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random name

Posted on 2011-09-21
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Last Modified: 2012-05-12
I am creating a random passwork.
I loop a list and want to create a new password for each. But the same password comes up. Do you know why that can is?

 foreach (User s in sList)
{
      if (s.Password1 == "")
      {
              s.Password1 = CreateRandomPassword();
       }
}

private static string CreateRandomPassword()
    {
        Random rd1 = new Random();
        Random rd2 = new Random();
        string password = "";
        string[] names = new string[] { "Car", "House", "Building" };
        for (int i = 0; i < 5; i++)
        {
            password = names[rd1.Next(names.Length - 1)] + rd2.Next(1, 99);
        }
        return password;
    }
0
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Question by:johnkainn
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3 Comments
 
LVL 3

Expert Comment

by:paradox_cla
ID: 36573098
The problem is call "new Random()".
The call should not be repeated.
Call "new random" use the PC timer.
0
 
LVL 2

Accepted Solution

by:
arundhaj earned 250 total points
ID: 36573130
If you wanted to generate random password, one possible approach is
public static String CreateRandomPassword()
{
    String possible = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
    StringBuilder password = new StringBuilder();
    Random rand = new Random();
    int lengthOfPassword = 5;

    for (int i = 0; i < lengthOfPassword; i++)
        password.Append(possible.Substring(rand.Next(possible.Length), 1));

    return password.ToString();
}

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0
 
LVL 86

Assisted Solution

by:Mike Tomlinson
Mike Tomlinson earned 250 total points
ID: 36574408
I agree with Paradox...MOVE your Random variable to outside your function at Class level like this:
*Note that you only need ONE instance*
private Random Rnd = new Random();

    private static string CreateRandomPassword()
    {
        string password = "";
        string[] names = new string[] { "Car", "House", "Building" };
        for (int i = 0; i < 5; i++)
        {
            password = names[Rnd.Next(names.Length - 1)] + Rnd.Next(1, 99).ToString();
        }
        return password;
    }

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