dynamic password creation

Question

i wana to create dynamic password as for some character with specified length, such as PERMUTATION,i think this is possible through recursion but i m unable to manage it dynamically

Purushottam Rathore · Answer

HTML clipboard asp : Label ID ="Label1" runat ="server" ForeColor ="navy" Text ="Enter Password Length:"> asp : Label > asp : TextBox ID ="TextBoxPasswordLength" runat ="server" Width ="100" MaxLength ="2"> asp : TextBox > asp : Label ID ="LabelShowPwd" runat ="server" ForeColor ="navy"> asp : Label > asp : Button ID ="ButtonOk" runat ="server" Text =" Ok " OnClick ="ButtonOk_Click" /> Partial Class _Default Inherits System.Web.UI.Page Protected Sub Page_Load( ByVal sender As Object , ByVal e As EventArgs) If IsPostBack Then LabelShowPwd.Text = "Please enter a password length." End If End Sub Public Shared Function CreateRandomPassword( ByVal PasswordLength As Integer ) As String Dim _allowedChars As String = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789" Dim randomNumber As New Random() Dim chars(PasswordLength - 1) As Char Dim allowedCharCount As Integer = _allowedChars.Length For i As Integer = 0 To PasswordLength - 1 chars(i) = _allowedChars.Chars( CInt (Fix((_allowedChars.Length) * randomNumber.NextDouble()))) Next i Return New String (chars) End Function Protected Sub ButtonOk_Click( ByVal sender As Object , ByVal e As EventArgs) If TextBoxPasswordLength.Text "" Then Dim password As String = TextBoxPasswordLength.Text.ToString() LabelShowPwd.Text = "Your generated password is: " & CreateRandomPassword( Integer .Parse(password)) End If End Sub End Class

Niradhip Chakraborty · Answer

You can use hash table. After creation of each password Check the value is available in hash table or not if not then add the value in hash table,If the value is already present(which means it is repeating) there then Increase the length. Note: If you are not generating the password in single shot,then you have to store the already created password in database or in xml file and before creating again the new password add all the password in hash table and compare it.

ishika singh · Answer

that's good, but suppose we have check password between min-5 and max- 10, so first we check all 5 length password than we go in 6 length password, but problem is that when we can determine that all 5 length password have checked, at which condition we have to move 5 to 6 length password

Niradhip Chakraborty · Answer

It will not repeat,as we are using Random RandChar = new Random(); constructor to generate password. you also can incresae the length by incresaing the calling parameter. say GenerateRandomPassword(5); or GenerateRandomPassword(7); First one will give password for 5 characters and second one will give password for 7 characters.

ishika singh · Answer

but i thinks it will repeat some string (password), and how can we create password with exact combination as there are a lot combination and also we have to increase also length, that result to slow speed of getting password as some will repeated thanks

Niradhip Chakraborty · Answer

private string GenerateRandomPassword ( int Length ) { // We want all popular characters except for the ones that can easily be confused (e.g., l and 1) string [ ] Characters = new string [ 56 ] ; Characters [ 0 ] = "a" ; Characters [ 1 ] = "b" ; Characters [ 2 ] = "c" ; Characters [ 3 ] = "d" ; Characters [ 4 ] = "e" ; Characters [ 5 ] = "f" ; Characters [ 6 ] = "g" ; Characters [ 7 ] = "h" ; Characters [ 8 ] = "j" ; Characters [ 9 ] = "k" ; Characters [ 10 ] = "m" ; Characters [ 11 ] = "n" ; Characters [ 12 ] = "p" ; Characters [ 13 ] = "q" ; Characters [ 14 ] = "r" ; Characters [ 15 ] = "s" ; Characters [ 16 ] = "t" ; Characters [ 17 ] = "u" ; Characters [ 18 ] = "v" ; Characters [ 19 ] = "w" ; Characters [ 20 ] = "x" ; Characters [ 21 ] = "y" ; Characters [ 22 ] = "z" ; Characters [ 23 ] = "A" ; Characters [ 24 ] = "B" ; Characters [ 25 ] = "C" ; Characters [ 26 ] = "D" ; Characters [ 27 ] = "E" ; Characters [ 28 ] = "F" ; Characters [ 29 ] = "G" ; Characters [ 30 ] = "H" ; Characters [ 31 ] = "J" ; Characters [ 32 ] = "K" ; Characters [ 33 ] = "L" ; Characters [ 34 ] = "M" ; Characters [ 35 ] = "N" ; Characters [ 36 ] = "P" ; Characters [ 37 ] = "Q" ; Characters [ 38 ] = "R" ; Characters [ 39 ] = "S" ; Characters [ 40 ] = "T" ; Characters [ 41 ] = "U" ; Characters [ 42 ] = "V" ; Characters [ 43 ] = "W" ; Characters [ 44 ] = "X" ; Characters [ 45 ] = "Y" ; Characters [ 46 ] = "Z" ; Characters [ 47 ] = "2" ; Characters [ 48 ] = "3" ; Characters [ 49 ] = "4" ; Characters [ 50 ] = "5" ; Characters [ 51 ] = "6" ; Characters [ 52 ] = "7" ; Characters [ 53 ] = "8" ; Characters [ 54 ] = "9" ; Random RandChar = new Random ( ) ; string Password = "" ; for ( int i = 0 ; i ) { Password += Characters [ RandChar. Next ( 0 , 55 ) ] ; } return Password; } You also can Check it: 1. http://www.obviex.com/Samples/Password.aspx

ishika singh · Answer

suppose i hav charset(a-z) and we have to create all combination of these character with length 5 to 10 such that suppose for 5 length string aaaaa aaaab aaaac aaaad ----- ----- ----- aaaaz aaaba aaaca aaada aaaea ----- aaaza aabaa aacaa aadaa aaeaa ----- aazaa ---- ---- baaaa caaaa daaaa eaaaa ----- ----- ----- zaaaa for 6 length string aaaaaa aaaaab aaaaac ------ aaaaaz similarly for next other could this possible by any loop i thinks that didn't repeat any string than thanks for all replying me......

Kirtan Patel · Answer

private void button2_Click(object sender, EventArgs e) { //Create Password of Random 5 Characters Random rnd = new Random(); //Set Range From ASCII (65 -90 ) ie (A-Z) and Create 5 Different Random Characters char char1 = (char)(rnd.Next(65, 90)); char char2 = (char)(rnd.Next(65, 90)); char char3 = (char)(rnd.Next(65, 90)); char char4 = (char)(rnd.Next(65, 90)); char char5 = (char)(rnd.Next(65, 90)); //Combine All the Randome Characters Created and Make One String string RandomPassword = char1.ToString() + char2.ToString() + char3.ToString() + char4.ToString() + char5.ToString(); textBox1.Text = RandomPassword; }

Niradhip Chakraborty · Answer

You can use random number to generate password dynamically. Check the code: protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { int intPassword = RandomNumber(100000, 999999); } } private int RandomNumber(int min, int max) { Random random = new Random(); return random.Next(min, max); } Note: You can change the minimum, maximum length.