password generator in c#

Question

I want to create a random password generator in my ASP.NET C# web application.

I have created that but it is applicable only for the integer format. If I give any characters it shows the debugging error as "my input format is wrong".

But i want to enter both string and integer. For example my textbox field may be

dft-001

I want a random password for this value and also my length should be restricted to 8 digits or characters only.

My current code:

public static string CreateRandomPassword(int PasswordLength)
    {

        string _allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789-";
        Random randNum = new Random();
        char[] chars = new char[PasswordLength];
        int allowedCharCount = _allowedChars.Length;

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

        return new string(chars);
    }


protected void Button1_Click(object sender, EventArgs e)
    {
        userid.Text = "Your id is:     " + id.Text;
        if(id .Text!="")
        {

            string myInt = id.Text.ToString(); 
            password.Text = "Your password is: " + CreateRandomPassword(int.Parse(myInt)); 
        }
       }

Can you elaborate on what you mean by "I want a password for this value"? What relationship are you looking for between the user id and the random generated password? Your code reads like the password length would be set by the user id (!?) but then you say you only want 8 characters. — Infotekka
– Infotekka, Commented Mar 11, 2011 at 5:12
maybe you should try using the built-in generator msdn.microsoft.com/en-us/library/… — phillip
– phillip, Commented Aug 9, 2013 at 18:10

Infotekka · Accepted Answer · 2013-08-09 18:03:15Z

4

There are a few things you can do to correct this based on what you have stated in your requirements.

public static string CreateRandomPassword()  //If you are always going to want 8 characters then there is no need to pass a length argument
    {
        string _allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789-";
        Random randNum = new Random((int)DateTime.Now.Ticks);  //Don't forget to seed your random, or else it won't really be random
        char[] chars = new char[8];
        //again, no need to pass this a variable if you always want 8

        for (int i = 0; i < 8; i++)
        {
            chars[i] = _allowedChars[randNum.Next(_allowedChars.Length)];
            //No need to over complicate this, passing an integer value to Random.Next will "Return a nonnegative random number less than the specified maximum."
        }
        return new string(chars);
    }


protected void Button1_Click(object sender, EventArgs e)
    {
        userid.Text = "Your id is:     " + id.Text;
        if(id .Text!="")
        {
            password.Text = "Your password is: " + CreateRandomPassword(); 
        }
    }

If I've understood what you want to do correctly, then this should get you there.

Ref: MSDN Random.Next

edited Aug 9, 2013 at 18:03

answered Mar 11, 2011 at 5:25

Infotekka

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4 Comments

jeni Over a year ago

hi thank u very much for your code..i hope it wil works surely..but i got two errors do u have any idea on that.. Error 1: The best overloaded method match for 'System.Random.Random(int)' has some invalid arguments Error 2: Argument '1': cannot convert from 'long' to 'int'

jeni Over a year ago

hey i found that..code is now working...thanks a lot...am so happy

Bahamut Over a year ago

you need to add an int cast to the DateTime.Now.Ticks to pass the ticks as ints instead of long

User970008 Over a year ago

There is also a typo in the var randNum and rndNum.

Stephen Chung · Accepted Answer · 2011-03-11 04:58:25Z

3

Shouldn't it be:

password.Text = CreateRandomPassword(int.Prase(myInt))

However, it is extremely strange that you are:

Taking a user ID and converting it to string (I assume it is a number)
Parsing that string back to a number
Using this number as the length of your password

If your user ID is, say, 345678, you are creating a password which contains 345,678 characters! Please check your program logic.

answered Mar 11, 2011 at 4:58

Stephen Chung

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2 Comments

rgmills Over a year ago

Based on the reqs, CreateRandomPassword should take a constant value for PASSWORD_LENGTH and really shouldn't be based on the value of id.Text at all.

Stephen Chung Over a year ago

Most definitely a program logic error. Agree totally with your answer.

Nikhil · Accepted Answer · 2012-11-05 04:35:47Z

1

Check out the following link for password generator in asp.net

Password generator in ASP.NET

edited Nov 5, 2012 at 4:35

Nikhil

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answered Dec 5, 2011 at 17:11

Rajesh

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Comments

ClotzA · Accepted Answer · 2013-08-09 22:54:12Z

1

Depending on how fast the function is called multiple times the passwords will look similar if you use Random as a local variable (even when using Ticks as seeds). A solution for that is to use it as a static variable to ensure the result is truly random.

private static Random rand = new Random();

Another one is to use .net built-in solutions like Path.GetRandomFileName() which will give you results like

// w143kxnu.idj

Use only the first 8 chars and arbitrary change some chars to Upper().

answered Aug 9, 2013 at 22:54

ClotzA

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Comments

rgmills · Accepted Answer · 2011-03-11 05:06:17Z

0

Changing your Button_Click method to the following will give you a random password of 8 characters:

static const int PASSWORD_LENGTH = 8;
protected void Button1_Click(object sender, EventArgs e)
{
    userid.Text = "Your id is:     " + id.Text;
    if(id .Text!="")
    {

        string myInt = id.Text.ToString(); 
        password.Text = "Your password is: " + CreateRandomPassword(PASSWORD_LENGTH); 
    }
   }

Otherwise if id.Text contains any letters at all, it will fail which is most likely the error you are getting.

edited Mar 11, 2011 at 5:06

answered Mar 11, 2011 at 5:01

rgmills

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Comments

Fun Mun Pieng · Accepted Answer · 2011-03-11 05:32:24Z

0

There is 2 problems. First id.Text.ToString() may contain some characters that make it fail to be parsed to int. "helo123" can't be parsed directly into int. This causes "input format is wrong". You should only do this if you are sure the ID contains only numbers.

Second, CreateRandomPassword(int.Parse(myInt)) creates a password that is as long as the value of the id. You should CreateRandomPassword(10) or CreateRandomPassword(randNum.Next(5, 12)). If your ID is 500, your password would be 500 characters long and if your field is supposed to be 32 characters, this could fail with "input format is wrong".

answered Mar 11, 2011 at 5:32

Fun Mun Pieng

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Comments

user3174290 · Accepted Answer · 2014-01-21 16:40:56Z

0

First we take 6 (any) random numbers from “0123456789”.

Then we create a function that returns the random number.

private string GetRandomNumber()
{
    var chars = "0123456789";
    var random = new Random();
    var result = new string(
        Enumerable.Repeat(chars, 6)
                  .Select(s => s[random.Next(s.Length)])
                  .ToArray());
    return result;
}

Visit Here to View full example.

answered Jan 21, 2014 at 16:40

user3174290

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