VB6 Encryption & Decryption

[aquriuz81]

Hi Members of VB forums,

I am new in the VB6 developping, and i am from a french spoken country so excuse my english if it's bad.
i came wanted to build a little program that can do encryption and decryption in VB6 and Asp but i have some problems to make it work. the promgrams uses ASCII switch to mixed caracters just like it is shown here : http://www.sourcecodester.com/tutori...using-vb6.html

        Private Sub Command1_Click()
        Text2.Text = Encrypt(Text1.Text, Len(Text1.Text))
    End Sub
     
    Public Function Encrypt(Name As String, Key As Long) As String
    Dim v As Long, c1 As String, z As String
        For v = 1 To Len(Name)
            c1 = Asc(Mid(Name, v, 1))
            c1 = Chr(c1 + Key)
            z = z & c1
        Next v
        Encrypt = z
    End Function
     
    Private Sub Command2_Click()
        Text2.Text = Decrypt(Text2.Text, Len(Text1.Text))
        Text1.Text = ""
    End Sub
     
    Public Function Decrypt(Name As String, Key As Long) As String
    Dim v As Long, c1 As String, z As String
        For v = 1 To Len(Name)
            c1 = Asc(Mid(Name, v, 1))
            c1 = Chr(c1 - Key)
            z = z & c1
        Next v
        Decrypt = z
    End Function

but i want to use my own encryption that will give :

123456789 = Qj>:i5&F:LaGYXuyxr

abcdefghij = Luvwi5zn>:Kw?6wtg:,,

0123456789 = A/uy&Fi5&F:LaGYXuyxr

1234567890 = ZV=?d@D)|i3kBo(l/e>a

i have seen from some websites that the encryption and decryption goes like here :

function encrypt(ecode)
Dim texts
dim i
for i=1 to len(ecode)
texts=texts & chr(asc(mid(ecode,i,1))+i)
next
encrypt = texts
end function 

function decrypt(dcode) 
dim texts
dim i
for i=1 to len(dcode)
texts=texts & chr(asc(mid(dcode,i,1))-i)
next
decrypt=texts
end function

function mistake(preString)
Dim texts
Dim seed
Dim i,length
prestring = trim(preString)
length = len(preString)
seed = length
Randomize(length)
texts = ""
for i = 1 to length
seed = int(94*rnd(-asc(mid(preString,i,1))-seed*asc(right(prestring,1)))+32)
texts = texts & chr(seed) & chr(int(94*rnd(-seed)+32))
next
dim dist
dist=""
for i = 1 to len(texts)
if mid(texts,i,1)<>"'" then
dist=dist+mid(texts,i,1)
end if
next 
mistake = dist
end function

So i am asking how can i put this into the codes i post above that will give encryption and decryption into a text box in vb6 so that when i type '123456' it will give this 'Nqo(&FZV|ixr' in the encrypted text box, and when i click decrypt it will give the '123456'
I have tried but now luck.
i am a computer techncian and i am new into programmming stuffs. any help is welcome
Thanks.

[Shaggy Hiker]

Welcome to the forum. I edited your post to add CODE tags. To do this yourself, just press the # button and paste your code between the [CODE][/CODE] tags.

[aquriuz81]

it does not work, i have tried it and it does not work. any help please?

[aquriuz81]

any help will be welcome. i Just want to do this encryption and decryption prog so that when i type into a text box in vb6 '123456' it will give this 'Nqo(&FZV|ixr' in the encrypted text box, and when i click decrypt it will give the '123456'
The encryption and decryption code is shown above.

[Billy Conner]

any help will be welcome. i Just want to do this encryption and decryption prog so that when i type into a text box in vb6 '123456' it will give this 'Nqo(&FZV|ixr' in the encrypted text box, and when i click decrypt it will give the '123456'
The encryption and decryption code is shown above.

Where are you getting the comparison value of 'Nqo(&FZV|ixr' ? It seems you are wanting us to break another encryption that is already present somewhere else. A normal programmer is not going to be able to just make an encryption program that jumbles 123456 to Nqo(&FZV|ixr. Although I haven't broke down the data myself for your comparison, I would assume it would possibly take much time/effort to recreate it, and even doing so would not be likely to recreate all the entries you have provided in the same algorithm. maybe if someone knew where you got the original data, they could find out what cypher the program is using and possibly link you something that is already written for it.

[irwingretczko]

Here is a text encryption program to beat all others!

https://www.dropbox.com/s/dq4fcp72yc...02017.zip?dl=0