Latest Announcemnent
New thread

Dcrews Combination Finder 1.1 PRO

Status
Not open for further replies.
D

dcrew

Banned
Banned
231
2010
51
0
Re-coded from scratch in C#


I recently got a private message:

hihotfile said:
Hi dcrew

i see your thread http://www.wjunction.com/92-development-area/116888-proffesional-combination-finder.html


your tool is amazing...i think that i ever think .... i know that is your educational project .

may be you written in dot net....

can you please give me hint how i can develop in php :) .... or can you release this ....
Click to expand...

dcrew said:
And was thinking about releasing the source codes for both VB.NET and C#. What do you think I should do?
Click to expand...
ARE RELEASING

Source-Code Download Links:
Mediafire Only
VB.NET - Click Here!
C# - Click Here!


Ever wondered how many possible combinations you can get using all letters and numbers, making 3 length words? 46656!

This programme can help you find a random password (and damn... I mean random!)

MPAPkZ.png



Suggestions and constructive critism are welcome!

Download Links [All Updated to 1.1 PRO]:

Main Mediafire: Click Here!

Mirror Uppit: Click Here!

Mirror Megaupload: Click Here!​
 
Last edited:
8 comments
Here's one of the main functions I wrote that the programme relies on:

VB.NET

Code: 
    Public Function GetNextChar(ByVal currentcharacter As String) As String
        If IncludeLetters = True Then
            Dim IsLetter, IsNumber As Boolean
            Dim CharIndexLetter As Integer = Array.IndexOf(Letters, currentcharacter)
            If CharIndexLetter = -1 Then IsLetter = False Else IsLetter = True
            Dim CharIndexNumber As Integer = Array.IndexOf(Numbers, currentcharacter)
            If CharIndexNumber = -1 Then IsNumber = False Else IsNumber = True
            If CharIndexLetter < LettersCount - 1 And currentcharacter <> Letters(LettersCount - 1) And IsLetter = True Then
                currentcharacter = Letters(CharIndexLetter + 1)
                Return currentcharacter
                Exit Function
            ElseIf IsNumber = True Or CharIndexLetter >= LettersCount - 1 Then
                If IncludeNumbers = True Then
                    If CharIndexNumber < NumbersCount - 1 Then
                        currentcharacter = Numbers(CharIndexNumber + 1)
                        Return currentcharacter
                        Exit Function
                    End If
                End If
            End If
            If currentcharacter = vbNullString Then
                currentcharacter = "done"
                Return currentcharacter
                Exit Function
            End If
        ElseIf IncludeLetters = False And IncludeNumbers = True Then
            Dim IsNumber As Boolean
            Dim CharIndexNumber As Integer = Array.IndexOf(Numbers, currentcharacter)
            If CharIndexNumber = -1 Then IsNumber = False Else IsNumber = True
            If CharIndexNumber < NumbersCount - 1 Then
                currentcharacter = Numbers(CharIndexNumber + 1)
                Return currentcharacter
                Exit Function
            End If
            If currentcharacter = vbNullString Then
                currentcharacter = "done"
                Return currentcharacter
                Exit Function
            End If
        End If
    End Function



C#

Code: 
        public string getnextchar(string currentchar)
        {
            if (IncludeLetters && IncludeNumbers)
            {
                Boolean IsLetter, IsNumber;
                int CharIndexLetter = Array.IndexOf(Letters, currentchar);
                if (CharIndexLetter == -1) { IsLetter = false; } else { IsLetter = true; }
                int CharIndexNumber = Array.IndexOf(Numbers, currentchar);
                if (CharIndexNumber == -1) { IsNumber = false; } else { IsNumber = true; }
                if (CharIndexLetter < LettersCount - 1 && currentchar != Letters[LettersCount - 1] && IsLetter)
                {
                    currentchar = Letters[CharIndexLetter + 1];
                    return currentchar;
                } else if (IsNumber || CharIndexLetter >= LettersCount-1)
                {
                    if (CharIndexNumber < NumbersCount-2)
                    {
                        currentchar = Numbers[CharIndexNumber + 1];
                        return currentchar;
                    }
                } else {
                    currentchar = null;
                }
                if (currentchar == null || currentchar == "")
                {
                    currentchar = "done";
                    return currentchar;
                }
            }
            return null;
        }
 
Status
Not open for further replies.
Back
Top