How to randomly generate different variable length string .the srting be unique.?

Question

3.33/5 (3 votes)

See more:

Hi i created one function for automatic random generation of string
The function i geiven below:

C#

public static string createRandomString(int Length)
       {
           string _allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ";
           Random randNum = new Random();
           char[] chars = new char[Length];
           int allowedCharCount = _allowedChars.Length;

           for (int i = 0; i < Length; i++)
           {
               chars[i] = _allowedChars[(int)((_allowedChars.Length) * randNum.NextDouble())];
           }

           return new string(chars);
       }

here in length i am passing the varchar length..ie varchar(50) mean i am passing lenght 50.it randomly generate string takes 50 whole length

it generate string upto 50 characters

but i want variable lenght characters.that is if i gave length 50 it want to produce string like different length within 50 ..not exceeds 50

that is
eg;

addncvu
axccdfgghh
wdedglikdfdf
adsd
asdfeddfgdfgdfsggggfgfd
advefcvv
adfsadfafccffgbbb

like this i want to vary the length of string..how to generate it..how to mdodify the function i have written already..
anybody know tell me.........

Posted 22-Jun-11 19:17pm

suryaswathi

Add a Solution

Comments

RakeshMeena 23-Jun-11 2:06am

I don't understand why somebody downvoted the question. My 5!

5 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

walterhevedeich · Answer 1 · 2011-06-22T19:28:00

Solution 1

To randomize the length of the string, you create a random number from 1 to 50. And to do that, you use Random[^] class. Please see example below.

System.Random RandNum = new System.Random();
int MyRandomNumber = RandNum.Next(1, 51);

Posted 22-Jun-11 19:28pm

walterhevedeich

RakeshMeena · Answer 2 · 2011-06-22T19:32:00

Solution 2

This article will give you a better approach. Further to generate random length string, the "length" i/p parameter should be provided by a "Random" class instance with max value as say 50.

Hope this helps!

Posted 22-Jun-11 19:32pm

RakeshMeena

Tarun Mangukiya · Answer 3 · 2011-06-22T19:34:00

You can use like this
This will generate a string which has a random length less than you have given in Length variable.

C#

public static string createRandomString(int Length)
       {
           string _allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ";
           Random randNum = new Random();
           char[] chars = new char[Length];
           int allowedCharCount = _allowedChars.Length;
           for (int i = 0; i < randNum.NextDouble(1,Length); i++)
           {
               chars[i] = _allowedChars[(int)((_allowedChars.Length) * randNum.NextDouble())];
           }
           return new string(chars);
       }

You can Also Use this code to generate Random String

C#

String RndString;
Int Zero, Nine, A, Z, Count, RandNum;
Random randNum;
Random oRandom = New Random(System.DateTime.Now.Millisecond);
Zero = Asc("0");
Z = Asc("Z");
While (Count < randNum.NextDouble(1,Length))
{
    RandNum = oRandom.Next(Zero, Z);
    RndString = (char)RandNum;
    Count++;
}

Thanks.

Peter_in_2780 · Answer 4 · 2011-06-22T19:58:00

Solution 4

Tarun Mangukiya almost got it. Calculating the output length in the loop test will change the target length every iteration of the loop, which is probably NOT what you want.

C#

public static string createRandomString(int Length)
       {
           string _allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ";
           Random randNum = new Random();
           int outputLength = randNum.NextDouble(1,Length);  //<<<<<<<<< moved out of loop
           char[] chars = new char[outputLength];  //<<<<<<<< use outputLength here
// not used           int allowedCharCount = _allowedChars.Length;
           for (int i = 0; i < outputLength ; i++)
           {
               chars[i] = _allowedChars[(int)((_allowedChars.Length) * randNum.NextDouble())];
           }
           return new string(chars);
       }

Peter

[edit] used outputLength for chars[] size [/edit]

Posted 22-Jun-11 19:58pm

Peter_in_2780

Updated 22-Jun-11 20:01pm

v2

Comments

suryaswathi 23-Jun-11 2:21am

I want every time the string want to be generate is different length..
in that i got an error

int outputLength = randNum.NextDouble(1,Length); radnum .NextDouble not takes two arguments nu

wat i want to do

Peter_in_2780 23-Jun-11 2:29am

int outputLength = (int)(randNum.NextDouble() * Length);

suryaswathi 23-Jun-11 2:55am

sir thanks lkot.i got different variable lenth string

but again one pblm..in that in a row three field mean the three column in a row i getting same string..the string is unique

how to do..plz reply

Tarun Mangukiya 23-Jun-11 3:07am

See My Edited Answer

Peter_in_2780 23-Jun-11 3:19am

Your second code segment has exactly the same problem. Every time round the character loop, you are changing the loop upper limit. As well as that, the original question wanted characters from [A-za-z], not numerics and certainly not from [:;<=>?@] which yours will generate.

Peter_in_2780 23-Jun-11 3:09am

I don't understand your last problem. Pleas explain in more detail.

Tarun Mangukiya 23-Jun-11 3:16am

It is according to ascii code.
We have taken a ascii of 0 to ascii of z.
Then we are taking the random ascii form it.
And Converting it to the char so that we can give our random string's next character.

This step is done for a random times so we can generate a random length string.

Peter_in_2780 23-Jun-11 3:19am

You replied to the wrong post. My comment is for the OP, not you.

suryaswathi 23-Jun-11 3:25am

your code is working good sir

in a table in row wise generated different records
in column wise generated same string ...

in single row has all are same string in all fields what i have in my table..

did u understood sir?

that want

Peter_in_2780 23-Jun-11 3:41am

The problem is that each call to the function starts an identical random generator. Either use
Random randNum = new Random(something) where "something" changes - like the time to a fine resolution (microseconds?)
or move randNum out as a global and call the same generator for all sequences.

i-balaji-mani · Answer 5 · 2011-06-22T20:21:00

Hi,

If the requirement is just a random generated unique string, probably you might think of using the below.

C#

public static string createRandomString(int Length)
        {

            char[] charsToIgnore = {'0', '1','2', '3', '4', '5', '6', '7', '8', '9', '-'};
            string uniqueString = Guid.NewGuid().ToString();
            for(int i=0; i<charsToIgnore.Length; i++)
            {
                uniqueString = uniqueString.Replace(charsToIgnore[i].ToString(), "");
            }

            return uniqueString.Substring(0, Length);
        }

The Guid class always generates unique but in hex dec. But, you may trim the unwanted chars as done above.
Cheers
- Balaji