guid.aspx

0
venkata kumar14y ago
hi
u can take one addtional column in table.
that column u make auto generated column.
when your generating new id. get last number of that column.
add with letter and by incrementing that number.

Thanks & Regards
Ravi
0