[RESOLVED] Remove double String from a String

[killer7k]

Hi ,

i'am trying to remove double String from
like in this example

steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:
987654321_qn�K.X1_qn�K.X1_qn�K.X1_

i'am trying to get in the first string steve and in the second string 987654321

i dont know how to compare or something to extract that

Thanks

[Code Doc]

Hi,
I'm trying to remove the double string from this example:

steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:
987654321_qn�K.X1_qn�K.X1_qn�K.X1_

I am trying to get in the first string, steve and in the second string 987654321. I don't know how to compare or something to extract that.

Thanks

Any idea what is inside the square boxes? That could be rather important before code is written to solve the problem. Also, if I did not translate your question correctly, please advise.

[killer7k]

Yes thanks i cant explain it well
i only got the same result posted there
i wanna get ride of those string after the original string
feel free to understand it on your way

wanna extract from the first string is gonna be unique always come on the first string and remove other duplicated

Thanks

[Code Doc]

This does it. Add a command button to a form and then click on it after adding this code:

Private Sub Command1_Click()
Dim Str1 As String, Str2 As String
Str1 = "steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:"
Str2 = "987654321_qn?K.X1_qn?K.X1_qn?K.X1_"
Str1 = Left$(Str1, InStr(Str1, ".") - 1)
Str2 = Left$(Str2, InStr(Str2, "_") - 1)
MsgBox Str1 & vbCrLf & Str2
End Sub

[killer7k]

This does it. Add a command button to a form and then click on it after adding this code:

Private Sub Command1_Click()
Dim Str1 As String, Str2 As String
Str1 = "steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:"
Str2 = "987654321_qn?K.X1_qn?K.X1_qn?K.X1_"
Str1 = Left$(Str1, InStr(Str1, ".") - 1)
Str2 = Left$(Str2, InStr(Str2, "_") - 1)
MsgBox Str1 & vbCrLf & Str2
End Sub

hmm Thanks ,
but i think you didnt understand me good cause i cant use only . or _ everytime i can have different string so i should make it dynamic
here you other example

brad123]slI,Z3]slI,Z3]slI,Z3] 
12345~mj!D2[5~mj!D2[5~mj!D2[5~mj
steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:
987654321_qn�K.X1_qn�K.X1_qn�K.X1_

must be like 
brad123
12345
steve
987654321

Thanks cause i really tried to get an algo for that but i couldnt get it

[killer7k]

oh i wanna just added that in each string some string double each other i think is the way to get ride off

Thanks

[Code Doc]

If the delimiter character changes continuously from one string to the next, then you are a dead duck because there is no consistency.

[killer7k]

If the delimiter character changes continuously from one string to the next, then you are a dead duck because there is no consistency.

yes but i was studying the output and found that the other string that i wanna remove come as double are same
look in this example

brad123]slI,Z3]slI,Z3]slI,Z3]

Now the true string is : brad123
next it come a serie of 3 doube of string ]slI,Z3]..

brad123]slI,Z3]slI,Z3]slI,Z3]
1 2 3
as you see the double are the same

brad123 ]slI,Z3 ]slI,Z3 ]slI,Z3]
'true str 1 double 2 double 3 double


what do you think ?

Thanks

[Code Doc]

Killer7k said, "what do you think?"
-------------
I think that you should mark your thread Resolved using the thread tools.

[killer7k]

but i didnt got the answer yet m8

[Billy Conner]

well i worked on it a bit. my instinct tells me i need to work from the right side of the string to the left to get true results but i haven't gotten it to work that way. this code here doesn't follow that pattern but it seems to work so far. from the way it's made i don't really see it being faultless, but you can check it out to see how it works out for you.

Private Function RemoveExtra(pStr As String) As String

    Dim Pos As Long, Cnt As Long, pLen As Long
    Dim Data() As String, mStr As String
    
    Pos = 1
    pLen = 4
    Cnt = 0
    Do Until Pos = Len(pStr)
        mStr = Mid$(pStr, Pos, pLen)
        Data() = Split(pStr, mStr)
        If UBound(Data) >= 3 Then
            Cnt = Cnt + 1
            pLen = pLen + 1
        Else
            If Cnt Then
                RemoveExtra = Left$(pStr, InStr(pStr, mStr))
                Exit Do
            Else
                Pos = Pos + 1
            End If
        End If
    Loop
    
End Function

edit: to call it just do something like MyRetStr = RemoveExtra(FullStr)

[Siddharth Rout]

I think it is possible, if my weird logic stands

Let me give it a try. But before I do I have couple of questions. If you find them funny, still answer them if you want me to help you...

1) What will be the minimum length of the "True" String in any given condition?
2) What will be the minimum and maximum length of the "Duplicate" String in any given condition?

[killer7k]

well i worked on it a bit. my instinct tells me i need to work from the right side of the string to the left to get true results but i haven't gotten it to work that way. this code here doesn't follow that pattern but it seems to work so far. from the way it's made i don't really see it being faultless, but you can check it out to see how it works out for you.

Private Function RemoveExtra(pStr As String) As String

    Dim Pos As Long, Cnt As Long, pLen As Long
    Dim Data() As String, mStr As String
    
    Pos = 1
    pLen = 4
    Cnt = 0
    Do Until Pos = Len(pStr)
        mStr = Mid$(pStr, Pos, pLen)
        Data() = Split(pStr, mStr)
        If UBound(Data) >= 3 Then
            Cnt = Cnt + 1
            pLen = pLen + 1
        Else
            If Cnt Then
                RemoveExtra = Left$(pStr, InStr(pStr, mStr))
                Exit Do
            Else
                Pos = Pos + 1
            End If
        End If
    Loop
    
End Function

edit: to call it just do something like MyRetStr = RemoveExtra(FullStr)

Thanks m8 your function work really good
it helped me a lot of and got my life easier with that

@koolsid,
Thanks m8 but seem Billy Conner function working good

Thanks again and cheerz

[Billy Conner]

i know its resolved but i'd like to expand on this further.
the problem with your pattern is that the last character of what you want to keep is always part of the pattern..(unless this is how it was meant to be designed)
for example..
(underlined and italic are the full pattern. bold is the partial pattern)
if you start from the right hand side the pattern is steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T: <---Start on this side
if you start from the left hand side the pattern is steve.K=T:qbe.K=T:qbe.K=T:qbe.K=T:
basically the same each way you look at it. the reason i start with the "e" on the left side is because if i make a program to look for the best matches the computer will find that as the starting point is its part of the pattern.

if you notice, your pattern on each of your examples starts with the last letter of what you want to return. in this case its the e in steve, so in my function i had to leave the first letter in of the pattern in my example to make it work.
that is the reason i don't believe this will work flawlessly is because i have not examined all possible outcomes that you could come by in whatever data your inputting from. the data your inputting (in theory) could use more of the pattern in what you wish to return than just one character. in that case you could end up with a lot less data returning than whats really wanted. so i suggest you do more research on the data input to make sure this doesn't happen.
but anyway, i cannot think of a better solution for this currently. if one magically pops in my head sometime i'll post it in here for you.

[killer7k]

Thanks m8 really appreciated
for now it work really good if somedays i got a string as different output i will try to see what we can do

Thanks & cheerz